1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A particle moves in a straight line with an acceleration 'a' m/s2 given as function of time, a=−1t2. At time t = 1 s the particle has velocity of 3 m/s, then find the velocity at t = 4 s.

A
2.25 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
5 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4.5 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 2.25 m/sa=dVdt=−1t2 ⇒dV=−1t2dt or integrating both sides w.r.t time with limits V = 3 m/s at t = 1 sec and V at t = 4 sec V∫3dV=4∫1−1t2dt ⇒[V]V3=−[t−2+1−2+1]41 orV−3=[1t]41 orV−3=14−1=−34 ∴V=3−34=2.25m/s Why this question? Tip: Acceleration is varrying with time here, hence apply a=dVdt and use method of integration to obtain velocity at time 't'.

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program