Question

A particle moves along $x-$ axis as $x=4\left(t - 2\right)+a\left(t - 2\right)^{2}$ , here $a$ is a constant. Which of the following is true?

• A. The initial velocity of the particle is $4ms^{- 1}$
• B. The acceleration of particle is $2a$
• C. The particle is at origin at $t=0$
• D. None of the above

Answer 1

Answer: (B)

Given, $ \, x=4\left(t - 2\right)+a\left(t - 2\right)^{2}$
$ \, \, v=\frac{d x}{d t}=4+2a\left(t - 2\right)$
At $t=0, \, \, \, v=4\left(1 - a\right)\text{and }x=\left(4 a - 8\right)$
Acceleration, $a=\frac{d^{2} x}{d t^{2}}=2a$