Find the distance traveled in 45 seconds by an object traveling at a velocity of v(t) = 20 + 9cos(t) feet per second.

Answer 1

Take a look at your text. There must be a problem similar to this.

Consider an object traveling at constant velocity v = k. Then, each second, it travels k feet, so after t seconds, the distance traveled is k*t.

Now, suppose its speed increases by 1 ft/sec each second. So, a ballpark estimate for distance traveled after t seconds is
1 + 2 + 3 + 4 + ... + t = t(t+1)/2, or approximately 1/2 t^2.

Do you see a pattern forming here? Given a velocity function v, the distance s is figured by taking the integral. In this case,

v = 20 + 9 cos(t)
s = 20t + 9 sin(t) from 0 to 45 = s(45)-s(0)

s(0) = 0 so, the distance is just s(45) = 900 + 9sin(45) = 900 + 9*0.85 = 907.65 feet.