First, we have a change in the velocity from 85 to 164 m/s in 10 sec.
Then, we calculate the acceleration as:
[tex]a=\frac{v_{f}-v_{i} }{t} =\frac{164-85}{10}=7.9 m/s^2[/tex]
Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:
[tex]v_{f}=v_{i}+at=85+7.9*2=100.8m/s[/tex]
Then, using the second equation of motion to calculate the distance:
[tex]d=v_{i} t+\frac{1}{2}at^2[/tex]
[tex]d=100.8*2+\frac{1}{2}*7.9*(2)^2=217.4m[/tex]
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 :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{164-85}{10}[/tex]
[tex]a=7.9\ m/s^2[/tex]
Distance covered by the vehicle at t = 2 s is :
[tex]s_1=ut+\dfrac{1}{2}at^2[/tex]
[tex]s_1=85(2)+\dfrac{1}{2}\times 85\times (2)^2[/tex]
[tex]s_1=340\ m[/tex]
Distance covered by the vehicle at t = 6 s is :
[tex]s_2=ut+\dfrac{1}{2}at^2[/tex]
[tex]s_2=85(6)+\dfrac{1}{2}\times 85\times (6)^2[/tex]
[tex]s_2=2040\ m[/tex]
So, the distance between t = 2 s to t = 6 s is :
[tex]s=s_2-s_1[/tex]
[tex]s=2040-340[/tex]
s = 1700 meters
Hence, this is the required solution.
