(C) Given acceleration $a = \frac{dv}{dt} = bt$.Integrating with respect to time $t$:$\int dv = \int bt \, dt \Rightarrow v = \frac{1}{2}bt^2 + C_1$.A

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(C) Given acceleration $a = \frac{dv}{dt} = bt$.Integrating with respect to time $t$:$\int dv = \int bt \, dt \Rightarrow v = \frac{1}{2}bt^2 + C_1$.At $t = 0$,$v = v_0$,so $C_1 = v_0$.Thus,the velocity is $v = \frac{1}{2}bt^2 + v_0$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \frac{1}{2}bt^2 + v_0$.Integrating with respect to time $t$:$x = \int (\frac{1}{2}bt^2 + v_0) dt = \frac{1}{2}b(\frac{t^3}{3}) + v_0t + C_2$.At $t = 0$,$x = 0$,so $C_2 = 0$.Therefore,the distance travelled is $x = v_0t + \frac{1}{6}bt^3$.

Class 11 Physics Motion in a Straight Line Uniformly Accelerated Motion

Medium

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be

A
${v_0}t + \frac{1}{3}b{t^2}$
B
${v_0}t + \frac{1}{3}b{t^3}$
C
${v_0}t + \frac{1}{6}b{t^3}$
D
${v_0}t + \frac{1}{2}b{t^2}$

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Motion in a Straight Line

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Uniformly Accelerated Motion

$A$ body starting from rest at $t=0$ moves along a straight line with a constant acceleration. At $t=2 \ s$,the body reverses its direction keeping the acceleration same. The body returns to the initial position at $t=t_0$,then $t_0$ is

MediumAP EAMCET 2019

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The position of a particle moving along the $x$-axis is given by $x = (-2t^3 + 3t^2 + 5) \ m$. The acceleration of the particle at the instant its velocity becomes zero is ....... $m/s^2$.

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The velocity of a body moving with a uniform acceleration of $2\,m/s^2$ is $10\,m/s$. Its velocity after an interval of $4\,s$ is...........$m/s$.

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$A$ lift is coming down from the $8^{th}$ floor and is just about to reach the $4^{th}$ floor. Taking the ground floor as the origin and the upward direction as positive for all quantities,which one of the following is correct?

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An automobile,travelling at $40\, km/h$,can be stopped at a distance of $40\, m$ by applying brakes. If the same automobile is travelling at $80\, km/h$,the minimum stopping distance,in metres,is (assume no skidding)..........$m$

MediumJEE MAIN 2018

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The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be

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$A$ body starting from rest at $t=0$ moves along a straight line with a constant acceleration. At $t=2 \ s$,the body reverses its direction keeping the acceleration same. The body returns to the initial position at $t=t_0$,then $t_0$ is

The position of a particle moving along the $x$-axis is given by $x = (-2t^3 + 3t^2 + 5) \ m$. The acceleration of the particle at the instant its velocity becomes zero is ....... $m/s^2$.

The velocity of a body moving with a uniform acceleration of $2\,m/s^2$ is $10\,m/s$. Its velocity after an interval of $4\,s$ is...........$m/s$.

$A$ lift is coming down from the $8^{th}$ floor and is just about to reach the $4^{th}$ floor. Taking the ground floor as the origin and the upward direction as positive for all quantities,which one of the following is correct?

An automobile,travelling at $40\, km/h$,can be stopped at a distance of $40\, m$ by applying brakes. If the same automobile is travelling at $80\, km/h$,the minimum stopping distance,in metres,is (assume no skidding)..........$m$

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