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

Step-by-step 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=(v-u)/(t)`

`a=(164-85)/(10)`

`a=7.9\ m/s^2`

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

`s_1=ut+(1)/(2)at^2`

`s_1=85(2)+(1)/(2)* 85* (2)^2`

`s_1=340\ m`

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

`s_2=ut+(1)/(2)at^2`

`s_2=85(6)+(1)/(2)* 85* (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=(v_(f)-v_(i) )/(t) =(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+(1)/(2)at^2`

`d=100.8*2+(1)/(2)*7.9*(2)^2=217.4m`