The velocity of a particle moving along a straight line increases according to the linear law $v = v_{0} + k x$ , where $k$ is a constant. Then

the acceleration of the particle is

k (v_{0} + k x)

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the particle takes a time

\frac{1}{k} l o g_{e} (\frac{v_{1}}{v_{0}})

to attain a velocity

v_{1}

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velocity varies linearly with displacement with slope of velocity displacement curve equal to

k

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the acceleration of the particle is zero.

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Solution

The correct options are
A the acceleration of the particle is $k (v_{0} + k x)$ .
B the particle takes a time $\frac{1}{k} l o g_{e} (\frac{v_{1}}{v_{0}})$ to attain a velocity $v_{1}$ .
C velocity varies linearly with displacement with slope of velocity displacement curve equal to $k$ .
$v = v_{0} + k x$
$\to a = \to v \frac{d \to v}{d x} = (v_{0} + k x) \frac{d}{d x} (v_{0} + k x)$
$\to a = (v_{0} + k x) k = k (v_{0} + k x)$
Hence, option A is correct .

$v = v_{0} + k x$
$\frac{d x}{d t} = v_{0} + k x$
$\Rightarrow \int_{v_{0}}^{v_{1}} \frac{d x}{v_{0} + k x} = \int_{t_{1}}^{t_{2}} d t$
$\Rightarrow \frac{1}{k} l o g_{e} (v_{0} + k x) = t$
$\Rightarrow \frac{1}{k} l o g_{e} (\frac{v_{1}}{v_{0}}) = t_{2} - t_{1} = Δ t$

$Δ t$ is time taken to reach $v_{1}$ .
Hence, option B is correct.

From given velocity equation: $v = v_{0} + k x$ velocity varies linearly with distance travelled and slope of the curve is equal to $k$ .
Hence, option C is correct.

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Similar questions

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.

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.

a = 2 s

a = - k y

a

y . (if v_{0}

)

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