The motion of a particle along a straight lin | vedclass

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(C) Given the position function: $x = (2t - 3)^2$.Expanding the expression: $x = 4t^2 - 12t + 9$.Velocity $v$ is the first derivative of position with

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$8$

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(C) Given the position function: $x = (2t - 3)^2$.Expanding the expression: $x = 4t^2 - 12t + 9$.Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(4t^2 - 12t + 9) = 8t - 12$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(8t - 12) = 8 \,m/s^2$.Since the acceleration is constant,at $t = 2 \,s$,the acceleration remains $8 \,m/s^2$.

The motion of a particle along a straight line is described by the function $x = (2t - 3)^2$,where $x$ is in metres and $t$ is in seconds. The acceleration of the particle at $t = 2 \,s$ is (in $\,m/s^2$)

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