Question

A train has a length of 102 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 = 18.3 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 38.0 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

atrain = 0.5658 m/s^2

vocar = 10.75 m/s

Step-by-step explanation:

we have the following data:

vcar =?

acar = 0

votrain = 0

atrain =?

t1 = 18.3 s

L = 102 m

t2 = 38 s

For time t1, the distance that the car advances is the same as the distance that the train advances, to this we must add the length of the train. For time t2, the distance the car travels is the same as the train travels. Therefore, an equation equals:

x = vo*t + (a*t^2)/2

for the car, we have:

dcar = vocar*t

for the train, we have:

dtrain = (atrain*t^2)/2

dcar = dtrain + L = (atrain*t1^2)/2 + L = vocar*t1 (For the time 1, t1)

For the time 2, t2:

dcar = dtrain = (atrain*t2^2)/2 = vocar*t2

Clearing vocar:

vocar = (atrain*t2)/2 (eq. 1)

(atrain*t1^2)/2 + L = vocar*t1 = (atrain*t2) * t1

Clearing atrain:

(atrain*t2) * t1 - (atrain*t1^2) = L

atrain = (2*L/(t2*t1 - t1^2) = (2*102)/(38*18.3 - 18.3^2) = 0.5658 m/s^2

In eq. 1:

vocar = (0.5658*38)/2 = 10.75 m/s

Answer 2

Step-by-step explanation:

We must analyze this problem carefully, they give us conditions for three different moments of time.

a) Let's start with the second time we have that the car that started in the final part of the train has advanced to the front, so the distance traveled is the length of the train (x = 102m) and as it goes at constant speed we calculate it

v = x / t

v = 102 / 18.3

v = 5.57 m / s

This is the speed of the car at all times

b) It is the second part we have the time and they tell us that the car is back at the back of the train, this is because the train has acceleration, we can calculate the total distance traveled by the car

x = V t

x = 5.57 38

x = 211.7 m

In this distance the car goes and returns to the back of the train, but the train has also advanced the same distance, since the car is at the same point. Therefore we can calculate the acceleration of the train

x = vo t + ½ a t²

The initial speed is zero for the train the distance is the same for the two vehicles

x = ½ to t²

a = 2x / t²

a = 2 211.7 / 38²

a = 0.29 m / s²

the train acceleration os 0.29 m/s²