For an object whose velocity is given by v(t)=6−2t, what is its distance on the interval t = 0 to t = 4?

Question

Do I distribute and then solve ?

Answer 1

Total distance traveled is given by the integral of speed, ie the absolute value of the velocity function, over the given time interval. This is relevant because the velocity is negative between t = 3 and t = 4. Thus displacement ≠ total distance. We could use a calculator and use the absolute value function to calculate ∫₀⁴ |6 - 2t| dt, or we can do it by hand by splitting up the integral into 2 separate ones as I will do here:

total dist = ∫₀³ (6-2t)dt + ∫₃⁴(2t-6)dt = (6t - t²) ]₀³ + (t² - 6t)]₃⁴ = 9 - 8 - (-9) = 10

Btw we can calculate this distance easily using geometry, given that the linear velocity function forms two triangles with the x-axis, the first with height = 6 and base = 3 so A = 9, and the second with height = -2 and base = 1 so A = 1.