Find the constant acceleration of the train

[Quang]

Homework Statement

Liz rushes down onto a subway platform to find her train
already departing. She stops and watches the cars go by.
Each car is 8.60 m long. The first moves past her in 1.50 s
and the second in 1.10 s. Find the constant acceleration of
the train.

Homework Equations

The Attempt at a Solution

vi = 8.6/1.5
vf = 8.6/1.1
a = 2(vf - vi)/2.6
?
any advice? I'm confused

 

[Doc Al]

You're given distance and time, so what kinematic formula relates those two for constant acceleration?

 

[nlightner]

First, we can calculate the average velocity of the train by dividing the distance traveled by the time it took for the first car to pass by. This would be vi = 8.6/1.5 = 5.73 m/s. Similarly, for the second car, the average velocity would be vf = 8.6/1.1 = 7.82 m/s.

Now, we can use the formula for average acceleration, a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time it took for the train to accelerate.

Plugging in the values, we get a = (7.82 - 5.73)/2.6 = 0.81 m/s^2. This is the constant acceleration of the train.

In order to confirm our answer, we can also use the formula for distance traveled with constant acceleration, d = vi*t + 1/2*a*t^2.

For the first car, we know that the distance traveled is 8.6 m, vi = 5.73 m/s, and t = 1.5 s. Plugging in these values, we get 8.6 = 5.73*1.5 + 1/2*a*1.5^2. Solving for a, we get a = 0.81 m/s^2, which confirms our previous answer.

Therefore, the constant acceleration of the train is 0.81 m/s^2.