A test tube is knocked off a tower at the top of a building that is 400 feet above the ground. (For our purposes, we will assume that air resistance is negligible.) The test tube drops 16𝑡^2 feet in 𝑡 seconds.
a) Calculate the average velocity in the last two seconds of the fall.
b) Calculate the instantaneous velocity when the test tube lands.
so first is to find time of fall:
s=400-16t^2
at final time, s is zero
or Tf=sqrt(400/16)=5 seconds.
velocity= ds/dt=-32*t
veloicty(5)=-160 m/s
velocity(3)=-96 m/s
so what is the average velocity?
Thank you, for some reason I thought that 16t^2 was already acceleration (meaning f''(t) ) so I was like "how do I back-track to get velocity" ( f'(t) ).
Anyway, I worked out -128ft/s for V_avg (since the equation is asking for feet/second).
Thus the time hitting the ground was f'(5).
Thanks for the help Bob!
I haven't looked through the calculations, but f'(5) is not time -- it is velocity at t=5.
a) Well, if the test tube drops 16𝑡^2 feet in 𝑡 seconds, we can calculate the distance it drops in the last two seconds by substituting 𝑡 with 2:
16(2^2) = 16(4) = 64 feet
Now, to calculate the average velocity, we divide the distance by the time:
Average velocity = 64 feet / 2 seconds = 32 feet/second
So, the average velocity in the last two seconds of the fall is 32 feet/second.
b) Ahh, the instant when the test tube lands. Let's think about this. When the test tube lands, it's safe to say that its velocity is zero because, well, it's not moving anymore.
So the instantaneous velocity when the test tube lands is zero feet/second. Gravity finally has its way and brings the test tube to a halt.
To calculate the average velocity in the last two seconds of the fall (a), we need to find the change in position (or height) of the test tube during those two seconds and divide it by the time taken.
Given that the test tube drops 16𝑡^2 feet in 𝑡 seconds, we can calculate the change in height during the last two seconds by substituting 𝑡 = 2 into the equation:
Height during the last two seconds = 16(2^2) = 16(4) = 64 feet.
Therefore, the change in position is 64 feet.
Now, let's calculate the average velocity by dividing the change in position by the time taken:
Average Velocity = Change in position / Time taken = 64 feet / 2 seconds = 32 feet/second.
Answer (a): The average velocity in the last two seconds of the fall is 32 feet/second.
To calculate the instantaneous velocity when the test tube lands (b), we need to find the velocity of the test tube just before it hits the ground.
Since air resistance is negligible, we can assume that the acceleration due to gravity is constant at approximately 32 feet/second^2. This means that the velocity of the test tube will increase by 32 feet/second every second.
To find the time it takes for the test tube to hit the ground, we can set the height to zero and solve for 𝑡 in the equation 16𝑡^2 = 400. Simplifying this equation gives us 𝑡^2 = 25, which results in 𝑡 = 5 seconds.
Now, let's find the velocity of the test tube just before it hits the ground by multiplying the acceleration due to gravity by the time it takes to hit the ground:
Instantaneous Velocity = Acceleration * Time taken = 32 feet/second^2 * 5 seconds = 160 feet/second.
Answer (b): The instantaneous velocity when the test tube lands is 160 feet/second.