If a ball is thrown into the air with a velocity of 12 m/s, its height (in meters) after t seconds is given by y = 12t − 4.9t2. Find the velocity when t = 1.
y = 12*1-4.9*1^2 = 12-4.9 = 7.1 m.
dy/dt = 12 - 9.8 t = 2.2 m/s
To find the velocity when t = 1, we need to calculate the derivative of the height equation with respect to time. The derivative of y = 12t − 4.9t^2 represents the rate at which the height is changing with respect to time, which is the velocity.
Step 1: Differentiate the height equation with respect to time.
The derivative of y = 12t − 4.9t^2 can be found by applying the power rule of differentiation.
dy/dt = d/dt(12t − 4.9t^2)
dy/dt = 12 - 9.8t
Step 2: Substitute t = 1 into the derivative equation.
To find the velocity at t = 1, substitute t = 1 into the derivative equation we obtained in the previous step.
dy/dt = 12 - 9.8(1)
dy/dt = 12 - 9.8
dy/dt = 2.2 m/s
Therefore, the velocity when t = 1 is 2.2 m/s.