Explanation:
Given that,
She walks in east,
Speed = 0.80 m/s
Time = 4.0 min
In north,
Speed = 0.50 m/s
Time = 5.5 min
In west,
Speed = 1.1 m/s
Time = 2.8 min
(a). We need to calculate the unit-vector velocities for each of the legs of her journey.
The velocity of her in east
[tex]\vec{v_{1}}=0.80\ \hat{x}\ m/s[/tex]
[tex]\vec{v_{2}}=0.50\ \hat{y}\ m/s[/tex]
[tex]\vec{v_{3}}=1.1\ \hat{-x}\ m/s[/tex]
(b). We need to calculate the unit-vector displacements for each of the legs of her journey
Using formula of displacement
[tex]\vec{d_{1}}=v_{1}\times t_{1}[/tex]
In east ,
[tex]\vec{d_{1}}=0.80\times4.0\times60[/tex]
[tex]\vec{d_{1}}=192\ \hat{x}\ m[/tex]
In north,
[tex]\vec{d_{2}}=0.50\times5.5\times60[/tex]
[tex]\vec{d_{2}}=165\ \hat{y}\ m[/tex]
In west,
[tex]\vec{d_{3}}=1.1\times2.8\times60[/tex]
[tex]\vec{d_{3}}=184.8\ \hat{-x}\ m[/tex]
(c). We need to calculate the net displacement from the postal truck after her journey is complete
[tex]\vec{d}=\vec{d_{1}}+\vec{d_{2}}+\vec{d_{3}}[/tex]
Put the value in the formula
[tex]\vec{d}=192\hat{x}+165\hat{y}+184.8\hat{-x}[/tex]
[tex]\vec{d}=7.2\hat{x}+165\hat{y}[/tex]
We need to calculate the magnitude of the displacement
[tex]d=\sqrt{(7.2)^2+(165)^2}[/tex]
[tex]d=165.16\ m[/tex]
The magnitude of the displacement is 165.16 m.
Hence, This is the required solution.
The radius of Mercury (from the centerto just above the atmosphere)
is 2440km (2440103
m),and its mass is 0.31024
kg.An object is launched straight up from just above the atmosphere
ofMercury.
(a) What initial speed is needed so that when the object is farfrom
Mercury its final speed is2000 m/s?
Answer:
u = 12962.11 m/s
Explanation:
Given that,
The radius of mercury, [tex]r=2440\ km=2440\times 10^3\ m[/tex]
Mass of Mercury, [tex]M=3\times 10^{24}\ kg[/tex]
Final speed of the object, v = 2000 m/s
Let u is its initial speed when the object is far from Mercury. It can be calculated by applying the conservation of energy as :
Initial kinetic energy + gravitational potential energy = final kinetic energy
[tex]\dfrac{1}{2}mu^2+(-\dfrac{GmM}{r})=\dfrac{1}{2}mv^2[/tex]
[tex]\dfrac{1}{2}u^2+(-\dfrac{GM}{r})=\dfrac{1}{2}v^2[/tex]
[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+\dfrac{GM}{r}[/tex]
[tex]u^2=2\times (\dfrac{1}{2}v^2+\dfrac{GM}{r})[/tex]
[tex]u^2=2\times (\dfrac{1}{2}(2000)^2+\dfrac{6.67\times 10^{-11}\times 3\times 10^{24}}{2440\times 10^3})[/tex]
u = 12962.11 m/s
So, the initial speed of the object is 12962.11 kg. Hence, this is the required solution.
An object object travels 24 meters in the +x direction for 3.5 seconds, and then immediately travels some distance in the -x direction. After a total time of 9.3 seconds, the object's average velocity is 0 m/s. What distance did the object travel in the -x direction?
Answer:
The distance in -x axis is 39.77 m
Explanation:
Given that,
Distance in +x axis = 24 m
Time = 3.5 sec
Total time = 9.3 sec
Average velocity = 0
We need to calculate the velocity in +x axis
[tex]v=\dfrac{d}{t}[/tex]
Put the value into the formula
[tex]v=\dfrac{24}{3.5}[/tex]
We need to calculate the velocity in -x axis
[tex]v=\dfrac{x}{9.3-3.5}[/tex]
We need to calculate the distance
Using formula of average velocity
[tex]\dfrac{\dfrac{24}{3.5}-\dfrac{x}{9.3-3.5}}{2}=0[/tex]
[tex]x=\dfrac{24}{3.5}\times5.8[/tex]
[tex]x=39.77\ m[/tex]
Hence, The distance in -x axis is 39.77 m.
At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead (from rest) with a constant acceleration of 2.9 m/s². At the same instant, a truck traveling with a constant speed of 36.3 mph passes the car. How much time in seconds does it take the car to catch up with the truck. Keep 2 decimal places.
Answer:
It take the car to catch up with the truck 111.91 s.
Explanation:
To solve this problem we have to use the formula for uniformly accelerated motion (for the car) and the formula for uniform rectilinear movement (for the truck).
We apply the corresponding formula for each vehicle, so we will have two equations. As the question is how much time, time is the unknown variable that we will call t from now on.
Equation for the car is:
[tex]x_{c}=\frac{1}{2}*a*t^{2}[/tex]
Equation for the truck is
[tex]x_{t} =v*t[/tex]
We know that t will be the same for the two vehicle on the instant the car catch up the truck.
On the time t the distance x traveled for both cars are the same, so we can equate the two formulas ans isolate t.
[tex]v*t=\frac{1}{2} *a*t^{2} \\t=(2*v)/a\\t=111,91s[/tex]
Note: all unit of measurement must be the same, for speed, we need to convert 36,3mph to m/s.
36,3mph=162.27m/s we use 162.27m/s in the formulas.
It will take the car approximately 11.17 seconds to catch up with the truck that passes it traveling at a constant speed when the car starts from rest and accelerates at 2.9 m/s².
