Question

If a ball is thrown into the air with a velocity of 38 ft/s, its height (in feet) after t seconds is given by y = 38t − 16t². Find the velocity when t = 1.

Answer 1

To find the velocity when t = 1, we need to find the derivative of the height function y = 38t - 16t² with respect to time t. The velocity when t = 1 is 6 ft/s.

Step-by-step explanation:

To find the velocity when t = 1, we need to find the derivative of the height function y = 38t - 16t² with respect to time t. The derivative represents the velocity at any given time.

y' =
`(dy)/(dt)`

= 38 - 32t

Now, substitute t = 1 into the derivative to find the velocity at t = 1 :

`\( y'_(t=1)` = 38 - 32(1)

= 38 - 32

= 6

Therefore, the velocity when t = 1 is 6 ft/s.