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Question

(D) Given the displacement equation: $x = c_0(t^2 - 2) + c(t - 2)^2$. To find the velocity $v$,we differentiate $x$ with respect to $t$: $v = \frac{dx

Vedclass · Accepted Answer

The acceleration of the particle is $2(c + c_0)$.

Vedclass · Answer

(D) Given the displacement equation: $x = c_0(t^2 - 2) + c(t - 2)^2$. To find the velocity $v$,we differentiate $x$ with respect to $t$: $v = \frac{dx}{dt} = \frac{d}{dt}[c_0(t^2 - 2) + c(t - 2)^2] = 2c_0t + 2c(t - 2)$. To find the acceleration $a$,we differentiate $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{d}{dt}[2c_0t + 2c(t - 2)] = 2c_0 + 2c = 2(c_0 + c)$. Thus,the acceleration of the particle is $2(c + c_0)$.

$A$ particle moves along the $x$-axis and its displacement $x$ varies with time $t$ according to the equation $x = c_0(t^2 - 2) + c(t - 2)^2$,where $c_0$ and $c$ are constants of appropriate dimensions. Which of the following statements is correct?

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