Explanation:The question asks how long it takes for a car that starts from rest with a constant acceleration to catch up with a truck traveling at a constant speed. The car accelerates from rest at 2.9 m/s², while the truck travels at a constant speed of 36.3 mph, which is approximately 16.2 m/s (1 mph equals approximately 0.44704 m/s).
To solve this problem, we need to consider that the car and truck will have traveled the same distance when the car catches up to the truck. The equations for distance for the car (with acceleration) and the truck (traveling at constant speed) are:
Distancecar = ½ * acceleration * time²Distancetruck = speed * timeWe set the distances equal to each other:
½ * 2.9 m/s² * time² = 16.2 m/s * time
This gives us a quadratic equation:
½ * 2.9 * time² - 16.2 * time = 0
Factoring out the common term 'time', we get:
time(½ * 2.9 * time - 16.2) = 0
Ignoring the solution where time equals zero (since we want to know how long it takes after they have started moving), we get:
½ * 2.9 * time = 16.2
time = (16.2 / (0.5 * 2.9)) seconds
Solving this, time ≈ 11.17 seconds
Therefore, it will take the car approximately 11.17 seconds to catch up with the truck.
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.923 g, q = 4.52 µC is located on the x axis at x = 22.6 cm, moving with a speed of 45.7 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Answer:
[tex]Q = -1.43\times 10^[-5} coulomb[/tex]
Explanation:
Given data:
particle mass = 0.923 g
particle charge is 4.52 micro C
speed of particle 45.7 m/s
In this particular case, coulomb attraction will cause centrifugal force and taken as +ve and Q is taken as -ve
[tex]-\frac{Qq}{4\pi \epsilon r^2} = \frac{mv^2}{r}[/tex]
solving for Q WE GET
[tex]Q = -\frac{mv^2}{r} \times r^2 \frac{4\pi \epsilon}{q}[/tex]
[tex]Q = -mv^2\times r \frac{4\pi \epsilon}{q}[/tex]
[tex]Q = - \frac{0.923\times 10^{-3} \times 45.7^2\times (22.6\times 10^{-2})} {4.52\times 10^{-6} \times 9\times 10^9}[/tex]
where[tex] \frac{1}{4\pi \epsilon} = 9\times 10^9[/tex]
[tex]Q = -1.43\times 10^[-5} coulomb[/tex]
Final answer:
To find the charge Q for circular motion, equate centripetal force m * v^2 / r with Coulomb's force k * |Q * q| / r^2, and solve for Q. Use m = 0.923 g, v = 45.7 m/s, q = 4.52 µC, and convert units accordingly.
Explanation:
To determine the value of charge Q that will allow the moving particle to execute circular motion, we use the concept of centripetal force. Centripetal force is the net force required to keep an object moving in a circle at a constant speed and is directed towards the center of the circle. For a charged particle moving in a circular path due to an electric force, the centripetal force is provided by the electric force between the charges.
The centripetal force (Fc) is equal to the mass (m) of the particle times the square of its speed (v) divided by the radius (r) of the circle:
Fc = m * v2 / r.
The electric force (Fe) acting on the particle is given by Coulomb's law:
Fe = k * |Q * q| / r2,
where k is Coulomb's constant (8.99 x 109 Nm2/C2), Q is the charge at the origin, q is the charge of the moving particle, and r is the separation between the charges.
Setting the centripetal force equal to the electric force yields:
m * v2 / r = k * |Q * q| / r2,
Solving for Q gives us:
Q = m * v2 / (k * q).
Plugging in the values:
Q = (0.923 g * 45.7 m/s2) / (8.99 x 109 Nm2/C2 * 4.52 µC)
Remembering to convert grams to kilograms and microcoulombs to coulombs, the final calculation will yield the required charge Q for circular motion.
Q = 1.03 mC
Aaron finds a pirate’s treasure map. The treasure map says to start at an oak tree and walks 111 feet East and 234 feet South from the tree. How far did the pirate walk? What is the displacement vector for the Pirate? What is the distance and direction relative to East?
Answer:
pirate walk 259 feet
displacement vector = 111 i - 234 j
direction is along south of east at angle 64.62° , anticlockwise
total travel 345 feet
Explanation:
given data
walk east = 111 feet
walk south = 234 feet
to find out
How far did the pirate walk and displacement vector and distance and direction relative to east
solution
we consider here distance AB is 111 feet and than he turn right i.e south distance BC is 234 feet so
so angle BAC will be
tan θ = [tex]\frac{234}{111}[/tex]
θ = 64.62
and AC distance will be
AC = [tex]\sqrt{234^{2} + 111^{2}}[/tex]
AC = 259 feet
so pirate walk 259 feet
and
displacement vector is express as
displacement vector = AC ( cosθ i + sinθ j )
displacement vector = 259 ( cos64.62 i + sin64.62 j )
displacement vector = 111 i - 234 j
and
so direction is along south of east at angle 64.62° , anticlockwise
Describe the difference between the velocity graph made walking at a steady rate and the velocity graph made at an increasing rate.
Answer:
The difference between the velocity graph made walking at a steady rate means that its the same value in time, that means there's no slope on the graph, so its acceleration is 0
On the other hand, if the velocity is increasing with time, the slope of the graph becomes positive, which means that the acceleration of the particle is positive.
Two 2.0 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge. PART A
What is the magnitude of the electric force on one of the masses?
Express your answer to two significant figures and include the appropriate units.
PART B
What is the initial acceleration of the mass if it is released and allowed to move?
Express your answer to two significant figures and include the appropriate units.
Answer:
A) Force = 0.69 N
B) Acceleration = 0.34 m/s^2
Explanation:
The electric force is given by the expression:
[tex]F_e= K *\frac{q_1*q_2}{r^2}[/tex]
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and 12 is the charge of the particles, and r is the distance:
[tex]F_e = 9*10^9 Nm^2/C^2 * \frac{(9.6*10^{-6}C)^2}{(1.1m)^2} = 0.69 N[/tex]
Part B.
