Suppose a coin is dropped from the top of the Empire State Building in New York, which is 1454 feet tall. The position function for free-falling objects is s(t) =−16t^2+v0t + s0.
Question: At what time is the instantaneous velocity of the coin equal to the average velocity of the coin found in Part B?
Part B and my answer was:
Determine the average velocity of the coin on the interval [1, 3].
s(3) = 16(3)^2+1454
= 1598
s(1)=16(1)^2+1454
= 1470
(s(3)-s(1))/(3-1)
(1598-1470/(3-1)=128/2=64ft/s
In order to solve the question, do i plug in different time values to find which is the closest? Any help would be appreciated.
find the velocity function from the derivative of s
v=s'=-32t+vo
set that equal to 64, solve for time t.
In your average velocity, you should have had a negative distance, which would have made a negative velocity (meaning downward). see the original equation for the negative sign.
Well, calculating the instantaneous velocity is quite a tall order! But don't worry, I'm here to help... or at least try my best.
First, let's recall that the instantaneous velocity is given by the derivative of the position function. So, let's find the derivative of s(t):
s(t) = -16t^2 + v0t + s0
Taking the derivative of both sides with respect to t:
s'(t) = -32t + v0
Now, we want to find when the instantaneous velocity is equal to the average velocity of the coin on the interval [1, 3]. The average velocity is given by the change in position divided by the change in time:
average velocity = (s(3) - s(1))/(3 - 1)
average velocity = (1598 - 1470)/2
average velocity = 64 ft/s (as you calculated correctly)
So, we want to find the time when s'(t) = 64 ft/s. Substituting s'(t) = -32t + v0 and 64 ft/s, we get:
-32t + v0 = 64
Now, unfortunately, we don't have the value of v0 (initial velocity) given in the question. So, without that information, we can't determine the exact time when the instantaneous velocity is equal to the average velocity.
However, we can still give it a shot and try some clownish tricks! How about we assume a value for v0? Let's say v0 = 0, just for fun.
-32t + 0 = 64
Solving this equation, we get:
t = -64/-32
t = 2
So, for the assumption that v0 = 0, the time when the instantaneous velocity would be equal to the average velocity would be around 2 seconds.
But remember, this value heavily relies on the assumption we made about v0. Without a given value, we can't provide an accurate answer. Keep that in mind, and always double-check your calculations!
To find at what time the instantaneous velocity of the coin is equal to the average velocity of the coin, you need to calculate the instantaneous velocity equation and set it equal to the average velocity. The instantaneous velocity function, v(t), is the derivative of the position function, s(t).
Taking the derivative of the position function s(t) = -16t^2 + v0t + s0, we get:
v(t) = -32t + v0
Where v0 is the initial velocity and is equal to the coefficient of t in the position function, which is v0 = 0 in this case as the coin is dropped from rest.
Now set the instantaneous velocity, v(t), equal to the average velocity, which is 64 ft/s as calculated in Part B:
-32t = 64
Solving this equation for t:
t = 64 / -32
t = -2
Since time cannot be negative in this scenario, we discard the negative solution.
Therefore, the instantaneous velocity of the coin is equal to the average velocity at t = 2 seconds.
To find the time at which the instantaneous velocity of the coin is equal to the average velocity of the coin, you can set up an equation between the two velocities and solve for time.
The instantaneous velocity at any time t is given by the derivative of the position function, which in this case is s(t) = -16t^2 + v0t + s0. Taking the derivative gives us v(t) = -32t + v0, where v0 is the initial velocity of the coin.
The average velocity on the interval [1, 3] is given by the change in position divided by the change in time:
Average velocity = (s(3) - s(1))/(3 - 1) = (1598 - 1470)/2 = 64 ft/s.
To find the time at which the instantaneous velocity is equal to the average velocity, we can set up the equation:
-32t + v0 = 64.
To solve for t, you'll need to know the initial velocity of the coin (v0). Unfortunately, it is not provided in the question. If you have the initial velocity, you can solve the equation by substituting the value of v0 and solving for t.
If you're looking to find the value of t without knowing v0, you can use a trial-and-error method. Plug different time values into the equation -32t + v0 = 64 and check if the instantaneous velocity equals the average velocity. Start with a reasonable range of time values that make sense. For example, try values between 1 to 3 since that's the interval given in Part B.
Note that without knowing the initial velocity (v0), you won't be able to find a specific answer for the time at which the instantaneous velocity equals the average velocity. You can only approximate it by trying different values within a reasonable range.