A particle moves along a line so that its velocity at time t is ( | |. Find the displacement of the particle during the time period .

Complete question is;

A particle moves along a line so that its velocity at time t is

v(t) = t² − t − 6

(measured in meters per second).

(a) Find the displacement of the particle during the time period 1 ⩽ t ⩽ 4.

(b) Find the distance traveled during this time period.

A) -4.5 m

B) 10.17 m

Explanation:

We are given;

v(t) = t² − t − 6

We know that v = ds/dt

Thus,

S = v dt

s(4) - s(1) = (4,1)∫t² − t − 6 dt

= (4,1)[(t³/3) - (t²/2) - 6t]

= ((4³/3) - (4²/2) - 6(4)) - ((1³/3) - (1²/2) - 6(1))

= 64/3 - 8 - 24 - 1/3 + 1/2 + 6

= -4.5 m

B) Since v(t) = t² − t − 6, then factorizing we can write it as;

v(t) = (t + 2)(t - 3)

Thus, v(t) ⩽ 0 at the interval (1, 3) and v(t) ≥ 0 at the interval (3, 4)

Thus;

-x1 = (1, 3)∫t² − t − 6 dt

-x1 = (1,3)(t³/3 - t²/2 - 6t)

-x1 = (3³/3 - 3²/2 - 6(3)) - (1³/3 - 1²/2 - 6(1))

-x1 = 9 - 9/2 - 18 - 1/3 + 1/2 + 6

-x1 = -3 - 4 - 1/3

-x1 = -22/3 m

x1 = 22/3 m

x2 = (3, 4)∫t² − t − 6 dt

x2 = (3, 4)(t³/3 - t²/2 - 6t)

x2 = ((4³/3) - (4²/2) - 6(4)) - (3³/3 - 3²/2 - 6(3))

x2 = 64/3 - 8 - 24 - 9 + 9/2 + 18

x2 = -23 + 64/3 + 9/2

x2 = (-138 + 128 + 27)/6

x2 = 17/6 m

Thus, total distance = x1 + x2 = 22/3 + 17/6 = 10.17m