Question

At time​ t, the position of a body moving along the​ s-axis is sequalsnegative t cubed plus 15 t squared minus 72 t m. a. Find the​ body's acceleration each time the velocity is zero. b. Find the​ body's speed each time the acceleration is zero. c. Find the total distance traveled by the body from tequals0 to tequals5.

Answer 1

Answer:

Step-by-step explanation:

Given that at time t, the position of a body moving along the​ s-axis is sequalsnegative t cubed plus 15 t squared minus 72 t m

i.e. $s(t) = -t^3+15t^2-72 t$

Velocity is nothing but s'(t) = derivative of s

and acceleration is s"(t) = derivative of v(t)

$v(t) = -3t^2+30t-72\\=-3(t^2-10t+24)\\= -3(t-6)(t-4)$

a) v(t) =0 when t = 4 or 6

b) $a(t) = -6t+30$

a(t) =0 when t =5

c) Distance travelled by the body from 0 to 5 would be

$s(5)-s(0)\\= -5^3+15(5^2)-72(5)\\= -125+375-360\\=-110$

i.e. 110 miles (distance cannot be negative)