# Help With finding displacement in uniform acceleration

mithilsheth

## Homework Statement

A subway train starting from rest leaves a station with a constant acceleration. At the
end of 7.8 s, it is moving at 11.778 m/s. What is the train’s displacement in the ﬁrst 5.6394 s of motion? Answer in units of m

## Homework Equations

a= velocity final - velocity initial / time

delta x = velocity initial(time) + 1/2 (a)(t)^2

## The Attempt at a Solution

delta x = 0 (7.8)+ 1/2 (1.51)(7.8)^2
delta x = 1/2 91.8684
delta x = 45.9342
that is delta x for 7.8 seconds. how do i find delta x for 5.6394 seconds?

## The Attempt at a Solution

Mentor
that is delta x for 7.8 seconds. how do i find delta x for 5.6394 seconds?
Use the same formula, but with the different time! (Why did you solve for x at t = 7.8 second?)

mithilsheth
i also did substitute 5.6394 s for the time rather than 7.8 seconds, and i got 66.4208532002 for delta x. but this was wrong. what did i do wrong?

Mentor
i also did substitute 5.6394 s for the time rather than 7.8 seconds, and i got 66.4208532002 for delta x. but this was wrong. what did i do wrong?
Show your calculation. (You know you must have done something wrong. This displacement is bigger than the one you found for the longer time. That can't be right.)

mithilsheth
delta x = 1/2 (1.51)(5.6394)^2
delta x = 24.0111384

that is what i got this time.....

Mentor
delta x = 1/2 (1.51)(5.6394)^2
delta x = 24.0111384

that is what i got this time.....
That's more like it.

1 person