(B) Given position: $x = t^3 - 6t^2 + 20t + 15 \ m$.Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = 3t^2 -

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(B) Given position: $x = t^3 - 6t^2 + 20t + 15 \ m$.Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = 3t^2 - 12t + 20 \ m/s$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 12 \ m/s^2$.Set acceleration to zero to find the time: $6t - 12 = 0 \implies t = 2 \ s$.Substitute $t = 2 \ s$ into the velocity equation: $v = 3(2)^2 - 12(2) + 20 = 12 - 24 + 20 = 8 \ m/s$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

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$A$ particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15) \ m$. The velocity of the body when its acceleration becomes zero is ........... $m/s$.

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A
$4$
B
$8$
C
$10$
D
$6$

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Motion in Straight Line

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Uniformly Accelerated Motion

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$A$ particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15) \ m$. The velocity of the body when its acceleration becomes zero is ........... $m/s$.

Similar Questions

In some appropriate units,the time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^2 + x$. The acceleration of the particle is

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