If velocity of a particle is v(t)=2t−4m/s, the average speed and magnitude of average velocity of the particle in 5sec respectively are
A
145m/s,95m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
135m/s,95m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
135m/s,1m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
145m/s,1m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C135m/s,1m/s Given, v(t)=2t−4, 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. ⇒2t−4=0⇒t=2sec Also, we know that xf−xi=∫t0v(t)dt=∫t0(2t−4)dt ⇒xf−xi=t2−4t Let particle starts from origin i.e xi=0 So, at t=0,xf=0, at t=2,xf=22−4×2=−4m at t=5,xf=52−4×5=5m The motion of the particle can be represented as shown
So, the average speed of the particle is |−4|+|−4|+55=135m/s and, average velocity is 55=1m/s