Question

# If velocity of a particle is v(t)=8−2t m/s, the average speed and magnitude of average velocity of the particle in 5 sec respectively are

A
3 m/s,175 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3 m/s,315 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
175 m/s,3 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
175 m/s,175 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 175 m/s,3 m/sGiven, v(t)=8−2t, we will check if the particle has changed its direction or not within the asked time interval, i.e. the time when velocity of the particle becomes zero. ⇒ 8−2t=0⇒ t=4 sec Also, we know that xf−xi=∫t0v(t)dt=∫t0(8−2t)dt ⇒ xf−xi=8t−t2 Let particle starts from origin i.e xi=0 So, at t=0,xf=0, at t=4,xf=8×4−42=16 m at t=5,xf=8×5−52=15 m The motion of the particle can be represented as shown So, the average speed of the particle is 16+15=175 m/s and, average velocity is 155=3 m/s

Suggest Corrections
0
Related Videos
Speed and Velocity
PHYSICS
Watch in App
Explore more