Prove magnitude of acceleration.

AI Thread Summary
To prove the magnitude of acceleration as (4S)/(T^2) for a car traveling distance S in time T, the Mean Value Theorem (MVT) is suggested as a potential method. The discussion highlights that if the car maintains a constant speed of S/T, the acceleration would be zero, indicating a need for clarification in the problem statement. It is proposed that the car must accelerate from rest to cover the distance, implying an initial velocity of zero. Suggestions include using MVT to establish that there exists a point where the acceleration meets the required magnitude. Overall, the conversation revolves around finding an appropriate mathematical approach to demonstrate the acceleration condition.
guipenguin
Messages
7
Reaction score
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!
 
Physics news on Phys.org
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.
 
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:
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!
 
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top