For the acceleration, you need Newton's second Law:
F = m*a
Then,
[tex]a = \frac{F}{m} = \frac{0.69 N}{2 kg} = 0.34 m/s^2[/tex]
The magnitude of the electric force on one of the masses is 102.71 N. The initial acceleration of the mass is 51.36 m/s^2.
Explanation:PART A:
To find the magnitude of the electric force on one of the masses, we can use Coulomb's Law.
The formula for the magnitude of the electric force is:
F = k * (|q1| * |q2|) / r^2
where F is the force, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of the charges (9.6 μC), and r is the distance between the charges (1.1 m).
Plugging in the values:
F = (9 x 10^9 Nm^2/C^2) * (9.6 μC * 9.6 μC) / (1.1 m)^2
F = 102.71 N
The magnitude of the electric force on one of the masses is 102.71 N.
PART B:
To find the initial acceleration of the mass when it is released and allowed to move, we can use Newton's second law.
The formula for the acceleration is:
a = F / m
where a is the acceleration, F is the force (102.71 N), and m is the mass (2.0 kg).
Plugging in the values:
a = 102.71 N / 2.0 kg
a = 51.36 m/s^2
A car that weighs 1.5 × 10^4 N is initially moving at a speed of 36 km/h when the brakes are applied and the car is brought to a stop in 17 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)
Answer:
Explanation:
Initial velocity u = 36 km/h = 10 m /s
v = 0 , accn = a , Time taken to stop = t , distance covered to stop = s
v² = u² - 2as
u² = 2as
a = u² / 2s
= 10 x 10 / 2 x 17
= 2.94 ms⁻²
Force applied = mass x acceleration
= 15000 / 9.8 x 2.94
= 4500 N
b )
v = u + at
0 = 10 - 2.94 t
t = 10 / 2.94
= 3.4 s
c )
from the relation
u² = 2as
it is clear that stopping distance is proportional to u², if acceleration a is constant .
If initial speed u is doubled , stopping distance d will become 4 times or 17 x 4 = 68 m .
d )
u = at
if a is constant time taken to stop will be proportional to initial velocity.
If initial velocity is doubled , time too will be doubled. Or time will become
3.4 x 2 = 6.8 s .
A square coil with 100 turns of wire has sides of length 5 cm. The resistance of the coil is 12.0 Ω. The coil is placed in a uniform magnetic field that is perpendicular to the face of the coil and whose magnitude is decreasing at a rate of 0.081 T/s. What is the magnitude of the current circulating through the coil?
Answer:
The magnitude of the current circulating through the coil is 1.68 mA.
Explanation:
Given that,
Number of turns = 100
Length = 5 cm
Resistance = 12.0 Ω
Rate of magnetic field = 0.081 T/s
We need to calculate the magnitude of the current circulating through the coil
Using formula of current
[tex]I=\dfrac{e}{R}[/tex]
[tex]I=\dfrac{-NA\dfrac{dB}{dt}}{R}[/tex]
Where, A = area
N = number of turns
R = resistance
Put the value into the formula
[tex]I=\dfrac{100\times(0.05)^2\times0.081}{12.0}[/tex]
[tex]I=0.0016875\ A[/tex]
[tex]I=1.68\times10^{-3}\ A[/tex]
[tex]I=1.68\ mA[/tex]
Hence, The magnitude of the current circulating through the coil is 1.68 mA.
The question involves Faraday's law, which relates changing magnetic fields to induced currents. Given the rate of change of the magnetic field and the resistance of the coil, the magnitude of the induced current is calculated to be 1.67 mA.
Explanation:This question involves the concept of electromagnetic induction in physics. According to Faraday's law, a changing magnetic field will induce an electromotive force (EMF) in a circuit. In this case, we're given that the magnetic field is decreasing at a rate of 0.081 T/s. This rate of change can be used to calculate the induced EMF, which is equal to the rate of change of the magnetic flux.
The magnetic flux (Φ) through the coil is given by Φ = NBA, where N is the number of turns in the coil, B is the magnetic field, and A is the area of the coil. So, the rate of change of magnetic flux (dΦ/dt) is N*d(BA)/dt. Because the area of the coil is constant, this simplifies to NAdB/dt. A is 0.0025 m^2, N is 100, and dB/dt is -0.081T/s, so dΦ/dt is -0.02 Wb/s.
Faraday's law tells us that the induced EMF (ε) is equal to -dΦ/dt, so ε has magnitude 0.02 V. If we define positive current as current that would generate a magnetic field opposing the decrease in the existing field (Lenz's Law), the induced EMF 'pushes' positive current in this direction. The current (I) can then be calculated using Ohm's Law (V = IR), giving I = ε/R = 0.02V / 12 Ω = 0.00167 A, or 1.67 mA.
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Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m, the tire's rotation has increased to 6.4 rev/s. What was the tire's angular acceleration? Give your answer in rad/s^2.
Answer:
0.76 rad/s^2
Explanation:
First, we convert the original and final velocity from rev/s to rad/s:
[tex]v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s[/tex]
[tex]v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s[/tex]
Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:
[tex]D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads[/tex]
Now, we just use the formula:
[tex]w_f^2-w_o^2=2\alpha*x[/tex]
[tex]\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2[/tex]
Explanation:
The given data is as follows.
