Question

The velocity of a ball thrown upward into the air is given by v(t)= -32t+96 where t is measured in seconds and velocity v in feet per second calculate the displacement of the ball between t=1 and t=4

Answer 1

The displac ment between t = 1 and t = 4 is 272 feet

Explanation:

To get the displacement of the ball, we first need to get the equation for the position of the ball with respect to time.

We can do this by differentiating the velocity equation given:

v(t) = -32t +96

integrating between the limits of t = 1 and t = 4, we have

`s = \int\limits^4_1 {-32t +96} \, dt`

`s(t) = -32t^2/2 +96t`

`s =[ -16t^2 +96t] t = 4, t =1`

where t = 4.

s = 256 + (96) =352 feet

Where t = 1

s = 80

Displacement of the ball = 352 -80 = 272ft

The displac ment between t = 1 and t = 4 is 272 feet