The velocity of a particle along a straight line increases according to the linear law `v=v_(0)+kx`, where k is a constant. Then

Options

A. the acceleration of the particle is `k(v_(0)+kx)`
B. the particle takes a time `(1)/(k)log_(2)((v_(1))/(v_(0)))` to attain a velocity `v_(1)`
C. velocity varies linearly with displacement with slope of velocity displacement curve equal to k.
D. data is insufficient to arrive at a conclusion.

Answer

A, B, C

Explanation

`"Acceleration"=(dv)/(dt)=v=0+kx`
`{becausex=(dx)/(dt)=v}impliesv=a=kv=k(v_(0)+kx)`
further, `a=(dv)/(dt)=kvimplies(dv)/(dt)=kvimplies(dv)/(v)=kdt`
`impliesint_(v_(o))^(v_(1))(dv)/(v)=kint_(0)^(t)dtimpliest=(1)/(k)log_(e)((v_(1))/(v_(0)))`
Since, `v=v_(0)+Kx.` hence slope of velocity displacement curve is `(dv)/(dx)=k`