The velocity of an object moving in a straight line path is given as a function of time by `v=6t-3t^2`, where v is in `ms^-1`, t is in s. The average velocity of the object between , t=0 and t=2 s is

Options

A. 0
B. `3ms^-1`
C. `2ms^-1`
D. `4ms^-1`

Explanation

Given velocity, `v=6t-3t^2`
As we know that,
`v= (dx)/(dt)`
Here, x is the displacement of the particle.
Now, dx= vdt
Integrate on the both sides, limit t=0 to t=2, we get
`thereforex=int_0^2v dt=int_0^2(6t-3t^2)dt`
`=[(6t^2)/2]_0^2-[(3t^3)/3]_0^2=[3t^2]_0^2-[t^3]_0^2`
`=[3(2)^2-3(0)^2]-[(2)^3-(o)^2]`
`[12-0]-[8-0]=12-8=4m`
Average velocity , `v_(avg)=(Total displacement)/(Total time taken)`
`=4/2=2m//s`
Hence, the average velocity of the object between t=0 to t=2s is `2m//s`.