Question

The velocity of particle is given by the equation v=2t²+5. Find the change in the velocity of the particle during the time interval t1=2s and t2=4s...2nd accelerate at same time

Answer 1

Answer:

• velocity =2t^2+5
• velocity at t =2 ,=13m/s
• velocity at t=4,=37m/s
• so change in velocity =37-13=24m/s

average acceleration = change in

velocity/time duration

=24÷2=12m/s^2

but absolute acceleration =4t(by differentiation)

so acceleration at t=2,=8m/s^2

and at t=4,= 16m/s^2