You visited us 1 times! Enjoying our articles? Unlock Full Access!

Motion Under Variable Acceleration

A body is mov...

Question

A body is moving in a straight line, starting its journey from origin. At any instant, its velocity is given by, $K_{1} t^{(K_{1} - 1)}$ when $K_{1}$ is a constant and t is the time. Find the acceleration of the particle, when it is at a distance s from the origin:

$K_{1} (K_{1} - 1) s^{(1 - 1 / K_{1})}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$K_{1} (K_{1} - 1) s^{(1 - 2 / K_{1})}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$K_{1} (K_{1} - 1) s^{(2 - 1 / K_{1})}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$K_{1} (K_{1} - 1) s^{(3 - 1 / K_{1})}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$K_{1} (K_{1} - 1) s^{(1 - 2 / K_{1})}$

$v = K_{1} t^{(K_{1} - 1)} = \frac{d x}{d t} \int_{o}^{s} d x = s = \int K_{1} t^{(K_{1} - 1)} . d t = t^{(K_{1})} t = (S)^{1 / K_{1}} a = \frac{d v}{d t} = K_{1} (K_{1} - 1) t^{K_{1} - 2} = K_{1} (K_{1} - 1) (S)^{(1 - 2 / K_{1})}$

Suggest Corrections

Similar questions

(4 t^{3} - 2 t)

x - a x i s

(4 t^{3} - 2 t)

t

m / s

2 m

4 t^{3} - 2 t

m / s

2 m

\frac{t^{4}}{4} - 2 t^{3} + 4 t^{2} - 7 .

t

s = \frac{1}{4} t^{4} - 2 t^{3} + 4 t^{2} - 7.

t_{v}

t_{a}

Join BYJU'S Learning Program