SOLUTION: Thanks for the previous questions answered. Pls, I need your help here, and it is urgent An object is thrown vertically upward where its height, in meters, is given by s(t) =

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Thanks for the previous questions answered. Pls, I need your help here, and it is urgent An object is thrown vertically upward where its height, in meters, is given by s(t) =       Log On


   

Question 136230: Thanks for the previous questions answered. Pls, I need your help here, and it is urgent
An object is thrown vertically upward where its height, in meters, is given by
s(t) = -5t2 + 25t and t is the time, in seconds. Answer the question below.

What is the velocity of the object at 1 second after it has been released?

Thanks
Found 2 solutions by solver91311, Earlsdon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Just substitute 1 for t in the formula and do the arithmetic.

s%28t%29=-5t%5E2%2B25t, so s%281%29=-5%281%29%5E2%2B25%281%29

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
s%28t%29+=+-5t%5E2%2B25t as the function of the height of an object propelled vertically upwards, find the velocity of the object after 1 second.
Velocity is rate of change of distance (height in this case) with respect to time, t.
To find the rate of change of the function, you would find the first derivative of s(t).
d%28s%28t%29%29%2Fdt+=+-10t%2B25 Now substitute t = 1 second.
d%28s%281%29%29%2Fdt+=+-10%281%29%2B25 = 15
The velocity at 1 second is 15 m/s.