(C) Given acceleration $a = \frac{dv}{dt} = bt$.Integrating with respect to time to find velocity $v$:$\int_{v_0}^{v} dv = \int_{0}^{t} bt \, dt$$v -

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(C) Given acceleration $a = \frac{dv}{dt} = bt$.Integrating with respect to time to find velocity $v$:$\int_{v_0}^{v} dv = \int_{0}^{t} bt \, dt$$v - v_0 = \frac{1}{2}bt^2$$v = v_0 + \frac{1}{2}bt^2$Now,since $v = \frac{ds}{dt}$,we integrate to find displacement $s$:$\int_{0}^{s} ds = \int_{0}^{t} (v_0 + \frac{1}{2}bt^2) dt$$s = [v_0t + \frac{1}{2}b \cdot \frac{t^3}{3}]_0^t$$s = v_0t + \frac{1}{6}bt^3$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Difficult

The acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.

A
$v_0t + \frac{1}{3}bt^2$
B
$v_0t + \frac{1}{3}bt^3$
C
$v_0t + \frac{1}{6}bt^3$
D
$v_0t + \frac{1}{2}bt^2$

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Motion in Straight Line

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Uniformly Accelerated Motion

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$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 6t^2 + 3t + 4) \ m$. Find the velocity of the particle in $m/s$ when its acceleration is zero.

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What is stopping distance for a vehicle? What will be the stopping distance if the initial velocity is doubled?

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$A$ body travels in a straight line from point $A$ to point $B$ with an initial velocity zero and uniform acceleration, covering $1 \,m$ during the first second and $39 \,m$ during the last second. The distance between $A$ and $B$ in metre is

EasyTS EAMCET 2018

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The acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.

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