Question

Velocity of a particle is given by v = 2t^2 + 5m/s (A) find the change in velocity between the time interval T1=2 s and t2 = 4 s (B)find the average acceleration in same time interval (c )find the instantaneous acceleration at t =4s PLZZ ANSWER

Answer 1

• since v is 2t²+5m/s

velocity at T1➡

2(2)²+5

2.4+5

13m/s

velocity at T2

2(4)²+5

2.16+5

37m/s

change in velocity ➡37-13=24m/s

• v=2t²+5m/s

differentiation aboub equation

dv/dt=a=d/dt(2t²+5)

a=4t...............( 5 become zero, it is constant)

acceleration at T1

a=4(2)

a=8m/s²

acceleration at T2

a=4(4)

a=16m/s²

avarage acceleration=total accelration/total time

a(avarage)=16+8/2+4

a(avarage)=24/6

a(avarage)=4m/s²

• instantaneous acceleration

instantaneous velocity is given by

v=2t²+5m/s

then instantaneous acceleration is

a=dv/dt=4t (as given aboub)