Equation of motion of a body is `(dv)/(dt) = -4v + 8,` where v is the velocity in `ms^-1` and t is the time in second. Initial velocity of the particle was zero. Then,

Options

A. the initial rate of change of acceleration of the particle is `8 ms^-2`
B. the terminal speed is `2 ms^-1`
C. Both (a) and (b) are correct
D. Both (a) and (b) are wrong

Answer

B

Explanation

`(dv)/(dt) =a =-44v+8`
`:. (da)/(dt)=-4 (dv)/(dt) `
`=-4(-4v+8)`
`=16v-32`
`:. ((da)/(dt))_i=16v_i-32`
`=(16)(0)-32`
`=-32 m//s^2`
Further, `int_0^v (dv)/(8-4v)=int_0^tdt`
Solving this equation we get,
`v=2(1-e^(-4t))`

Hence, v-t graph is exponentially incresing graph,
terminating at 2 m//s.