Question

What would you conjecture about Raoul’s instantaneous velocity at t = 3s?

Raoul accelerates up the hill. After moving 10 meters, the engine expires. Raoul coasts up the hill, comes to a stop, and then begins to roll back down the hill. Joe calculates that Raoul’s distance from the bottom of the hill is given by the equation D=-t^2+10t+10 where t is the time in seconds since the engine expired and D is the distance in meters.

1) What would you conjecture about Raoul’s instantaneous velocity at t = 3s?
2) Why can we not find the instantaneous velocity at t = 3s by letting the time interval shrink to zero?

Answer 1

`4 m//s`

Explanation:

velocity = distance/time

velocity = `D/t = (Deltay)/(Deltax)`

power rule: (Deltay)/(Deltax) (x^n) = nx^(n-1)#

`x^0 = 1`

`(Deltay)/(Deltax) (-t^2+10t+10) = -2t^1 + 10t^0 + 0`

`= -2t + 10`

at `t = 3s, (Deltay)/(Deltax) = -6 + 10`

`= 4`

`(Deltay)/(Deltax) = v`

therefore, you can conjecture that the instantaneous velocity at `t = 3s` is `4 m//s.`

while 'average' velocity require a time interval, instantaneous velocity must be defined at a specific value of time.

average velocity is found by dividing total displacement by total time.

changes in instantaneous velocity at certain points would not directly affect the average velocity.

instantaneous velocity is velocity at a single instant in time. over a longer time interval, this can only be the same as the average velocity if a constant velocity is kept throughout.
otherwise, average and instantaneous velocity will be different.