Suppose a coin is dropped from the top of the Empire State building in New York, which is 1,454 feet tall. The position function for free-falling objects is:
s(t) = −16t^2 + v_0t + s_0,
where v_0 is the initial velocity and s_0 is the initial position.
Find the instantaneous velocities when t =1 and t = 3.
Since it is "dropped" , the initial velocity is zero
and the initial height is 1454 ft
s(t) = -16t^2 + 0t + 1454
v(t) = -32t
v(1) = -32 ft/sec
v(3) = -96 ft/sec
Suppose a coin is dropped from the top of the Empire State building in New York, which is 1,454 feet tall. The position function for free-falling objects is:
s(t) = −16t^2 + v_0t + s_0, where v_0 is the initial velocity and s_0 is the initial position.
1. Find the instantaneous velocities when t =1 and t = 3 (I've already solved)
2. What is the name of the theorem that says there must be at least one solution to Part 1?
3. Find the velocity of the coin just before it hits the ground.
To find the instantaneous velocities when t = 1 and t = 3, we need to differentiate the position function, s(t), with respect to time, t, to get the velocity function, v(t).
Given:
s(t) = -16t^2 + v₀t + s₀
To differentiate this function, we apply the power rule and the constant rule of differentiation.
Power Rule: If f(x) = x^n, then f'(x) = nx^(n-1)
Constant Rule: If f(x) = c, then f'(x) = 0
Differentiating s(t):
s'(t) = d/dt (-16t^2) + d/dt (v₀t) + d/dt (s₀)
Applying the power rule and the constant rule to each term:
s'(t) = -32t + v₀ + 0
Therefore, the velocity function is:
v(t) = -32t + v₀
Now, we can find the instantaneous velocities when t = 1 and t = 3 by substituting these values into the velocity function, v(t):
When t = 1:
v(1) = -32(1) + v₀ = -32 + v₀
When t = 3:
v(3) = -32(3) + v₀ = -96 + v₀
So, the instantaneous velocities when t = 1 and t = 3 are -32 + v₀ and -96 + v₀, respectively.