(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 Straight Line Uniformly Accelerated Motion

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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

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 Straight Line

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Uniformly Accelerated Motion

If the velocity of a particle is given by $v = (180 - 16x)^{1/2} \text{ m/s}$,then its acceleration will be ....... $\text{m/s}^2$.

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Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

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$A$ body starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $v = (4t^3 - 2t)$,where $t$ is in $s$ and velocity is in $m/s$. What is the acceleration of the particle when it is $2\, m$ from the origin? (in $m/s^2$)

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$A$ particle experiences a constant acceleration for $20\, s$ after starting from rest. If it travels a distance $s_1$ in the first $10\, s$ and a distance $s_2$ in the next $10\, s$,then:

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$A$ body of mass $4 \,kg$ is accelerated by a constant force. It travels a distance of $5 \,m$ in the first second and a distance of $2 \,m$ in the third second. The force acting on the body is: (in $\,N$)

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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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If the velocity of a particle is given by $v = (180 - 16x)^{1/2} \text{ m/s}$,then its acceleration will be ....... $\text{m/s}^2$.

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$A$ body starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $v = (4t^3 - 2t)$,where $t$ is in $s$ and velocity is in $m/s$. What is the acceleration of the particle when it is $2\, m$ from the origin? (in $m/s^2$)

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