If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t^2. Find the velocity when t = 1.
v = dy/dt = 34 - 32 t
if t = 1
v = 2
To find the velocity when t = 1, we need to find the derivative of the given position function with respect to time. The derivative of y = 34t − 16t^2 will give us the velocity function.
Differentiating the position function y = 34t − 16t^2 with respect to time t:
dy/dt = d/dt (34t − 16t^2)
= 34 - 32t
This gives us the velocity function: v(t) = 34 - 32t.
To find the velocity when t = 1, substitute t = 1 into the velocity function:
v(1) = 34 - 32(1)
= 34 - 32
= 2 ft/s
Therefore, the velocity when t = 1 is 2 ft/s.
To find the velocity when t = 1, we need to calculate the derivative of the function y = 34t - 16t² with respect to time (t).
The derivative of a function gives us the rate of change of the function with respect to the independent variable (in this case, time).
The derivative of y with respect to t can be found using differentiation rules. In this case, we can differentiate each term of the function separately. The derivative of 34t with respect to t is 34, and the derivative of -16t² with respect to t is -32t.
So, the derivative of y = 34t - 16t² with respect to t is dy/dt = 34 - 32t.
Now, we can substitute t = 1 into the derivative to find the velocity at t = 1:
dy/dt = 34 - 32(1)
= 34 - 32
= 2
Therefore, the velocity when t = 1 is 2 ft/s.