A particle in rectilinear motion follows the path s(t) = t^3 − 6t^2 + 9t, where s is in meters, and t is in seconds.

Question

a. What is the average velocity over [0, 2]?

b. What is the instantaneous velocity at t = 2?

Answer 1

The position of the particle at time t = 0 is s(0):

s(t) = t³ - 6t² + 9t

s(0) = (0)³ - 6(0)² + 9(0) = 0

And the position of the particle at time t = 2 is s(2):

s(2) = (2)³ - 6(2)² + 9(2) = 8 - 24 + 18 = 2

So the average velocity is (s₂ - s₁)/(t₂ - t₁) = [s(2) - s(0)]/(2 - 0) = 2/2 = 1 m/s

The instantaneous velocity is the derivative of the position function. Not knowing where you are in your class (whether you've been taught to use the limit definition to find the instantaneous velocity or if you've been taught the power rule) it's hard to know what method I should address this in.

Instantaneous velocity is defined as the limit of [s(t + Δt) - s(t)]/Δt as Δt approaches zero. You can find this by trying smaller and smaller values of Δt near t = 2 like this:

if Δt = 0.1 then s(2.1) = 1.701 so v = (1.701 - 2)/0.1 = -2.99 m/s

if Δt = 0.01 then s(2.01) = 1.970001 so v = (1.970001 - 2)/0.01 = -2.9999 m/s

if Δt = 0.001 then s(2.001) = 1.997000001 so v = (1.997000001 - 2)/0.1 = -2.999999 m/s

So obviously the limit as Δt approaches zero is -3 m/s

Using the power rule, you just take the derivative of s(t) to get s'(t) = 3t² - 12t + 9 and plug in t = 2 and you get the same answer.