If a ball is thrown into the air with a velocity of 10 m/s, its height (in meters) after t seconds is given by y = 10t − 4.9t2. Find the VELOCITY when t = 1.
To find the velocity of the ball at a specific time, you need to find the derivative of the function representing the height with respect to time.
In this case, the function representing the height of the ball is given by y = 10t − 4.9t². To find the velocity at t = 1, we need to find the derivative of this function with respect to t.
The derivative of y with respect to t is denoted as dy/dt or y'.
Using the power rule of differentiation, the derivative of 10t is simply 10, and the derivative of -4.9t² is -9.8t.
So, the derivative of y = 10t − 4.9t² is dy/dt = 10 - 9.8t.
To find the velocity at t = 1, substitute t = 1 into the derivative equation:
dy/dt = 10 - 9.8(1)
Calculating this, we get:
dy/dt = 10 - 9.8 = 0.2 m/s
Therefore, the velocity when t = 1 second is 0.2 m/s.
To find the velocity when t = 1, we need to calculate the derivative of the height equation y = 10t - 4.9t^2 with respect to time (t).
Taking the derivative of the equation y = 10t - 4.9t^2:
dy/dt = d/dt(10t - 4.9t^2)
To differentiate the equation, we use the power rule for differentiation:
dy/dt = 10 - 2 * 4.9t^(2-1)
dy/dt = 10 - 9.8t
Now we can substitute t = 1 into this equation to find the velocity when t = 1:
v = dy/dt, t = 1
v = 10 - 9.8 * 1
v = 10 - 9.8
v ≈ 0.2 m/s
Therefore, the velocity of the ball when t = 1 second is approximately 0.2 m/s.