Question

A space vehicle accelerates uniformly from 85 m/s at t = 0 to 164 m/s at t = 10.0 s .How far did it move between t = 2.0 s and t = 6.0 s ?

Answer 1

Explanation:

Given that,

Initial speed of the vehicle, u = 85 m/s

Final speed of the vehicle, v = 164 m/s

Time, t = 10 s

Firstly, we will find the acceleration of the vehicle using first equation of motion as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{164-85}{10}$

$a=7.9\ m/s^2$

Distance covered by the vehicle at t = 2 s is :

$s_1=ut+\dfrac{1}{2}at^2$

$s_1=85(2)+\dfrac{1}{2}\times 85\times (2)^2$

$s_1=340\ m$

Distance covered by the vehicle at t = 6 s is :

$s_2=ut+\dfrac{1}{2}at^2$

$s_2=85(6)+\dfrac{1}{2}\times 85\times (6)^2$

$s_2=2040\ m$

So, the distance between t = 2 s to t = 6 s is :

$s=s_2-s_1$

$s=2040-340$

s = 1700 meters

Hence, this is the required solution.

Answer 2

First, we have a change in the velocity from 85 to 164 m/s in 10 sec.

Then, we calculate the acceleration as:

$a=\frac{v_{f}-v_{i} }{t} =\frac{164-85}{10}=7.9 m/s^2$

Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:

$v_{f}=v_{i}+at=85+7.9*2=100.8m/s$

Then, using the second equation of motion to calculate the distance:

$d=v_{i} t+\frac{1}{2}at^2$

$d=100.8*2+\frac{1}{2}*7.9*(2)^2=217.4m$