Question

A subway train starting from rest leaves a

station with a constant acceleration. At the
end of 8.06 s, it is moving at 21.0366 m/s.
What is the train’s displacement in the first
5.41632 s of motion?
Answer in units of m

Answer 1

38.2842m

Explanation:

Answer 2

38.2842 m

Explanation:

The acceleration is given by ...

a = ∆v/∆t

and the distance traveled is given by ...

d = 1/2at^2

So the distance we're looking for is ...

d = (1/2)(21.0366 m/s)/(8.06 s)(5.41632 s)^2 = 32.2842 m