I was learning to make a car game and came across this situation where my car is on an inclined plane. It's initial velocity is 0. Now, the problem is that my memory serves me this formula to calculate the final velocity
$$v_f^2 – v_i^2 = 2as$$
Since, $v_i = 0$, the equation would be
$$v_f^2 = 2as$$
For an inclined plane $a = g \sin \theta$,
$$ \Rightarrow v_f^2 = 2 g s \sin \theta $$
However, my memory doesn't serve me what to do when you do not know how far the car would have to travel on this inclined plane.
My problem in short:
$\bullet$ I need to add Velocity to the car which is on the inclined plane
$\bullet$ What is the formula I should use? does the formula mentioned above of any good in this case?
$\bullet$ If yes, how would I come over this displacement drawback (drawback in the sense, I would not know how far the car would travel)
$\bullet$ If no, what would you use to add the velocity to the car?
Best Answer
You can cast your problem in terms of what is the velocity of the car at any instant of time after it started. The answer is $v = a_{\rm eff} t$ from the first kinematical equation. In the same time, the car would've traveled a distance $s = (1/2)a_{\rm eff} t^2$ from the second kinematical equation.
The other option is - if your inclined plane is of any fixed length $L$ and the time $t$ of concern to you is the time in which it has completely traversed the plane, you can cast everything in terms of this distance, by using the second of these equations to write $t_f = (2L/a_{\rm eff})^{1/2}$, and $v_f = a_{\rm eff} t_f$.
Thus, in order to model everything perfectly, you need to have one of these as inputs - either $L$, or the instant of time $t$.