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Average and instantaneous velocity(Orginal right?)

1. Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t) = bt^2 - ct^3, where b = 2.40 m/s^2 and c = 0.100 m/s^3.

1. Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s.

2. Calculate the instantaneous velocity of the car at t=0s.

3. Calculate the instantaneous velocity of the car at t=5.00 s.

4. Calculate the instantaneous velocity of the car at t=10.0 s.

5. How long after starting from rest is the car again at rest?

2. Homework Equations

x2-1/t2-t1



3. The Attempt at a Solution

1. I was wondering what the x-values were since I think change in time is just x2-x1/10.0 seconds.

2,3,and 4, do I just plug t into the equation bt^2 - ct^3?

5. No idea.
 
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Avg. Velocity is just the slope of the line between those two points when t=1 and t=10.

For 2,3, and 4, you want the instantaneous velocity. You need to take the derivative of the function that is giving you distance, and plug the time values into that to get the instantaneous velocity at those times.

For number 5 you need to think what "at rest" is. When something is "at rest" is has no velocity right? If the original function is giving you distance, the max/min points (and any constant line) is where the car or whatever it is, is not moving. Doesn't the first derivative test in calculus handle this?
 
Avg. Velocity is just the slope of the line between those two points when t=1 and t=10.

For 2,3, and 4, you want the instantaneous velocity. You need to take the derivative of the function that is giving you distance, and plug the time values into that to get the instantaneous velocity at those times.

For number 5 you need to think what "at rest" is. When something is "at rest" is has no velocity right? If the original function is giving you distance, the max/min points (and any constant line) is where the car or whatever it is, is not moving. Doesn't the first derivative test in calculus handle this?
Ok thanks, I will try to work it out.

I take calculus simultaneously with physics, as advised by my adviser. Needless to say, physics moves much faster. ¯\_(ツ)_/¯
 
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Oh I see.

You might want to just watch the videos on finding the derivative through use of the "power rule" on khan academy or something real fast. That will allow you to solve this problem. The derivative as a function will give you the instantaneous change in y over the change in x. So really, after differentiating the function you are left with another function that will tell you the slope of the original function at time t. In the first derivative test, you simply set the first derivative to 0 and solve for t. This will tell you the time when velocity is 0, (when the thing is at rest).
 
Oh I see.

You might want to just watch the videos on finding the derivative through use of the "power rule" on khan academy or something real fast. That will allow you to solve this problem. The derivative as a function will give you the instantaneous change in y over the change in x. So really, after differentiating the function you are left with another function that will tell you the slope of the original function at time t. In the first derivative test, you simply set the first derivative to 0 and solve for t. This will tell you the time when velocity is 0, (when the thing is at rest).
Yes, I was successful at finding all of the instantaneous velocities, so thanks for that!

So I have to use the derivative 2bt-3ct2 and differentiate it for the slope(average velocity)?

(t=5, x=16.5) and (t=10, x= 18), so would a point-intercept slope equation not find me the slope(cant because its a curve right)?
 
Last edited:
I received 0.3 for the slope(average velocity). Is that right?
 

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