Question

5. Brandon and Chloe ride their bikes for 4 hours along a flat, straight road. Brandon’s velocity, in miles per

hour, at time t hours is given by a differentiable function B for 0 st 54. Values of B(t) for selected times t
are given in the table above. Chloe's velocity, in miles per hour, at time t hours is given by the piecewise
function C defined by
C(t)
tep
for 0 12 – 3t – 12 for 2 <154.
(a) How many miles did Chloe travel from time t = 0 to time t = 2 ?
Do not write beyond this border.
Do not write beyond this border.

Answer 1

Explanation:

a) ∫from 0 to 2 of te^(4-t^2) dt. Let u = 4-t^2. du= -2dt. And, tdt= -dy/2.

If x=0, u= 4 and if x=2, u=0.

Therefore our new integral will be, -(1/2) ∫ from 0 to 4 of e^u du.

= -(1/2)(e^u) | 0 to 4

= -(1/2)(e^0 - e^4)

= -(1/2) + (1/2)e^4

= (1/2)(e^4 -1) miles

Therefore, from t=0 to t=2, Chloe has travelled a distance of (1/2)(e^4-1) miles.

b) At t=3, C(t)=12-3t-t^2

C(t) is already a function of velocity so simply plug in C(3), which is -6 miles/hour.

We also need to know acceleration.

C'(t)= -3 - 2t

C(3)= - 9 miles/hour squared.

Therefore, since velocity and acceleration are both negative, Chloes speed is increasing at t =3 hours.