Prove magnitude of acceleration.

In summary: Integrating the acceleration given and showing that if the acceleration is less then that value then there is not enough acceleration to cover the distance in time.
  • #1
guipenguin
7
0
Going a distance S in time T, prove at some instant the magnitude of the acceleration of the car is (4S)/(T^2)

I have thought a lot about the problem and how I would go about proving it. Would using the Mean Value Theorem be the best way?

I was thinking that I could take two times A and B during the trip, and according to the MVT there exists another time c such that v'(c) <= (v(b) - v(a) ) / (b - a )

I also thought about doing an epsilon-delta proof to state that as (b - a) approaches T, (v(b) - v(a) ) / (b - a ) would approach a value that was equal to or greater than (4S)/(T^2).

I am not sure what to do to prove this. Any suggestions would be great!
 
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  • #2
There is something missing in the statement of the problem. For example if the car is going at a constant speed S/T, then the acceleration will always be 0.
 
  • #3
mathman said:
There is something missing in the statement of the problem. For example if the car is going at a constant speed S/T, then the acceleration will always be 0.

The problem is given in its entirety. My thought is that a car can't go from point A to point B without ever accelerating, so initial velocity = 0. I believe I would do it using the Mean Value Theorem, to prove that there exists a mag of acceleration equal to 4S/T^2 somewhere between point A and B
 
Last edited:
  • #5
John Creighto said:

I know, I just can't seem to apply it in the context of this problem...

I know that a(t)=v'(t) and there is some point is [0,S] where (S - 0 ) / (T - 0 ) = (4S) / (T^2 )

I know the magnitude of acceleration is the slope of the tangent at a point on a speed-time graph.

Any other hints for setting up a proof using the MV Theorm? Or maybe a better way? I can't have someone work it all out for me, I just ask for any hints or suggestions.

Thanks!
 
  • #6
Are we assuming it starts from rest? How about just integrate the acceleration given and show that if the acceleration is less then that value then there is not enough acceleration to cover the distance in time.
 

Related to Prove magnitude of acceleration.

1. What is the formula for calculating magnitude of acceleration?

The formula for calculating magnitude of acceleration is acceleration (a) = change in velocity (Δv) / time interval (Δt), or a = Δv / Δt.

2. How is magnitude of acceleration different from average acceleration?

Magnitude of acceleration is the absolute value of acceleration, while average acceleration takes into account the direction of the acceleration. Average acceleration is calculated as the change in velocity divided by the change in time, taking into account the direction of the velocity change.

3. How do you measure the magnitude of acceleration in an experiment?

To measure the magnitude of acceleration in an experiment, you will need to measure the change in velocity and the time interval. This can be done using tools such as a stopwatch or a velocity sensor.

4. What are the units for magnitude of acceleration?

The units for magnitude of acceleration are distance per time squared, such as meters per second squared (m/s^2) or feet per second squared (ft/s^2).

5. Can magnitude of acceleration be negative?

Yes, magnitude of acceleration can be negative. This indicates that the object is slowing down in the direction of motion, or accelerating in the opposite direction.

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