If a ball is thrown into the air with a velocity of 41 ft/s, its height (in feet) after t seconds is given by y = 41t − 16t2. Find the velocity when t = 1.
y = 41t - 16 t^2
differentiating with respect to t,
dy/dt = 41 - 32 t
dy/dt is the velocity of the ball
at t = 1,
dy/dt = 41 - 32 = 9
so the velocity is 9 ft/s (upwards)
To find the velocity when t = 1, we need to differentiate the equation for height y(t) with respect to time t.
Given that y = 41t - 16t^2, we can differentiate both sides of the equation with respect to t:
d/dt (y) = d/dt (41t - 16t^2)
The first term, 41t, has a derivative of 41.
The second term, -16t^2, has a derivative of (-16) * (2t), which simplifies to -32t.
Hence, the derivative of y(t) with respect to t is:
dy/dt = 41 - 32t
Now, we can substitute t = 1 into the derivative equation to find the velocity at t = 1:
dy/dt = 41 - 32(1)
= 41 - 32
= 9
Therefore, the velocity when t = 1 is 9 ft/s.
To find the velocity when t = 1, we need to differentiate the height function y = 41t - 16t^2 with respect to time (t).
Differentiating y = 41t - 16t^2:
dy/dt = 41 - 32t
Now we can substitute t = 1 into the derivative to find the velocity when t = 1:
dy/dt = 41 - 32(1)
dy/dt = 41 - 32
dy/dt = 9 ft/s
Therefore, the velocity when t = 1 is 9 ft/s.