Q. A particle is moving with constant initial velocity 4 $ms^{- 1}$ till $t = 1.5$ s. Then it accelerates at 10 $ms^{- 2}$ till $t = 3.$ The distance covered is (Take $g = 10ms^{- 2}$)
A) 17.25 m
B) 36.25 m
C) 40 m
D) 23.25 m
Explanation
(d) $S = \left( 4 \times 1.5 + 4 \times 1.5 + \frac{1}{2} \times 10 \times 1.5 \times 1.5 \right)$m
$= (6 + 6 + 5 \times 2.25)$m = 23.25m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.