Velocity question. How do I solve it

A ball is thrown vertically upward from the ground at a velocity of 98 feet per second. It's distance from the ground after T seconds is given by S(t)= -16t^2 + 98t. How fast is the ball moving 4 seconds after being thrown?
A. 136 ft per sec
B. -42 ft per sec
C. 34 ft per sec
D. -30 ft per sec
How do I solve it?

Take the derivative of the position function, which gives the velocity. s'(t) = v(t)

v(t) = -32t + 98
v(4) = -32(4) + 98
v(4) = -128 + 98
v(4) = -30

ANSWER: D. -30 feet per second

s(t) = - 16t^(2) + 98t ---> Position Function

Take the derivative with respect to t to the position function in order to get the velocity function.

d/dt [ s(t) ] = d/dt [ -16t^(2) ] + d/dt [ 98t ]

s ' (t) = v(t) = - 32t + 98

Plug in t = 4 seconds to determine how fast the ball is moving 4 seconds after being thrown.

s ' (4) = v(4) = - 32(4) + 98 = - 30 ft / sec -----velocity is decreasing at a rate of 30 ft/sec
------> ANSWER

Taking the derivate of a function of position yields a function of velocity.

We derive S(t) and get S'(t) = V(t) = -32t + 98

Simply insert t=4 and you get V(4) = -32 * 4 + 98 = -30 ft per sec

S(t)= -16t^2 + 98t

Differentiate with respect to time

Velocity= -32t+98
After 4 s
Vel = -32(4)+98
Vel = 98-128= -30ft/s
Ball is moving downwards @ 30 ft/s

Hope you understood...!