Q. A particle starts from rest, acceleration at $2\ m/s^{2}$ for $10\ s$ and then goes with constant speed for $30\ s$ and then decelerates at $4\ m/s^{2}$ till it stops. What is the distance travelled by it
A) $750\ m$
B) $800\ m$
C) $700\ m$
D) $850\ m$
Explanation
(a) Velocity required by body in $10\ sec$
$v = 0 + 2 \times 10 = 20\ m/s$
And distance travelled by it in $10\ sec$
$S_{1} = \frac{1}{2} \times 2 \times (10)^{2} = 100\ m$
Then it moves with constant velocity $(20\ m/s)$ for $30\ sec$
$S_{2} = 20 \times 30 = 600\ m$
After that due to retardation $(4\ m/s^{2})$ it stops
$S_{3} = \frac{v^{2}}{2a} = \frac{(20)^{2}}{2 \times 4} = 50m$
Total distance travelled $S_{1} + S_{2} + S_{3} = 750m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.