Answer to Question #148135 in Calculus for Sean

Question

An object moves along a straight line so that after 𝑡 minutes, its distance from its starting point

is 𝑠(𝑡) = 10t + (5/t+1) - 5 meters.

(a) At what speed is the object moving at the end of 4 minutes?

(b) How far does the object actually travel during the 5th minute?

(c) How fast is its velocity changing at 𝑡 = 6 𝑚𝑖𝑛?

Answer 1

$S(t)=10t+\frac{5}{t+1}-5\\ V(t)=S'(t)=10-\frac{5}{(t+1)^2}\\ \text{(a)}\\ V(4)=10-\frac{5}{(4+1)^2}\\ V(4)=10-\frac{5}{(5)^2}\\ V(4)=10-0.2\\ V(4)=9.8m/min.\\ \text{(b)}\\ S(5)-S(4)\\ (10(5)+\frac{5}{(5)+1}-5)-(10(4)+\frac{5}{(4)+1}-5)\\ =45.83-36\\ =9.83m.\\ \text{(c)}\\ V(t)=10-\frac{5}{(t+1)^2}\\ V(0)=10-\frac{5}{(0+1)^2}\\ =10-5\\ V(0)=5m/min\\ V(6)=10-\frac{5}{(6+1)^2}\\ V(6)=10-\frac{5}{(7)^2}\\ V(6)=10-0.10\\ V(6)=9.9m/min.$