Initial velocity, u = [tex]3.3 rev/s \times 2 \pi rad/rev[/tex]
= 20.724 rad/sec
Final velocity, v = [tex]6.4 rev/s \times 2 \times pi rad/rev[/tex]
= 40.192 rad/sec
Now, we will calculate the value of angular rotation (d) as follows.
d = No. of revolutions × [tex]2 \pi rad/rev[/tex]
d = [tex](\frac{250 m}{2 \pi r}) \times 2 \pi[/tex]
= [tex]\frac{250}{0.32}[/tex]
= 781.25 rad
Also, we know that,
[tex]v^{2} = u^{2} + 2ad[/tex]
or, a = [tex]\frac{(v^{2} - u^{2})}{2d}[/tex]
= [tex]\frac{(40.192)^{2} - (20.724)^{2})}{2 \times 781.25}[/tex]
= [tex]\frac{1615.39 - 429.48}{1562.5}[/tex]
= 0.7589 [tex]rad/s^{2}[/tex]
Thus, we can conclude that the tire's angular acceleration is 0.7589 [tex]rad/s^{2}[/tex].
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11-0m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
Answer:
ΔH = 33.17m
Explanation:
By knowing the amount of time it takes the rocket to travel the horizontal 27m, we will be able to calculate the height when x=27m. So:
[tex]X = V_{o}*cos(60)*t[/tex] where [tex]X=27m; V_{o}=75m/s[/tex]
Solving for t:
t=0.72s
Now we calculate the height of the rocket:
[tex]Y_{f}=V_{o}*sin(60)*t-\frac{g*t^{2}}{2} = 44.17m[/tex]
If the wall was 11m-high, the difference is:
ΔH = 33.17m
Using the equation of projectile motion, the rocket cleared the top of the wall by 38.305m.
The equation of projectile motion is given by:
y= xtanθ +(1/2)gx²/(u²cos²θ )
Given information:
Initial velocity, u=75m/s
The angle of projection, θ = 60°
The horizontal distance, x=27m
Vertical distance will be calculated as:
y= xtanθ +(1/2)gx²/(u²cos²θ )
y=27 tan 60° + 0.5×9.8×27²/(75²cos²60°)
y=49.305m
So, the rocket cleared the top of the wall by
49.305-11 =38.305m
Therefore, using the equation of projectile motion the rocket cleared the top of the wall by 38.305m.
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By evaluating their dot product, find the values of the scalar s for which the two vectors ~b = ˆx + syˆ and ~c = ˆx−syˆ are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.
Answer:
[tex]s=\pm 1[/tex]
Explanation:
The dot product of two vectors [tex]\vec{a}=a_1\vec{x}+b_1\vec{y}[/tex] and [tex]\vec{b}=a_2\vec{x}+b_2\vec{y}[/tex] is given by
[tex]\vec{a}\cdot\vec{b}=a_1\cdot a_2+b_1\cdot b_2[/tex]
The dot product of two orthogonal vector is always zero thus if [tex]\vec{a}=\vec{x}+s\vec{y}[/tex] and [tex]\vec{b}=\vec{x}-s\vec{y}[/tex], their dot product would be
[tex]\vec{a}\cdot\vec{b}=1\times1+s\times-(s)=1-s^2=0[/tex]
[tex]\implies 1-s^2=0[/tex]
[tex]\implies s^2=1[/tex]
[tex]\implies s=\pm 1[/tex]
A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocity of 5.47 m/s. She then continues running at this constant velocity. How long does the runner take to travel 165 m?
Answer:32.24 s
Explanation:
Given
mass of runner (m)=51.8 kg
Constant acceleration(a)=[tex]1.31 m/s^2[/tex]
Final velocity (v)=5.47 m/s
Time taken taken to reach 5.47 m/s
v=u+at
[tex]5.47=0+1.31\times t[/tex]
[tex]t=\frac{5.47}{1.31}=4.17 s[/tex]
Distance traveled during this time is
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m[/tex]
So remaining distance left to travel with constant velocity=153.57 m
thus time [tex]=\frac{distance}{speed}[/tex]
[tex]t_2=\frac{153.57}{5.47}=28.07 s[/tex]
Total time=28.07+4.17=32.24 s
Which of the following is always true about the magnitude of a displacement? A. It is less than the distance traveled. B. It is equal to the distance traveled. C. It is less than or equal to the distance traveled. D. It is greater than the distance traveled. E. It is greater than or equal to the distance traveled.
Final answer:
The magnitude of displacement is always less than or equal to the distance traveled. Displacement is the straight line distance between the starting and ending points, while distance is the total path length traveled. So the correct option is C
Explanation:
The magnitude of displacement is always less than or equal to the distance traveled. This is because displacement is the straight-line measurement from the initial point to the final point, regardless of the path taken, and it's a vector quantity with both magnitude and direction. On the other hand, distance is a scalar quantity that represents the total path length traveled without regard to direction. Therefore, if a path is straight and in one direction, the distance and the magnitude of displacement are equal. However, if the path involves changes in direction, the distance will be greater than the magnitude of displacement.
The correct answer to the question 'Which of the following is always true about the magnitude of a displacement?' is C. It is less than or equal to the distance traveled.
If a 10kg block is at rest on a table and a 1200N force is applied in the eastward direction for 10 seconds, what is the acceleration on the block?
Answer:
a = 120 m/s²
Explanation:
We apply Newton's second law in the x direction:
∑Fₓ = m*a Formula (1)
Known data
Where:
∑Fₓ: Algebraic sum of forces in the x direction
F: Force in Newtons (N)
m: mass (kg)
a: acceleration of the block (m/s²)
F = 1200N
m = 10 kg
Problem development
We replace the known data in formula (1)
1200 = 10*a
a = 1200/10
a = 120 m/s²
A rock is dropped from the top of a tall building. How high is the building if it takes 3.25seconds for the rock to strike the ground? Ignore the effects of air resistance. A) 169m B)15.9m C)51.8m D) 32.2m
Answer:
C)51.8m
Explanation:
Kinematics equation:
[tex]y=v_{oy}*t+1/2*g*t^2[/tex]
in this case:
[tex]v_{oy}=0[/tex]
[tex]y=1/2*9.81*3.25^2=51.8m[/tex]
(5%) Problem 17: A solid aluminum sphere of radius r1 = 0.105 m is charged with q1 = +4.4 μC of electric charge. It is surrounded concentrically by a spherical copper shell of inner radius r2 = 0.15 m and outer radius r3 = 0.242 m. An electric charge of q2 = -1.1 μC is placed on the copper shell. The situation is static.
