A subway train starting from rest leaves a
station with a constant acceleration. At the
end of 7.09 s, it is moving at 20.1356 m/s.
What is the train’s displacement in the first
4.67231 s of motion?
Answer in units of m

Question

A subway train starting from rest leaves a
station with a constant acceleration. At the
end of 7.09 s, it is moving at 20.1356 m/s.
What is the train’s displacement in the first
4.67231 s of motion?
Answer in units of m

danjs · Answer

Constant acceleration starting from rest.

$$v _{i}=0$$
$$v _{f}=20.1356 ~~~when~~~t=7.09$$
$$x _{0}=0$$
$$x _{f}=?$$

$$\large a =\frac{ \Delta v }{ \Delta t }=\frac{ 20.1356 }{ 7.09 }=2.84 ~~m/s^2$$

There is the acceleration, you can use a kinematic equation to geta function for displacement..

$$\large x _{f}-x _{0}=v _{0}*t + \frac{ 1 }{ 2 }*a*t^2$$
use original velocity is zero, and acceleratipon is 2.84, initial position is 0.
$$\large x _{f}=\frac{ 1 }{ 2 }*2.84*t^2$$

danjs · Answer

That is it. You can put in a value for time t and get the displacement.