Calculating Time with Forces Acting on a Block

[tryingtolearn1]

Homework Statement

A block starts at rest and slides down a frictionless plane inclined at an angle ##\theta##. What should ##\theta## be so that the block travels a given horizontal distance in the minimum amount of time?

Relevant Equations

$$F=ma$$

When drawing a diagram of the forces acting on the block, I have the following forces: $$\sum F_x = a_x = (g \sin\theta) \cos \theta .$$

Now, I can use the following kinematic equation $$x=vt+\frac{a_xt^2}{2}$$, where $$v=0$$ and $$a_x = (g \sin\theta) \cos \theta$$ $$\therefore \frac{2x}{t^2} = (g \sin\theta) \cos \theta .$$

Now in order to obtain the minimum amount of time, I will need to take the derivative and set it equal to zero but this is where I get stuck and don't know how to proceed?

 

[PeroK]

What if you simply maximise the horizontal acceleration? Does that work?

 

[Steve4Physics]

Hi.

To travel a given horizontal distance (x) in the minimum amount of time (t) requires you to maximise horizontal acceleration. You have already found that [tex]a_x = (g \sin\theta) \cos \theta[/tex]Hint: [itex]sin(2A) = sinA cos(A) [/itex] and ask yourself for what value of A is sin(2A) maximum? No calculus needed!

Alternatively, if you are required to provide a calculus-based solution, express t as a function of [itex]\theta[/itex], treating x as a constant (since it is a 'fixed distance'). Then differentiate and solve [itex] \frac {dt}{d\theta} = 0[/itex] as usual.

 

[tryingtolearn1]

Ah, I see thank you @PeroK and @Steve4Physics. The derivative expressing ##t## as a function of ##\theta## turned out to be messy where I had to solve the following: $$\frac{dt}{d\theta}=\sqrt{\frac{4x}{g\sin2\theta}}=0$$ so instead I just took the hint you provided of maximizing the horizontal acceleration by focusing on values that maximizes the function ##\sin2\theta## which is ##\frac{\pi}{4}##.

 

[PeroK]

Here's another tip: maximising or minimising ##t## is the same as maximising or minimising ##t^2##. That can be very useful. Same for ##v## and ##v^2##.

The same if you have to maximise ##\theta## and you have a formula for ##\tan \theta##. Just let ##u = \tan \theta## and maximise ##u##.

 

[haruspex]

Ah, I see thank you @PeroK and @Steve4Physics. The derivative expressing ##t## as a function of ##\theta## turned out to be messy where I had to solve the following: $$\frac{dt}{d\theta}=\sqrt{\frac{4x}{g\sin2\theta}}=0$$ so instead I just took the hint you provided of maximizing the horizontal acceleration by focusing on values that maximizes the function ##\sin2\theta## which is ##\frac{\pi}{4}##.

I assume you mean $$\frac{dt}{d\theta}=\frac{d}{d\theta}\sqrt{\frac{4x}{g\sin2\theta}}=0$$