A particle moves along the x-axis as x = 4(t- | vedclass

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(B) Given the position equation: $x = 4(t-2) + a(t-2)^2$.To find the velocity $v$,we differentiate $x$ with respect to time $t$:$v = \frac{dx}{dt} = \

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The acceleration of the particle is $2a$.

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(B) Given the position equation: $x = 4(t-2) + a(t-2)^2$.To find the velocity $v$,we differentiate $x$ with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt}[4(t-2) + a(t-2)^2] = 4 + 2a(t-2)$.To find the acceleration $a_{acc}$,we differentiate the velocity $v$ with respect to time $t$:$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}[4 + 2a(t-2)] = 2a$.Checking the options:$(a)$ Initial velocity at $t=0$ is $v = 4 + 2a(0-2) = 4 - 4a$. This is not $4$ unless $a=0$.$(b)$ The acceleration is $2a$,which matches our derivation.$(c)$ At $t=0$,$x = 4(0-2) + a(0-2)^2 = -8 + 4a$. This is not $0$ unless $a=2$.Therefore,the correct statement is that the acceleration of the particle is $2a$.

$A$ particle moves along the $x$-axis as $x = 4(t-2) + a(t-2)^2$. Which of the following statements is true?

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