suppose a ball is thrown straight up into the air, and the height of the ball above the ground is given by the function h(t) = 6 + 37t - 16t^2, where h is in feet and t is in seconds. What is the velocity of the ball at time t = 3.2
-65.4 ft/sec
nDeriv(6 + 37x - 16x^2, x, 3.2)
this gives you the derivative of the height function with respect to time (the velocity) at the time t = 3.2
To find the velocity of the ball at time t = 3.2 seconds, we need to find the derivative of the function h(t) with respect to time.
The derivative of h(t) with respect to t gives us the rate of change of the height function, which represents the velocity of the ball at any given time.
First, let's find the derivative of h(t):
h'(t) = d/dt (6 + 37t - 16t^2)
To find the derivative, you can apply the power rule and the constant rule:
h'(t) = 0 + 37 - 32t
Now, let's substitute t = 3.2 into the velocity function:
h'(3.2) = 37 - 32(3.2)
= 37 - 102.4
= -65.4
Therefore, the velocity of the ball at t = 3.2 seconds is -65.4 ft/s (negative because the ball is moving downward in a straight-up motion).
To find the velocity of the ball at time t = 3.2 seconds, we need to find the derivative of the height function h(t) with respect to time t, and then substitute t = 3.2 into the derivative.
The derivative of h(t) can be found by differentiating each term of the function separately.
The derivative of 6 is 0, as the constant term does not change with time.
To differentiate the term 37t, we treat t as a variable and differentiate it with respect to t. The derivative of 37t with respect to t is simply 37.
To differentiate the term -16t^2, we use the power rule for differentiation. The power rule states that if we have a function of the form f(x) = a*x^n, then the derivative of f(x) is f'(x) = a*n*x^(n-1). In this case, a = -16 and n = 2. Therefore, the derivative of -16t^2 with respect to t is -16*2*t^(2-1) = -32t.
Now, we can assemble the derivative of h(t):
h'(t) = 0 + 37 - 32t
To find the velocity at t = 3.2 seconds, we substitute t = 3.2 into h'(t):
h'(3.2) = 0 + 37 - 32(3.2)
Now we can calculate the value of h'(3.2):
h'(3.2) = 37 - 32(3.2)
= 37 - 102.4
= -65.4
Therefore, the velocity of the ball at t = 3.2 seconds is -65.4 feet per second.
t = 3.2 seconds I presume
h(t) = 6 + 37t - 16t^2
velocity = dh / dt
v = 0 + 37 - 32 t which is initial velocity - g t where g = 32 ft/s^2 (old text:)
at t = 3.2
v = 37 - 32 (3.2) = 37 - 102.4 = - 70.4 ft/second
by the way at t = 3.2
h = 6 + 37(3.2) - 16(3.2)^2 = 6 + 118.4 - 163.8 = -39.4 feet
below ground :)