At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.60 m/s^2 . At the same instant a truck, traveling with a constant speed of 23.5 m/s , overtakes and passes the car. a. How far beyond its starting point does the car overtake the truck?b. How fast is the car traveling when it overtakes the truck?


Answer 1


306.8264448 m

47.0016 m/s


t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled by car

s_c=ut+(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)at^2

Distance traveled by truck


In order to overtake both distances should be equal

(1)/(2)at^2=ut\n\Rightarrow (1)/(2)at=u\n\Rightarrow t=(2u)/(a)\n\Rightarrow t=(2* 23.5)/(3.6)\n\Rightarrow t=13.056\ s

s_c=(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)3.6* 13.056^2\n\Rightarrow s_c=306.8264448\ m

The distance the car has to travel is 306.8264448 m

v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 3.6* 306.8264448+0^2)\n\Rightarrow v=47.0016\ m/s

The speed of the car when it overtakes the truck is 47.0016 m/s

Related Questions

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The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?



A. 2.083 MV/m from anode to cathode.

B. 93648278.15 m/s

C. 2.5x10^-5 C and there are about 1.56x10^14 electrons

D. 4x10^-15 Joules


Voltage V across plate is 25 kV = 25x10^3 V

Distance apart x = 1.2 cm = 1.2x10^-2 m

A. Electric field strength is the potential difference per unit distance

E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m

= 2.083 MV/m

B. Energy of electron is electron charge times the voltage across

i.e eV

Charge on electron = 1.6x10^-19 C

Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules

Mass of electron m is 9.12x10^-31 kg

Kinetic energy of electron = 0.5mv^2

Where v is the speed

4x10^-15 = 0.5 x 9.12x10^-31 x v^2

v^2 = 8.77x10^15

v = 93648278.15 m/s

C. From Q = CV

Q = charge

C = capacitance = 1 nF 1x10^-9 F

V = voltage = 25x10^3 V

Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C

Total number of electrons = Q/e

= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons

D. To push electron from cathode to anode, I'll have to do a work of about

4x10^-15 Joules

The tension in a string is 15 N, and its linear density is 0.85 kg/m. A wave on the string travels toward the -x direction; it has an amplitude of 3.6 cm and a frequency of 12 Hz. What are the: a. Speed
b. Wavelength of the wave?
c. Write down a mathematical expression for the wave, substituting numbers for the variables



(a) The speed of the wave, v is 4.2 m/s

(b) Wavelength of the wave, λ is 0.35 m

(c) mathematical expression of the wave, Y = 0.036sin(5.71πx - 24πt)



tension on the string, T = 15 N

Linear density, μ = 0.85 kg/m

amplitude of the wave, A = 3.6 cm = 0.036 m

frequency of the wave, f = 12 Hz

(a) The speed of the wave, v is calculated as;

v = \sqrt{(T)/(\mu) } \n\nv = \sqrt{(15)/(0.85) }\n\nv = 4.2 \ m/s

(b) Wavelength of the wave, λ

v = fλ

λ = v / f

λ = 4.2 / 12

λ = 0.35 m

(c) mathematical expression of the wave;

Y = Asin((2\pi x)/(\lambda) -\omega t)\n\nY = Asin((2\pi x)/(\lambda) -2\pi f t)\n\nY = 0.036sin ((2\pi )/(0.35)x -2\pi *12 t)\n\nY = 0.036sin (5.71 \pi x - 24 \pi t)

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?



a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s


a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_(f) = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_(cm )^(2) + ½ I_(cm)

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g


let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

Explain why it is dangerous to jump from a fast moving train



When you jump off a train, you jump off a certain height and your downwards (vertical) velocity is zero. But your forward (horizontal) velocity is not. You will hit the ground on split second with your horizontal velocity practically the same as the train.


you be in serious injury.

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplished by spinning a sample around in a circle with a large angular speed. Suppose that after a centrifuge in a medical laboratory is turned off, it continues to rotate with a constant angular deceleration for 10.0s before coming to rest.Part A

If its initial angular speed was 3890rpm , what is the magnitude of its angular deceleration? (|?| in revs/s^2 )

Part B

How many revolutions did the centrifuge complete after being turned off?



a_r=389\ rev.s^(-2)

n=58350 rev



time of constant deceleration, t=10\ s


initial angular speed, N_i=3890\ rpm\

Using equation of motion:


0=3890+a_r* 10

a_r=389\ rev.s^(-2)


Using eq. of motion for no. of revolutions, we have:

n=N_i.t+(1)/(2) a_r.t^2

n=3890* 10+0.5* 389* 100

n=58350\ rev

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.