Question

Velocity of a particle moving along x-axis is given as $v=t^2-4t$

• A. Particle accelerates for $0\le t<2$
• B. Particle accelerates for $2\le t\le 4$
• C. Maximum speed attained by the body between 0 to 4 sec is 4 m/s
• D. Particle retards for $2<t\le 4$

Answer 1

Answer: (C), (D)

Particle retards for $2<t\le 4$

$a=\frac{dv}{dt}=2t-4$
$a<0from0sto2s,$ therefore it is retarding
$a>0from2sto4s$ therefore it is accelerating
As it is retarding for the first 2 seconds, the minimum velocity occurs at t = 2 sec and it is equal to $v\left(2\right)=2^2-4\times 2=-4m/s$. Miniumum negative velocity means maximum speed.
$\therefore$ maximum speed is 4 $m/s$