Question

The time variation of the position of a particle in rectilinear motion is given by $x=2t^3+t^2+2t.$ If v is the velocity and a the acceleration of the particle in consistent units, the motion started with

• A. $v=0,a=0$
• B. $v=0,a=2$
• C. $v=2,a=0$
• D. $v=2,a=2$

Answer 1

The correct option is D

$v=2,a=2$

Given:
$x=2t^3+t^2+2t$

Velocity: $V=\frac{dx}{dt}=6t^2+2t+2$

$Att=0$ (start of motion)
$V=0+0+2$
$V=2$
Acceleration: $a=\frac{dv}{dt}=12t+2$

$Att=0$
$a=0+2$
$\therefore a=2$