(C) Given the position function: $x = at^2 - bt^3$Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = 2at - 3b

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(C) Given the position function: $x = at^2 - bt^3$Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = 2at - 3bt^2$Acceleration $a_{acc}$ is the derivative of velocity with respect to time: $a_{acc} = \frac{dv}{dt} = 2a - 6bt$To find the time when acceleration is zero,set $a_{acc} = 0$:$0 = 2a - 6bt$$6bt = 2a$$t = \frac{2a}{6b} = \frac{a}{3b}$

Class 11 Physics Motion in Straight Line Acceleration and its graph

Easy

The position $x$ of a particle varies with time as $x = at^2 - bt^3$,where $a$ and $b$ are constants. The acceleration of the particle will be zero at:

MHT CET 2023 All MHT CET

A
$\frac{2a}{3b}$
B
$\frac{a}{b}$
C
$\frac{a}{3b}$
D
zero

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Motion in Straight Line

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Acceleration and its graph

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The acceleration-time graph for a particle moving along the $x$-axis is shown in the figure. If the initial velocity of the particle is $-5 \, m/s$, the velocity at $t = 8 \, s$ is ....... $m/s$.

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$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 3t^2 + 2) \, m$. The displacement when the acceleration becomes zero is ........ $m$.

Medium

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The displacement is given by $x = 2t^2 + t + 5$,the acceleration at $t = 2 \ s$ is ......... $m/s^2$.

Easy

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What will be the change in the speed of a moving object if both speed and acceleration are positive or both are negative?

Medium

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$A$ particle has a velocity in the negative direction and a constant acceleration in the positive direction. Match the following columns:

Column $I$ Column $II$

$(A)$ Velocity-time graph $(p)$ Slope $\rightarrow$ negative

$(B)$ Acceleration-time graph $(q)$ Slope $\rightarrow$ positive

$(C)$ Displacement-time graph $(r)$ Slope $\rightarrow$ zero

$(s)$ $|\text{Slope}| \rightarrow$ increasing

$(t)$ $|\text{Slope}| \rightarrow$ decreasing

$(u)$ $|\text{Slope}| \rightarrow$ constant

Column $I$	Column $II$
$(A)$ Velocity-time graph	$(p)$ Slope $\rightarrow$ negative
$(B)$ Acceleration-time graph	$(q)$ Slope $\rightarrow$ positive
$(C)$ Displacement-time graph	$(r)$ Slope $\rightarrow$ zero
	$(s)$ $\|\text{Slope}\| \rightarrow$ increasing
	$(t)$ $\|\text{Slope}\| \rightarrow$ decreasing
	$(u)$ $\|\text{Slope}\| \rightarrow$ constant

Medium

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The position $x$ of a particle varies with time as $x = at^2 - bt^3$,where $a$ and $b$ are constants. The acceleration of the particle will be zero at:

Similar Questions

The acceleration-time graph for a particle moving along the $x$-axis is shown in the figure. If the initial velocity of the particle is $-5 \, m/s$, the velocity at $t = 8 \, s$ is ....... $m/s$.

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 3t^2 + 2) \, m$. The displacement when the acceleration becomes zero is ........ $m$.

The displacement is given by $x = 2t^2 + t + 5$,the acceleration at $t = 2 \ s$ is ......... $m/s^2$.

What will be the change in the speed of a moving object if both speed and acceleration are positive or both are negative?

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