Ryan Y. answered • 07/04/24
Univ. level Calculus lecturer. Experienced former AP Calc/Phys inst.
Greetings,
I think the key is to use the fact that the velocity in the x-direction is constant.
Thus, x(t) = (v_x)t .
Draw a triangle to find v_x as a function of theta (which is a constant).
Once you have done that you can plug x(t) into the given equation for v(t).
I'm intentionally not writting the answer out step by step as that would defeat the purpose of the problem and would not help you to learn. I hope you understand that.
Have a good day,
Ryan
Ryan Y.
Hi Andrew, My assumption that the velocity in the x is based on the level of the topic (ie no air resistance). The magnitude of the velocity vector can still be written as a function of x (as presented). After all x is a function of t (even if we neglect air resistance) and we would have no problem with v(t).07/04/24
Ryan Y.
By the level of the topic I mean the title of Simple 2D motion. I doubt very much that air resistance is involved. If that were the case then you would get a diabolical answer, as you found out.07/04/24
Divyanshu K.
Hi Ryan, It is said in question that velocity is not constant as it's depends on displacement x. We just need to convert the velocity function of x to velocity function of t . In simple terms it would give how velocity vary with time.07/05/24
Ryan Y.
Yes the total velocity is not constant. The component of the velocity in the y-direction changes. However, (unless explicitly stated) the velocity in the x - direction does not change. Draw the velocity vector at an angle with the horizontal of theta. Use trigonometry to find the horizontal component (which is constant). Use the usual formula x = (v_xo)t . Replace the x in the given formula with (v_xo)t. That's it :)07/05/24
Andrew L.
Hi Ryan, I wanted to ask you about your initial assumption of the velocity in the x-direction being constant- I interpreted the question as the velocity being dynamic, as it relies on it's position on the x-axis; it increases exponentially as x increases. My initial idea was to set up velocity as dx/dt dx/dt = x^2 +10x +3 then separate the variables and integrate to solve for t in terms of x. Then I could go backwards and solve for x in terms of t. But going down that route, I got a really messy calculation involving a lot of square roots and logs. It was doable but I didn't think it was typical of a 2D kinematics problem. I saw your solution and was immediately intrigued by the new method, but I wasn't so sure about the initial assumption. Could you please elaborate more on it? Thanks, Andrew07/04/24