Confusion about when condition of constant acceleration applies

[deleted647690]

There was a question in TBR that asked, "What force is required to give the cannonballs an exit speed of 25 m/s?"

They gave the distance of the barrel of the cannon, so I solved for the acceleration by
using v = delta x / delta T, and plugging in 25 m/s for the velocity and 2.5 meters for the displacement. This gave me the time

I then just used a=delta v / delta t to get the acceleration, and used F=ma.

I was supposed to use one of the kinematics equations...why do the basic relationships between velocity, acceleration, and displacement not apply in this case?

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[aldol16]

Because using v = dx/dt will not give you the correct time. dt = v*dx only works when v is constant and isn't changing. Here, v is only 25 m/s when it exits. It starts a 0 m/s and goes through all the intermediate speeds before it exist. If you use v = 25 here, what you're saying is that it was 25 m/s at all points in the barrel which is not true.

 

[deleted647690]

Because using v = dx/dt will not give you the correct time. dt = v*dx only works when v is constant and isn't changing. Here, v is only 25 m/s when it exits. It starts a 0 m/s and goes through all the intermediate speeds before it exist. If you use v = 25 here, what you're saying is that it was 25 m/s at all points in the barrel which is not true.

Thanks!

 

[aldol16]

Thanks!

And something else I thought of, in case you are mathematically inclined, this doesn't mean that v = dx/dt isn't true. In fact, your confusion may stem from the fact that v = change in x/change in t is what is taught in algebra-based physics when the more useful form is v = dx/dt. Basically, this is a really simple differential equation in which separate and solve:

v = dx/dt
dt = 1/v*dx

Now, integrate both sides. The integral of dt is easy - it's just t - t0. On the right side, if v is constant, you can just move it out of the integral and integrate dx to get v*(x - x0). But here, v is not constant due to the acceleration. In other words, v can be expressed as a function of x, or its position within the cannon barrel. If you had a function for v in terms of x, you would integrate that and derive your expression for time.

 

[deleted647690]

And something else I thought of, in case you are mathematically inclined, this doesn't mean that v = dx/dt isn't true. In fact, your confusion may stem from the fact that v = change in x/change in t is what is taught in algebra-based physics when the more useful form is v = dx/dt. Basically, this is a really simple differential equation in which separate and solve:

v = dx/dt
dt = v*dx

Now, integrate both sides. The integral of dt is easy - it's just t - t0. On the right side, if v is constant, you can just move it out of the integral and integrate dx to get v*(x - x0). But here, v is not constant due to the acceleration. In other words, v can be expressed as a function of x, or its position within the cannon barrel. If you had a function for v in terms of x, you would integrate that and derive your expression for time.

Shouldn't it be dt = dx/v?

 

[aldol16]

Shouldn't it be dt = dx/v?

Yes, my mistake. Corrected above. It can also be v*dt = dx, if you can express v in terms of t instead of x.

 

[deleted647690]

Yes, my mistake. Corrected above. It can also be v*dt = dx, if you can express v in terms of t instead of x.

And yeah, I took algebra based physics, so calculus was never really emphasized. I did take calc a looong time ago though.

Also, I've always wondered, when you answer these questions, do you reference stuff alot? Your responses are amazing and filled with so much helpful information. I imagine you as a walking textbook haha

 

[aldol16]

Also, I've always wondered, when you answer these questions, do you reference stuff alot? Your responses are amazing and filled with so much helpful information. I imagine you as a walking textbook haha

Thanks for the compliment! Sometimes I return to the fundamentals and I look up derivations to save time. But for this particular question, I didn't look anything up - I just explained it how I think about it and hoped that would make sense!