A ball is thrown into the air with a velocity of 34 ft/s. Its height, in feet, after t seconds is given by s(t) = 34t − 16t2. Find the velocity (in ft/s) of the ball at time t = 1 second.

Question

Can you give a detailed explanation in how to do it? I've been doing it repeated times now, and I still do not know what I did wrong.

Answer 1

velocity is the derivative of the distance, instantaneous change in distance over change in time

s(t) = distance as a function of time = 34t-16t^2

take the 1st derivative, multiply the exponent by the coefficient to get the new coefficient

and reduce the exponent by 1

s'(t) = velocity = 34t^0-32t^1

= 34-32t

s'(1) = 34-32(1) = 2 feet per second