(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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DifficultJEE MAIN 2024

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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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The displacement $(x)$ of a particle depends on time $t$ as $x = \alpha t^2 - \beta t^3$. Choose the incorrect statements from the following.

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