(a) Find the magnitude of the electric field, in newtons per coulomb, inside the aluminum ball.
(b) What is the magnitude of the electric field, in newtons per coulomb, inside the copper of the shell, i.e., at a radius r from the center such that r2 < r < r3?
Answer:
Electric field at radius r inside the solid sphere is
[tex]E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:
Given
The radius of the solid sphere is [tex]r_{i}=0.105 \ m\\[/tex]
The charge on the solid sphere is [tex]q_{1}=+4.4\ \mu C[/tex]
The inner radius of the shell is [tex]r_{3}=0.15\ m[/tex]
The outer radius of the shell is [tex]r_{3}=0.242\ m[/tex]
The total charge on the shell is [tex]q_{2}=-1.1\ \mu C[/tex]
PART(A)
The magnitude of electric field at radius r where [tex] r<r_{1}[/tex] \\The volumetric charge density of the solid sphere will be
[tex]\rho=\dfrac{q_{1}}{V}\\ \rho=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\\[/tex]
The charge enclosed by the radius r inside the solid sphere is
rho=[tex]q_{enc}=\rho\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}r^3}{r_{1}^3}\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\E=\dfrac{\dfrac{q_{1}r^3}{r_{1}^3}}{\epsilon_{o}\times 4\pi r^2}\\E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
PART(B)
The electric field at radius r where [tex]r_{2}<r<r_{3}[/tex]
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
[tex]q_{i}=-q_{1}\\q_{i}=-4.4\ \mu C\\[/tex]
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
[tex]q_{o}=q_{2}-q_{i}\\q_{o}=-1.1\ \mu C-(-4.4\ \mu C)\\q_{o}=-1.1\ \mu C+4.4\ \mu C\\q_{o}=3.3\ \mu C\\[/tex]
The charge enclosed by the radius r where [tex]r_{1}<r<r_{2}[/tex]
[tex]q_{enc}=q_{1}+q_{i}\\q_{enc}=4.4\times 10^{-6}-4.4\times 10^{-6}\\ q_{enc}=0\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\ E=0\ N/C\\[/tex]
(a) Inside the aluminum sphere, the electric field is 0 N/C due to electrostatic equilibrium.
(b) Inside the copper shell, for any radius between r2 and r3, the electric field is given by (3.95 x 10⁴) / r² N/C.
To solve this problem, we need to understand the behavior of the electric field in different regions around charged spherical conductors.
(a) Electric Field Inside the Aluminum Sphere
Inside a conductor in electrostatic equilibrium, the electric field is zero. Since the aluminum sphere is a conductor, the electric field inside it (for any radius r < r1) is zero.
Result: The magnitude of the electric field inside the aluminum sphere is 0 N/C.
(b) Electric Field Inside the Copper Shell
For the region inside the copper shell but outside the aluminum sphere (r2 < r < r3), we can apply Gauss's Law. According to Gauss's Law, the electric field at a distance r from the center due to a symmetric charge distribution can be calculated as follows:
Define the Gaussian surface: Here, it will be a spherical surface of radius r where r2 < r < r3.
Calculate the enclosed charge: The total charge enclosed within the Gaussian surface is only the charge of the aluminum sphere, q1 = +4.4 μC, because the copper shell's net charge is enclosed separately.
Apply Gauss's Law: Gauss's law states that the electric flux through a surface is proportional to the enclosed charge:
Equation: E * 4πr² = q1/ε₀
Solve for E:
E = q1 / (4πε₀ r²)
Substitute the values: q1 = 4.4 x 10⁻⁶ C, ε₀ = 8.854 x 10⁻¹² C²/(N·m²), and r is the distance within the copper shell.
[tex]E = \frac{4.4 \times 10^{-6} \, \text{C}}{4\pi \cdot (8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2) \cdot r^2}[/tex]
Simplifying gives:
E = (4.4 x 10⁻⁶) / (1.112 x 10⁻¹⁰ * r²)
E = (3.95 x 10⁴) / r² N/C
Hence, the magnitude of the electric field inside the copper shell, at any point where r2 < r < r3, is (3.95 x 10⁴) / r² N/C.
Water leaks out of a 3,200-gallon storage tank (initially full) at the rate V '(t) = 80 -t, where t is measured in hours and V in gallons. a. How much water leaked out between 10 and 20 hours? b. How long will it take the tank to drain completely?
Answer:
water leak is 650 gallons
time required to full drain is 80 hrs
Explanation:
given data
volume V = 3200 gallon
rate = V(t) = 80 - t
to find out
how much water leak between 10 and 20 hour and drain complete
solution
we know here rate is 80 - t
so here rate will be
[tex]\frac{dV(t)}{dt}[/tex] = 80 - t
and for time 10 and 20 hour
take integrate between 10 and 20
so water leak = [tex]\int\limits^ {20}_ {10} {(80-t)} \, dt[/tex] .....................1
water leak = ( 80t - [tex]\frac{t^{2} }{2} )^{20}_{10}[/tex]
water leak = 650
so water leak is 650 gallons
and
we know here for full tank drain condition
water leak full = 80 t - [tex]\frac{t^{2} }{2} [/tex]
3200 = 80 t - [tex]\frac{t^{2} }{2} [/tex]
6400 = t² - 160 t
t = 80
so time required to full drain is 80 hrs
A hot-air balloon has just lifted off and is rising at the constant rate of 2.0m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 12m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?
To find out how high the passenger is when the camera reaches her, we use kinematic equations, taking into account the initial speed of the camera, the constant rise speed of the passenger, and gravity's acceleration. The solution requires equating the displacements of both camera and passenger to solve for time and therefore the height.
