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A particle moves along $x$-axis as
$x=4(t-2)+a(t-2)^2$
Which of the following is true?
Options:
$x=4(t-2)+a(t-2)^2$
Which of the following is true?
Solution:
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Verified Answer
The correct answer is:
The acceleration of particle is $2 a$
$\begin{aligned} & x=4(t-2)+a(t-2)^2 \\ & \text { At } t=0, x=-8+4 a=4 a-8\end{aligned}$
$\begin{aligned} & v=\frac{d x}{d t}=4+2 a(t-2) \\ & \text { At } t=0, v=4-4 a=4(1-a)\end{aligned}$
At $t=0, v=4-4 a=4(1-a)$
But acceleration, $a=\frac{d^2 x}{d t^2}=2 a$
$\begin{aligned} & v=\frac{d x}{d t}=4+2 a(t-2) \\ & \text { At } t=0, v=4-4 a=4(1-a)\end{aligned}$
At $t=0, v=4-4 a=4(1-a)$
But acceleration, $a=\frac{d^2 x}{d t^2}=2 a$
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