Question

Given below are the equations of motion of four particles $A,B,C$ and $D$.
$x_A=6t-3$;
$x_B=4t^2-2t+3$;
$x_C=3t^3-2t^2+t-7$;
$x_D=7\cos 60°-3\sin 30°$
Which of these four particles move with uniform non-zero acceleration?

• A. $A$
• B. $B$
• C. $C$
• D. $D$

Answer 1

The correct option is B $B$

$x_A=6t-3$
$v_A=\frac{d}{dt}(6t-3)=6$
Particle $A$ moves with constant velocity
$v_B=\frac{d}{dt}(4t^2-2t+3)=8t-2,a_B=\frac{d}{dt}(8t-2)=8$
Particle $B$ moves with constant acceleration
$v_C=\frac{d}{dt}(3t^3-2t^2+t-7)=9t^2-4t+1,a_C=\frac{d}{dt}(9t^2-4t+1)=18t-4$
So, the particle moves with variable acceleration.
$x_D=7\cos 60°-3\sin 30°$
Since $x_D$ does not depend upon time therefore particle is at rest.
Thus, only particle $B$ moves with uniform acceleration.