Question

How did I get b wrong?

Answer 1

47

Explanation:

Distance is the integral of the absolute value of velocity.

d = ∫ |v(t)| dt

To integrate this, we find where v(t) is positive and where it's negative.

0 = t² − 2t − 15

0 = (t + 3) (t − 5)

t = -3 or 5

1 ≤ t ≤ 5, v(t) is negative.

5 ≤ t ≤ 6, v(t) is positive.

Therefore:

d = -∫₁⁵ v(t) dt + ∫₅⁶ v(t) dt

d = ∫₅¹ v(t) dt + ∫₅⁶ v(t) dt

d = (⅓ t³ − t² − 15t)₅¹ + (⅓ t³ − t² − 15t)₅⁶

d = (⅓ (1)³ − (1)² − 15(1)) − (⅓ (5)³ − (5)² − 15(5)) + (⅓ (6)³ − (6)² − 15(6)) − (⅓ (5)³ − (5)² − 15(5))

d = (⅓ − 1 − 15) + (72 − 36 − 90) − 2 (⅓ (125) − 25 − 75)

d = ⅓ − 70 − ⅓ (250) + 200

d = 130 − ⅓ (249)

d = 130 − 83

d = 47

Answer 2

131/3

Explanation:

Using definite integral

from 1-5 = - 128/3 (make it + because distance isn't vector quantity)

from 5-6 = 13/3

so distance = 128/3 + 13/3 = 131/3 m