Physics word problem

Question

A train has a length of 109 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just reaches the end

of the train. The car is moving with a constant velocity. At a time t = 10.0 s, the car just reaches the front of the train. Ultimately,

however, the train pulls ahead of the car, and at time t = 32.1 s, the car is again at the rear of the train. Find the magnitudes of (a) the

car's velocity and (b) the train's acceleration.

Answer 1

The general equation for displacement with constant acceleration at various time is:

x = x₀ + v₀ t + (1/2) a t²

where x₀ is the initial position and v₀ is the initial velocity.

We have to choose a place to measure the location of the train. Since the car starts even with the end of the train, let's use the back of the train as it location. At the start, both the car and the train have location x = 0 at t = 0.

Then we know the following:

1. at time t = 0, the position of the car and the train are zero: x_C = x_T = 0
2. at time t = 10 s, the car is at the front of the train. We don't know the position of either relative to the starting location, but we know the position of the car must be 109 m more than the position of the back of the train, so x_C = x_T + 109
3. at time t = 32.1 s, the car is again even with the back of the train so x_C = x_T

Since the car is not accelerating and we set its original position to be zero, the position of the car at any other time is

x_C = v_C t

Since the train starts from rest, its initial velocity is zero, its original position is also zero. The position of the train (again, the end of the train) is given by:

x_T = (1/2) a_T t²

32.1 s

at 32.1s we know that x_C = x_T using the above equations, this means:

v_C (32.1) = (1/2) a_t (32.1)²

or

v_c = 16.05 a_t

10s

at 10 s we have:

x_C = x_T + 109

so

v_c 10 = (1/2) a_t (10)² + 109

substituting from what we found above; this is:

(16.05 a_T ) (10) = (1/2) a_T(10)² + 109

160.5 a_T = 50 a_T +109

or

110.5 a_T = 109

so

a_T = 0.986 m/s²

and using the equation relating v_C and a_T

v_C = 16.05 a_T = 16.05 (0.986) = 15.83 m/s

Check your answer

It's always a good idea to check that the values make sense. The speed of the car is about 35 mph, which make sense. Also double check that the positions agree with the problem:

at t = 10 s:

15.832 (10) = 158.3 m for the car

and

(1/2) 0.986 (10)² = 49.3 m for the train

so the car is 109 ahead of the end of the train (which means its even with the front of the train)

and at t = 32.1 s

15.832 (32.1) = 508.2 m for the car

and

(1/2) 0.986 (32.1)² = 508.2 m for the train

so the car is again even with the end of the train.

Answer 2

Using the kinematic equation: x_f = x_o + v_ot + (1/2)at², we can find positions for both the train and car in terms of their velocity and acceleration:

For the car:

x_C = v_Ct, since its initial position is zero, its initial velocity is v_C, and its acceleration is zero

For the train:

x_T = (1/2)a_Tt², since its initial position is zero, its initial velocity is zero, and its acceleration is a_T

At t = 10 seconds, the front of the car and the front of the train are a the same position:

x_C = x_T

_so...

v_Ct = (1/2)a_Tt², then plug in t = 10 sec

10v_C = 50a_T

v_C = 5a_T

At t = 32.1 seconds, the front of the train is 109 m ahead of the front of the car

x_C = x_T - 109m

_so...

v_Ct = (1/2)a_Tt² - 109, then plug in v_C = 5a_T

5a_T*t = (1/2)a_Tt² - 109, then solve for a_T

(1/2)a_Tt² - 5a_T*t = 109

a_T((1/2)t² - 5t) = 109

a_T = 109 / ((1/2)t² - 5t), then plug in t = 32.1 sec

a_T = 109 / ((1/2)(32.1)² - 5(32.1))

a_T = 0.3073 m/s²

then use v_C = 5a_T to solve for v_C:

v_C = 5 * 0.3073

v_C = 1.536 m/s