Explanation:A hot-air balloon is rising at a constant rate of 2.0m/s when a passenger's camera is tossed straight upward with an initial speed of 12m/s from a position 2.5m below her. To determine how high the passenger is when the camera reaches her, we can apply kinematic equations of motion, incorporating the constant acceleration due to gravity (approximately 9.81m/s² downwards).
For the camera: Its initial upward velocity is 12 m/s, and it is subject to gravity's acceleration. For the passenger: Rising at a constant 2.0 m/s, not accelerating since the rate is constant. Since the initial distance between them is 2.5 m, we need to calculate when the camera, starting from a lower point but moving faster, reaches the vertically moving passenger.
Using the formula s = ut + 0.5at² for both camera and passenger, where s is the displacement, u is initial velocity, a is acceleration, and t is time, we can set the equations equal to solve for t, then determine the height by applying it to the passenger's motion equation.
Due to the mathematical complexity and potential for variability in solving these equations, the exact numerical solution isn't presented here. However, the approach involves determining the time it takes for the camera to reach the same height as the passenger and using that to find her height at that moment.
A fireworks shell is accelerated from rest to a velocity of 55.0 m/s over a distance of 0.210 m. (a) How long (in s) did the acceleration last? s
(b) Calculate the acceleration (in m/s2). (Enter the magnitude.) m/s^2
Answer:
a) The acceleration took 0.0076s
b) The aceleration was of 7202.4 m/s^2
Explanation:
We need to use the formulas for acceleration movement in straight line that are:
(1) [tex]a = \frac{V}{t}[/tex] and (2)[tex]x=x_{0} +V_{0}t + \frac{1}{2} at^2[/tex]
Where
a = acceleration
V = Velocity reached
Vo = Initial velocity
t = time
x = distance
xo = initial distance.
We have the following information:
a = We want to find V = 55.0 m/s
Vo = 0m/s because it starts from rest t = we want to find
x = 0.210 m xo= 0 m we beging in the point zero.
We have to variables in two equations, so we are going to replace in the second equation (2) the aceleration of the first one(1):
[tex]x=x_{0} +V_{0}t + \frac{1}{2} ( \frac{V}{t})t^2[/tex] We can cancel time because it is mutiplying and dividing the same factor so we have
[tex]x=x_{0} +V_{0}t + \frac{1}{2} Vt[/tex]
In this equation we just have one variable that we don't know that is time, so first we are going to replace the values and after that clear time.
[tex]0.210=0 +0*t + \frac{1}{2} 55t[/tex]
[tex]0.210=27.5t[/tex]
[tex]\frac{0.21}{27.5} = t\\[/tex]
t = 0.0076s
a) The acceleration took 0.0076s
Now we replace in the (1) equation the values of time and velocity
[tex]a = \frac{V}{t}[/tex]
[tex]a = \frac{55}{0.0076}[/tex]
a = 7202.4 m/s^2
b) The aceleration was of 7202.4 m/s^2
Does percent error give indication of accuracy or precision? Discuss.
Answer:
Accuracy
Explanation:
Percent error is the ratio of the difference of the measured and actual value to the actual value multiplied by 100.
It gives the percent deviation of the value obtained from the actual value.
Accuracy is the measure of how close the readings are to the actual value or set standard and can be improved by increase the no. of readings in an experiment.
Precision is the measure of the closeness of the obtained values to one another.
Thus accuracy of the reading can be sensed by the percent error.
What is the minimum magnitude of an electric field that balances the weight of a plastic sphere of mass 14.2 g that has been charged to 7.7 nC. Give your answer to the nearest 0.1 MN/C (mega Newton per Coulomb)
Answer:
18.1 MN/C
Explanation:
The gravitational force of the plastic sphere is in equilibrium with the electric force.
Mass of the plastic sphere = m = 14.2 g = 0.0142 kg
Force of gravity = F = mg = (0.0142)(9.81) = 0.139 N
This force is balanced by the electric force due to the charge 7.7 nC
Charge = q = 7.7 x 10⁻⁹ C
Electric field = E = F / q = (0.139) /(7.7 x 10⁻⁹) = 18.1 MN/C
Express the following angles in radians: (a) 16.8°, (b) 53.7°, (c) 94.2°. Convert the following angles to degrees: (d) 0.258 rad, (e) 1.01 rad, (f) 7.51 rad.
Explanation:
(a) We need to convert following angles in radians. The conversion formula from degrees to radian is given by :
[tex]Radians=(\dfrac{\pi}{180})\times degrees[/tex]
1. If angle is 16.8°
[tex]Radians=(\dfrac{\pi}{180})\times 16.8[/tex]
16.8 degrees = 0.29 radians
2. If angle is 53.7°
[tex]Radians=(\dfrac{\pi}{180})\times 53.7[/tex]
53.7 degrees = 0.937 radians
3. If angle is 94.2°
[tex]Radians=(\dfrac{\pi}{180})\times 94.2[/tex]
53.7 degrees = 1.64 radians
(b) We need to convert following angles to degrees. The conversion formula from radian degrees to is given by :
[tex]Degrees=(\dfrac{180}{\pi})\times radians[/tex]
1. If angle is 0.258 radian
[tex]Degrees=(\dfrac{180}{\pi})\times 0.258[/tex]
0.258 radian = 14.78 degrees
2. If angle is 1.01 radian
[tex]Degrees=(\dfrac{180}{\pi})\times 1.01[/tex]
0.258 radian = 57.86 degrees
3. If angle is 7.51 radian
[tex]Degrees=(\dfrac{180}{\pi})\times 7.51[/tex]
0.258 radian = 430.29 degrees
Hence, this is the required solution.
One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 atm. Determine the value of q, w.
(b) Now assume the gas expands by the same amount again isothermally but now irre-
versibly against 1 atm pressure (instead of reversible expansion) and calculate again q, w, delta U and delta H.
