Question

A space vehicle accelerates uniformly from 65 m/s at t= 0
to 162 m/s at t=10.0s. How far did it move between t=2.0s and
t=6.0 s?​

Answer 1

Answer:

The uniform acceleration (a) is from initial velocity (65 m/s) to final velocity (162 m/s) over a span of 10 seconds.

a = (162–65)/10= 97/10= 9.7 m/s^2

Distance traveled (s) at any time (t) with an initial velocity (u) and acceleration (a) is:

s = u t + 1/2 a t^2

So, the distance at 2 seconds is:

s = 65 (2) + 1/2 (9.7) (2)^2

s = 130 + 19.4 = 149.4 m

And the distance at 6 seconds is:

s = 65 (6) + 1/2 (9.7) (6)^2

s = 390 + 174.6 = 564.6 m

Distance traveled between 2 seconds (149.4m) and 6 seconds (564.6 m) is:

564.6 - 149.4 = 415.2 meters