Word problem involving instantanous velocity and acceleration

Question

I'll try to explain this clear enough:

Let's say a baseball is "popped up" into the air, and it's height (meters) after t minutes is represented by the function $f(t) = -4.9t^2+25t+3$.

I have to find the instantaneous velocity and the acceleration of $t$ when $t = 2$.

Would I be able to use the power rule to get the derivative of this function, and then plug in 2 to find the instantaneous velocity? Then would I be able to find the second derivative to find the acceleration?

Answer 1

Yes, $v(t) = -9.8t + 25$ and $a(t) = -9.8$.

Then $v(2) = 5.4$ and $a(2) = -9.8$.

To solve for when the ball hits the ground we simply set $f(t) = 0$ and solve for $t$.

Answer 2

When you find first derivative you can instantly find second. So you will have 2 formulas and by plugging in $t$ in two of them you can get separately instantaneous velocity and acceleration. $\frac{d}{dt}f=-9.8t + 25$ is first derivative, $\frac{d^2}{dt^2}f=-9.8$ is second one. So instantaneous acceleration is seen without plugging any $t$ and equals $-9.8\frac{m}{sec^2}$