Calculus and Physics with Velocity

[pippintook]

A coin is dropped from a height of 750 feet. The height, s, (measured in feet), at time, t (measured in seconds), is given by s= -16t² + 750.

a) Find the average velocity on the interval [1, 3].
b) Find the instantaneous velocity when t = 3.
c) How long does it take for the coin to hit the ground?
d) Find the velocity of the coin when it hits the ground.


For (a), I graphed the equation on my calculator and found y = 686 when x = 2 (the average of 1 and 3). I don't know if that's the right thing to do though.

For (b), I simply plugged t = 3 into the equation given and got 606. Again, I don't know if that's the right thing to do.

For (c), I found when y = 0, x was 6.84. That doesn't seem like a logical time though.

I wasn't quite sure how to do (d), and I have the sneaking suspicion I'm not doing any of these right. Any help with any of the sections is appreciated!

 

[VeeEight]

I have not done these things in a while so excuse me if I make a mistake..
To find the velocity as it hits the ground, you want to measure the velocity at position 0. Since you have a position function, differentiating it with respect to time will give you distance over time which is velocity. So find the derivative function and input the position as 0.

 

[HallsofIvy]

A coin is dropped from a height of 750 feet. The height, s, (measured in feet), at time, t (measured in seconds), is given by s= -16t² + 750.

a) Find the average velocity on the interval [1, 3].
b) Find the instantaneous velocity when t = 3.
c) How long does it take for the coin to hit the ground?
d) Find the velocity of the coin when it hits the ground.


For (a), I graphed the equation on my calculator and found y = 686 when x = 2 (the average of 1 and 3). I don't know if that's the right thing to do though.

No, that is not the right thing to do. For one thing, I don't know what you mean by x and y! There are no "x" and "y" in the problem. Are you saying you graphed s= -16t²+ 750 as "y= -16x²+ 750"? For another, that is height, not velocity! "Average velocity" is "distance traveled divided by time". Surely you knew that! When t= 1, s= -16(1)²+ 750= 734 and when t= 3, s= -16(3)²+ 750= -16(9)+ 750= -144+ 750= 606. The coin has moved downward 606 feet in 2 seconds. What was its aveage velocity?

For (b), I simply plugged t = 3 into the equation given and got 606. Again, I don't know if that's the right thing to do.

And again, it's not! Don't just plug numbers into whatever formula you have without thinking. You were told, because you say it here, "the height is given by ...". That formula is height, not velocity. Velocity is the rate of change of distance or height: the derivative.

For (c), I found when y = 0, x was 6.84. That doesn't seem like a logical time though.

Why not? This is the one you got right!

I wasn't quite sure how to do (d), and I have the sneaking suspicion I'm not doing any of these right. Any help with any of the sections is appreciated!

You titled this "Calculus and Physics with Velocity". Didn't you expect it to involve Calculus? The velocity when the coin hits the ground is the derivative of the height function at the time the coin hits the ground- which you just calculated.