A block is moving along the x-axis and its position varies in time according to the expression x = (3.00 m/s^2)t^2, where x is in meters and t is in seconds.

Question

a) Determine its position (in m) at t = 2.60 s

b) Determine its position (in m) at t = (2.60 s) + Δt.

 (Use the following as necessary: Δt. Do not include units in your answer.)

c) Evaluate the limit of Δx divided by Δt  as Δt approaches zero to find the velocity (in m/s) at t = 2.60 s.

Answer 1

x(t) = 3t²

a) Position at t = 2.60, i.e. x(2.60) = 3(2.60)² = 20.28

b) Position at t = 2.60 + Δt, i.e. x(2.60 + Δt) = 3(2.60 + Δt)² = 3[(2.60)² + 2(2.60)Δt + (Δt)²]

= 3(2.60)² + 6(2.60)Δt + 3(Δt)² = 20.28 + 15.6Δt + 3(Δt)²

c) Δx = x(2.60 + Δt) - x(2.60) = 15.6Δt + 3(Δt)²

Δx/Δt = [15.6Δt + 3(Δt)²]/Δt = 15.6 + 3Δt

Limit of Δx/Δt  as Δt approaches zero = 15.6 + 3(0) = 15.6, i.e. velocity at t = 2.60 s is 15.6 m/s