a train at the end of 8.68 s is moving at 15.0164 m/s what is the displacement in the first 6.52736 s of motio

Answer 1

I assume you mean that the train is starting from rest. Also, it has a constant acceleration to acheive the velocity it has at the end of it's motion. Therefore, we have two equations that we need to use in order to solve for the distance:

First, we need to solve for the acceleration, using the velocity and total time we were given:

v = at
==> divide by t on both sides
v / t = a
==> now, plug in our given values to find a
a = (15.0164 m/s) / (8.68 s) = 1.73 m/s²

Now, we can use the position equation for motion that begins from rest. We just need to plug in our acceleration that we solved and the time over which we are interested:

x = ½at²
==> plug in values
x = ½(1.73 m/s²)(6.52736 s) = 36.85456 m

With correct significant digits, your answer is 36.9 meters.

Answer 2

98.0174 m