A particle moves along X-axis as x = 4(t - 2) | vedclass

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

Vedclass · Accepted Answer

The acceleration of the particle is $2a$.

Vedclass · Answer

(B) Given the position equation: $x = 4(t - 2) + a(t - 2)^2$.To find the velocity,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,we differentiate the velocity 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 $v = 4 + 2a(0 - 2) = 4 - 4a$,which is not $4$ unless $a = 0$.$(b)$ The acceleration is $2a$,which is constant and matches the derived result.$(c)$ At $t = 0$,$x = 4(0 - 2) + a(0 - 2)^2 = -8 + 4a$,which is not $0$ unless $a = 2$.Therefore,option $(b)$ is correct.

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

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