The velocity function (in meters per second) for a certain particle, moving in a straight line, is v(t)=t^2-2t-8

Question

The velocity function (in meters per second) for a certain particle, moving in a straight line, is v(t)=t^2-2t-8 for 1≤t≤6

 

A) Find the displacement of the particle over this period

 

B) Find the total distance by the particle over the time period

 

This question is from chapter 5, integrals of calculus 1

Answer 1

v(t) = t²-2t-8 = (t-4)(t+2) = 0 when t = 4 or -2

 

s(t) = position of particle at time t.   s'(t) = v(t)

 

When 0 ≤ t <4, v(t) < 0 (so, the particle is moving to the left

When 4 < t ≤ 6, v(t) > 0 (so, the particle is moving to the right

 

A. Displacement = change in position = s(6) - s(0) 

 

                                                       = ∫_{(from 0 to 6)} s'(t)dt

 

                                                       = ∫_{(from 0 to 6)} (t²-2t-8)dt

 

                                                       = [(1/3)t³ - t² - 8t]_{(from 0 to 6)}

 

                                                       = (1/3)(6)³ - 36 - 48 = -12 meters

 

            So, from t = 0 to t = 6, the particle moved 12 m to the left of its initial position.

 

B.  Total distance = ∫_{(from 0 to 6)} l t²-2t-8 l dt

 

                          = -∫_{(from 0 to 4)}(t² - 2t - 8)dt + ∫_{(from 4 to 6)}(t²-2t-8)dt

 

     I'll let you finish up.