The velocity of a particle moving along a straight line increases according to

Question -

The velocity of a particle moving along a straight line increases according to the linear law v = $v_{0}$ + kx

where k is a constant . Then

1) the accelearation of the particle is k( $v_{0}$ + kx)

2) the particle takes a time $\frac{1}{k}$ $l o g_{e}$ ( $\frac{v_{1}}{v_{0}}$ ) to attain a velocity $v_{1}$

3) velocity varies linearly with displacement with slope of velocity displacement curve equal to k

4) data is insufficient to arrive at a conclusion.

Options:

A . 1,2,3 are correct

B . 2,3 are correct

C . all of them are correct

D . 1 and 2 are correct

Answer: Option A
:
A

Options (1), (2), (3) are correct.

Accelearation = $\frac{d v}{d t}$ =kv

$\Rightarrow$ a = k ( $v_{0}$ + kx)

Further

$\Rightarrow$ $\frac{d v}{d t}$ = kv

$\Rightarrow$ $\frac{d v}{v}$ = kdt

$v_{1} \int v_{0}$ $\frac{d v}{v}$ = k $t \int 0$ ft

t = $\frac{1}{k}$ $l o g_{e}$ $\frac{v_{1}}{v_{0}}$

Since , v = $v_{0}$ + kx . Hence slope of velocity displacement curve is $\frac{d v}{d x}$ = k

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