(C) The position of the particle is given by the equation $x = 5t^2 - 4t + 5$.To find the velocity $v$,we differentiate the position $x$ with respect

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(C) The position of the particle is given by the equation $x = 5t^2 - 4t + 5$.To find the velocity $v$,we differentiate the position $x$ with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt}(5t^2 - 4t + 5)$.Applying the power rule,we get $v = 10t - 4$.Now,substitute $t = 2 \, s$ into the velocity equation:$v = 10(2) - 4 = 20 - 4 = 16 \, ms^{-1}$.Therefore,the magnitude of the velocity at $t = 2 \, s$ is $16 \, ms^{-1}$.

Class 11 Physics Motion in a Straight Line Instantaneous Velocity and Speed and Velocity-time Graph

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The position of a particle related to time is given by $x = (5t^2 - 4t + 5) \, m$. The magnitude of velocity of the particle at $t = 2 \, s$ will be $........... \, ms^{-1}$.

JEE MAIN 2023 All JEE MAIN

A
$10$
B
$14$
C
$16$
D
$06$

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Motion in a Straight Line

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Instantaneous Velocity and Speed and Velocity-time Graph

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The position of a particle related to time is given by $x = (5t^2 - 4t + 5) \, m$. The magnitude of velocity of the particle at $t = 2 \, s$ will be $........... \, ms^{-1}$.

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