Answer:
Explanation:
Given
1 mole of perfect, monoatomic gas
initial Temperature[tex](T_i)=300 K[/tex]
[tex]P_i=10 atm[/tex]
[tex]P_f=2 atm[/tex]
Work done in iso-thermal process[tex]=P_iV_iln\frac{P_i}{P_f}[/tex]
[tex]P_i[/tex]=initial pressure
[tex]P_f[/tex]=Final Pressure
[tex]W=10\times 2.463\times ln\frac{10}{2}=39.64 J [/tex]
Since it is a iso-thermal process therefore q=w
Therefore q=39.64 J
(b)if the gas expands by the same amount again isotherm-ally and irreversibly
work done is[tex]=P\Delta V[/tex]
[tex]V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L[/tex]
[tex]V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L[/tex]
[tex]\Delta W=1\times (12.315-2.463)=9.852 J[/tex]
[tex]\Delta q=\Delta W=9.852 J[/tex]
[tex]\Delta U=0[/tex]
A 100 meter rope is 20 kg and is stretched with a tension of 20 newtons. If one end of the rope is vibrated with small amplitude at 10Hz, what would the velocity of waves traveling down it be? What would the velocity be if it rained and the rope soaked up 5 kg of water?
Answer:
The velocity waves before rain is 10 m/s
The velocity of wave after the rope soaked up 5 kg more is 8.944 m/s
Solution:
As per the question:
Length of the rope, l = 100 m
Mass of the rope, m = 20 kg
Force due to tension in the rope, [tex]T_{r} = 20 N[/tex]
Frequency of vibration in the rope, f = 10 Hz
Extra mass of the rope after being soaked in rain water, m' = 5 kg
Now,
In a rope, the wave velocity is given by:
[tex]v_{w} = \sqrt{\frac{T_{r}}{M_{d}}}[/tex] (1)
where
[tex]M_{d}[/tex] = mass density
Mass density before soaking, [tex]M_{d} = \frac{m}{l} = \frac{20}{100} = 0.20[/tex]
Mass density after being soaked, [tex]M_{d} = \frac{m + m'}{l} = \frac{25}{100} = 0.25[/tex]
Initially, the velocity is given by using eqn (1):
[tex]v_{w} = \sqrt{\frac{20}{0.20}} = 10 m/s[/tex]
The velocity after being soaked in rain:
[tex]v_{w} = \sqrt{\frac{20}{0.25}} = 8.944 m/s[/tex]
A common parameter that can be used to predict turbulence in fluid flow is called the Reynolds number. The Reynolds number for fluid flow in a pipe is a dimensionless quantity defined as Re = rhovd μ where rho is the density of the fluid, v is its speed, d is the inner diameter of the pipe, and μ is the viscosity of the fluid. Viscosity is a measure of the internal resistance of a liquid to flow and has units of Pa · s. The criteria for the type of flow are as follows. • If Re < 2,300, the flow is laminar. • If 2,300 < Re < 4,000, the flow is in a transition region between laminar and turbulent. • If Re > 4,000, the flow is turbulent. (a) Let's model blood of density 1.06 103 kg/m3 and viscosity 3.00 10-3 Pa · s as a pure liquid, that is, ignore the fact that it contains red blood cells. Suppose it is flowing in a large artery of radius 1.25 cm with a speed of 0.0650 m/s. Show that the flow is laminar. (State the Reynolds number of the flow, which will be less than 2,300, indicating laminar flow.) (b) Imagine that the artery ends in a single capillary so that the radius of the artery reduces to a much smaller value. What is the radius of the capillary that would cause the flow to become turbulent? (Use the minimum Reynolds number for which flow is fully turbulent.)
Answer:
a) Re= 574.17
b) r= 8.71 cm
Explanation:
In order to solve this problem, we will need to use the formula for the Reynolds number:
Re=ρ*v*d/μ
All of the required data is already given in the problem, but before we use the above-mentioned formula, we need to convert the data to SI units, as follows:
The density already has SI units ( 1.06 *10³ kg/m³)The radius is 1.25 cm, which converts to 0.0125 m. Thus, the inner diameter is 0.025 mThe viscosity already has SI units (3.00 * 10⁻³ Pa · s)The speed already has SI units (0.0650 m/s)a) Now we proceed to calculate Reynolds number:
Re=1.06 *10³ kg/m³ * 0.0650 m/s * 0.025 m / (3.00 * 10⁻³ Pa · s)
Re=574.17
Re<2,300 ; thus the flow is laminar.
b) To answer this question we use the same equation, and give the Reynolds number a value of 4,000 in order to find out d₂:
4,000= 1.06 *10³ kg/m³ * 0.0650 m/s * d₂ / (3.00 * 10⁻³ Pa · s)
We solve for d₂:
d₂=0.174 m
Thus the radius of the capillary that would cause the flow to become turbulent is 0.174 m / 2= 0.0871 m or 8.71 cm, given that neither the speed nor the viscosity change.
However, in your question you wrote that the artery ends in a capillary so that the radius reduces its value. But the lower the radius, the lower the Reynolds number. And as such, it would not be possible for the flow to turn from laminar to turbulent, if the other factors (such as speed, or density) do not change.
(a) what wavelength photon would you need to ionize a hydrogen atom (ionization energy = 13.6 eV)? (b) Compute the temperature of the blackbody whose spectrum peaks at wavelength you found in (a).
Answer:
(a). The wavelength of photon is 914 A.
(b). The temperature of the black body whose spectrum peaks at wavelength is 31706.78 K.
Explanation:
Given that,
Ionization energy = 13.6 eV
(a). We need to calculate the wavelength
Using formula of wavelength
[tex]E=\dfrac{hc}{\lambda}[/tex]
[tex]\lambda=\dfrac{hc}{E}[/tex]
Where, h = Planck constant
c = speed of light
E = energy
Put the value into the formula
[tex]\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{13.6\times1.6\times10^{-19}}[/tex]
[tex]\lambda=9.14\times10^{-8}\ m[/tex]
[tex]\lambda=914\ \AA[/tex]
The wavelength of photon is 914 A.
