(C) Given the linear momentum $p = 5t^2 + t + 5$ and mass $m = 5 \,kg$.Since $p = mv$,we have $v = \frac{p}{m} = \frac{5t^2 + t + 5}{5} = t^2 + 0.2t +

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variable acceleration which is increasing with time

(C) Given the linear momentum $p = 5t^2 + t + 5$ and mass $m = 5 \,kg$.Since $p = mv$,we have $v = \frac{p}{m} = \frac{5t^2 + t + 5}{5} = t^2 + 0.2t + 1$.The acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 + 0.2t + 1) = 2t + 0.2$.As time $t$ increases,the acceleration $a = 2t + 0.2$ also increases linearly with time.Therefore,the body is moving with variable acceleration which is increasing with time.

Class 11 Physics Newton's Laws of Motion and Friction Second Law of Motion

Medium

The linear momentum $p$ of a body of mass $5 \,kg$ varies with time $t$ as $p = 5t^2 + t + 5$. It follows that the body is moving with:

A
constant acceleration
B
constant speed
C
variable acceleration which is increasing with time
D
variable retardation which is decreasing with time

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Newton's Laws of Motion and Friction

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Second Law of Motion

$A$ body of mass $5 \ kg$ is moving in a straight line. The relation between its displacement and time is $x = (t^3 - 2t - 10) \ m$. What is the force acting on it at the end of $5 \ s$ (in $N$)?

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At any instant,the velocity of a particle of mass $500 \, g$ is $(2t \hat{i} + 3t^2 \hat{j}) \, ms^{-1}$. If the force acting on the particle at $t = 1 \, s$ is $(\hat{i} + x \hat{j}) \, N$,then the value of $x$ will be:

MediumJEE MAIN 2023

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$A$ force $F = Be^{-Ct}$ acts on a particle of mass $m$,which is at rest at $t = 0$. Its terminal velocity is:

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$A$ force on an object of mass $100\, g$ is $(10 \hat{i}+5 \hat{j})\ N$. The position of that object at $t= 2\ s$ is $(a \hat{i}+b \hat{j})\, m$ after starting from rest. The value of $\frac{a}{b}$ will be..........

MediumJEE MAIN 2022

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$A$ body of mass $10 \,kg$ is acted upon by a force given by the equation $F = (3t^2 - 30) \,N$. The initial velocity of the body is $10 \,m/s$. What is the velocity of the body after $5 \,s$ (in $\,m/s$)?

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The linear momentum $p$ of a body of mass $5 \,kg$ varies with time $t$ as $p = 5t^2 + t + 5$. It follows that the body is moving with:

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$A$ body of mass $5 \ kg$ is moving in a straight line. The relation between its displacement and time is $x = (t^3 - 2t - 10) \ m$. What is the force acting on it at the end of $5 \ s$ (in $N$)?

At any instant,the velocity of a particle of mass $500 \, g$ is $(2t \hat{i} + 3t^2 \hat{j}) \, ms^{-1}$. If the force acting on the particle at $t = 1 \, s$ is $(\hat{i} + x \hat{j}) \, N$,then the value of $x$ will be:

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$A$ body of mass $10 \,kg$ is acted upon by a force given by the equation $F = (3t^2 - 30) \,N$. The initial velocity of the body is $10 \,m/s$. What is the velocity of the body after $5 \,s$ (in $\,m/s$)?

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