Answer to Question #248408 in Calculus for Bman

Question

The position of a particle moving in a straight line during a 5-second trip is

s(t) = t2 − t + 10 cm. Find a time t at which the instantaneous velocity is

equal to the average velocity for the entire trip.

Answer 1

$s = t² - t + 10$

Duration of trip is 5 second

Distance traveled in 5 seconds

= s(5) - s(0) cm

= (5² - 5 + 10) - (0² - 0 + 10) cm

= 25 - 5 +10 - 10 cm

= 20 cm

So average velocity = 20/5 = 4 cm/s

Given that

instantaneous velocity = average velocity

=> $\frac{ds}{dt} = 4$

=> 2t - 1 = 4

=> 2t = 4+1

=> 2t = 5

=> t = 5/2 = 2.5

So after 2.5 second of start instantaneous velocity is equal to the average velocity.