(C) Given the velocity equation: $v = \frac{t^2}{10} + 20$.To find the acceleration $a$,we differentiate the velocity with respect to time $t$:$a = \f

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(C) Given the velocity equation: $v = \frac{t^2}{10} + 20$.To find the acceleration $a$,we differentiate the velocity with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{t^2}{10} + 20 \right)$.Applying the power rule:$a = \frac{2t}{10} + 0 = \frac{t}{5}$.Since the acceleration $a$ depends on time $t$ $(a \propto t)$,it is not constant. Therefore,the body is undergoing non-uniform acceleration.

Class 11 Physics Motion in a Straight Line Acceleration and its graph

Medium

The velocity of a body depends on time according to the equation $v = \frac{t^2}{10} + 20$. The body is undergoing

A
Uniform acceleration
B
Uniform retardation
C
Non-uniform acceleration
D
Zero acceleration

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Motion in a Straight Line

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Acceleration and its graph

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The velocity of a body depends on time according to the equation $v = \frac{t^2}{10} + 20$. The body is undergoing

Similar Questions

$A$ particle starts from rest. Its acceleration $a$ versus time $t$ is shown in the figure. The maximum speed of the particle will be (in $m/s$)

$Assertion$ : Retardation is directly opposite to the velocity. $Reason$ : Retardation is equal to the time rate of decrease of speed.

If the speed of a moving object decreases,what are the directions of its velocity and acceleration?

$A$ particle is moving with constant acceleration,whose direction is the same as that of its instantaneous motion. The distance $(s)-$velocity $(v)$ graph for the particle is:

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:

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$Assertion$ : Retardation is directly opposite to the velocity.
$Reason$ : Retardation is equal to the time rate of decrease of speed.