Find the time during the first 8 seconds of fall at which the instantaneous velocity equals the average velocity.

Question

The height of an object t seconds after it is dropped from a height of 250 meters is s(t)=-4.9t^2+250. Find the time during the first 8 seconds of fall at which the instantaneous velocity equals the average velocity.

Answer 1

The height of an object t seconds after it is dropped from a height of 250 meters is s(t)=-4.9t^2+250. Find the time during the first 8 seconds of fall at which the instantaneous velocity equals the average velocity.

s = 250 - 4.9t^2

velocity = ds/dt = 9.8t

After 8 secs, velocoty = ds/dt(8) = -9.8*8 = -78.4 m/s

average velocity = Vav = -39.2 m/s

Using ds/dt = -9.8t      (instantaneous velocity)

Vav = -39.2 = -9.8t

t = 4 s       (time at which instanteous velocity reaches average velocity)