If a ball is thrown into the air with a velocity ofc 40ft/s,its height in feet after t seconds is given by y=40t-16t^2. Find the velocity when t =2. Could you please show the steps
To find the velocity when t = 2, we need to find the derivative of the height function y with respect to time (t) and then substitute t = 2 into the derivative.
Given: y = 40t - 16t^2.
Step 1: Take the derivative of y with respect to t using the power rule for derivatives.
- For the term 40t, its derivative is simply 40 since the derivative of a constant times t is the constant.
- For the term -16t^2, we apply the power rule: the derivative of t^n is n*t^(n-1). In this case, n = 2, so the derivative is -32t.
Therefore, the derivative of y with respect to t is dy/dt = 40 - 32t.
Step 2: Substitute t = 2 into the derivative to find the velocity.
- Plug in t = 2 into the derived expression: dy/dt = 40 - 32(2) = 40 - 64 = -24 ft/s.
So, the velocity when t = 2 is -24 ft/s.
Velocity is the derivative of y(t)
V = dy/dt = 40 - 32 t
Plug in t = 2 and calculate the velocity. It will be negative, meaning that the ball is on the way down at that tiome.