Question

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

• A. The particle is at origin at $t=0$
• B. The initial velocity of particle is$4$
• C. The acceleration of particle is$2a.$
• D. None of these

Answer 1

The correct option is C The acceleration of particle is$2a.$

Given,

$x=4(t-2)+a(t-2{)}^2$

The particle at $t=0s$

$x=4(0-2)+a(0-2{)}^2$

$x=-8+4a$

$x=(4a-8)m$

The initial velocity at $t=0s$ is,

$v=\frac{dx}{dt}$

$v=4(1-0)+2a(t-2)\times (1-0)$

$v=4+2a(t-2)$

$v{|}_{t=0s}=4+2a(0-2)$

$v{|}_{t=0s}=(4-4a)m/s$

Acceleration $=\frac{dv}{dt}$

Acceleration $=2am/s^2$

The correct option is C.