(b). We need to calculate the temperature of the black body whose spectrum peaks at wavelength
Using Wien's displacement law
[tex]\lambda_{max} T=2.898\times10^{-3}[/tex]
[tex]T=\dfrac{2.898\times10^{-3}}{\lambda}[/tex]
Put the value of wavelength
[tex]T=\dfrac{2.898\times10^{-3}}{914\times10^{-10}}[/tex]
[tex]T=31706.78\ K[/tex]
The temperature of the black body whose spectrum peaks at wavelength is 31706.78 K.
Hence, This is the required solution.
Initially at rest a single-stage rocket is launched vertically from the ground. The rocket’s thrust overcomes gravity and provides the rocket a constant upward acceleration a. The fuel is exhausted 10 seconds after launch and then the motion of the rocket is free fall only due to gravity. If the total flight time is 30 s when the rocket strikes the ground, determine (a) the initial acceleration a, (b) the rocket’s impact speed as it hits the ground, (c) the height h from the ground the rockets reaches
Answer:
Explanation:
Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds
s = 1/2at²
= .5 x a x 100
=50a
Velocity after 10 s
u = at = 10a
Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.
s = ut + 1/2 gt²
50a = -10ax20 + .5 x 10x400
250a = 2000
a = 8 m s⁻² .
Velocity after 10 s
= at = 80 m/s
further height reached with this speed under free fall
h = v² / 2g
= 80 x 80 / 2 x 10
= 320 m
height achieved under acceleration
= 50a
= 50 x 8 = 400m
Total height
= 320 + 400= 720 m
velocity after falling from 720 m
v² = 2gh = 2 x10 x 720
v = 120 m/s
Answer:
a) The initial acceleration is 7.84 m/s²
b) The impact speed is 117.6 m/s
c) The height is 705.6 m
Explanation:
a) The speed from A to B is:
v = u + at
Where
u = initial speed = 0
t = 10 s
Replacing:
v = 10t (eq. 1)
The vertical distance between A to B is:
[tex]h=\frac{1}{2} at^{2} +ut=\frac{1}{2} a*(10)^{2}+0=50a[/tex] (eq. 2)
From B to C, the time it take is equal to 20 s, then:
[tex]h=vt+\frac{1}{2} at^{2}[/tex]
Replacing eq. 1 and 2:
[tex]-50a=(10a*20)-\frac{1}{2} *g*20^{2} \\-250a=-\frac{1}{2} *9.8*20^{2} \\a=7.84m/s^{2}[/tex]
b) The impact speed is equal:
[tex]v_{i} ^{2} =v^{2} +2gs[/tex]
Where
s = h = -50a
[tex]v^{2} _{i} =(10a)^{2} +2*(-9.8)*(-50a)\\v=\sqrt{100a^{2}+980a } \\v=\sqrt{(100*7.84^{2})+(980*7.84) } =117.6m/s[/tex]
c) The height is:
[tex]v_{i} ^{2} =v^{2} +2gs\\0=(10a)^{2} -2gs\\(10a)^{2} =2gs\\s=\frac{(10a)^{2} }{2g} \\s=\frac{(10*7.84)^{2} }{2*9.8} =313.6m[/tex]
htotal = 313.6 + 50a = 313.6 + (50*7.84) = 705.6 m
A tortoise and a hare are competing in a 1200-meter race. The arrogant hare decides to let the tortoise have a 580-meter head start. When the start gun is fired the hare begins running at a constant speed of 9 meters per second and the tortoise begins crawling at a constant speed of 5 meters per second. Let t t represent the number of seconds that have elapsed since the start gun was fired. Write an expression in terms of t t that represents the hare's distance from the starting line (in meters).
Answer:
Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]
Explanation:
Tortoise travels an additional distance 580 m .
Since there is no acceleration, the distance traveled by any object is d = v t , where v is the total distance and t is the total time taken.
Speed of hare = [tex]v_{h}[/tex] = 9 m/s
Speed of tortoise = [tex]v_{t}[/tex] =5 m/s
Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]
Distance traveled by tortoise = d +580 m = 5 [tex]t_{t}[/tex]
Final answer:
The distance that the hare travels from the starting line after time t is represented by the expression 9t meters.
Explanation:
To write an expression that represents the hare's distance from the starting line in meters as a function of time, we use the formula for distance traveled at constant speed d = vt, where d is distance, v is speed, and t is time. Since the hare starts running at 9 meters per second, the distance the hare travels from the starting line after t seconds can be represented by the expression 9t meters. Note that we do not include the 580-meter head start given to the tortoise in the hare's expression since we are only considering the hare's distance from the starting line and not its position relative to the tortoise.
A car starts from rest and accelerates at 4 m/s^2 for 8 s, after which it continues at constant speed for 19 s. How far would it have traveled from its starting point by that time.
Answer:
The total distance traveled is 736 m
Solution:
According to the question:
Initial velocity, v = 0
(since, the car is starting from rest)
[tex]acceleration, a = 4 m/s^{2}[/tex]
Time taken, t = 8 s
Now, the distance covered by it in 8 s is given by the second eqn of motion:
[tex]d = vt + \farc{1}{2}at^{2}[/tex]
[tex]d = 0.t + \farc{1}{2}4\times 8^{2} = 128 m[/tex]
Now, to calculate the velocity, we use eqn 1 of motion:
v' = v + at
v' = 0 + 4(8) = 32 m/s
Now, the distance traveled by the car with uniform velocity of 32 m/s for t' = 19 s:
distance, d' = v't'
[tex]d' = 32\times 19 = 608 m[/tex]
Total distance traveled = d + d' = 128 + 608 = 736 m