Solving problems in Kinematics with simple calculus

[snath_98]

Hi,

I have a generic question about solving kinematic problems using simple calculus. For eg.
A variable force F = bv (v is instantaneous speed at some time t) acts on a vehicle moving with speed v. Suppose the initial speed at t = 0 of the vehicle is v0. Calculate the distance through which it moves before stopping.
By work energy theorem, (1/2) m (V₀)² = ∫ F. ds
Now how do we integrate the right side and move forward? That's my main question. Any help is appreciated

Thanks
Sanjay

 

[snath_98]

b above is a constant.

 

[Philip Wood]

I wouldn't start from here. I think (though I may be wrong) that the theorem in this form is not much use when the force is velocity-dependent.

Instead, assuming that the vehicle is confined to move in a straight line, you can use Newton's second law in the form F= ma. Substitute bv for F (remembering that b is negative - I'd write it as -β] and substitute dv/dt for a. This gives you a differential equation, which is easily solved to give v as a function of t. Integrate this wrt time, between zero and infinity to give you the distance gone.

This is quite interesting, as it shows that the vehicle goes a finite distance, but takes an infinite time to do it! (Well, to do every last nanometre). This is not what a real vehicle would do, because F = -βv does not model the usual rolling resistance, bearing friction etc. very well. These will not tend to zero as the vehicle's speed goes to zero.

 

[voko]

A note on terminology. Any problem involving forces is dynamical, not kinematic.

 

[PJC01]

Hello Sanjay,

Thank you for your question about solving kinematic problems using simple calculus. Calculus is a powerful tool for solving problems in physics, especially in kinematics. In order to solve this particular problem, we need to use the fundamental theorem of calculus, which states that the integral of a function can be calculated by finding the antiderivative of that function and evaluating it at the upper and lower limits of integration. In this case, the function we are integrating is F, the variable force acting on the vehicle.

To find the antiderivative of F, we need to first determine the function that gives us the force at any given time t. Since F = bv, we can rewrite this as F = b(v(t)), where v(t) is the function that gives us the instantaneous speed of the vehicle at time t. Then, we can use the power rule for integration to find the antiderivative of F: ∫ F dt = ∫ b(v(t)) dt = b ∫ v(t) dt.

Now, we need to determine the limits of integration for this problem. The initial speed of the vehicle is given as v0 at t = 0, and we are trying to find the distance it moves before stopping, so we can set our upper limit of integration as the time when the vehicle comes to a stop, which we can denote as tf. Our lower limit of integration will be t = 0.

Putting this all together, we have ∫ F dt = b ∫ v(t) dt = b ∫ v0 dt = b(v0)(tf - 0) = b(v0)(tf). This gives us the final equation for calculating the distance the vehicle moves before stopping: (1/2) m (v0)^2 = b(v0)(tf). Solving for tf, we get tf = (1/2) m (v0)/b.

This means that the vehicle will stop after a time of (1/2) m (v0)/b. To find the distance it moves before stopping, we can use the equation d = v0(tf) = (1/2) m (v0)^2 / b.

I hope this helps answer your question and gives you a better understanding of how to use simple calculus to solve kinematic problems. If you have any further questions, please don't hesitate to ask. Good luck with your studies!

Best