Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
Step-by-step 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.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.