How do I calculate the time an object accelerating down a slope of changing gradient will needs to reach from point A to point B.

Question

I was wondering how I would go about calculating the time a object needs to travel from point $A(0,0)$ to point $B(1,-1)$ while traveling along the line $-x^{1/2}$.

Graph for $-x^{1/2}$

The object is accelerating in a uniform gravitational field of strength $9.81$m/s and experiences no friction or air resistance.

Thanks for any help.

Answer 1

The vertical speed is

$$\frac{dy}{dt}=v \sin\theta$$

where $v = \sqrt{-2gy}$ (derived from $\frac12mv^2=-mgy $) and $\tan\theta = \frac{dy}{dx} = \frac1{2y}$. Plug them into the above expression to get,

$$\frac{dy}{dt}= \sqrt{\frac{-2gy}{4y^2+1}}$$

Thus, the time is obtained from,

$$T = \int_0^{-1} \sqrt{\frac{4y^2+1}{-2gy}}dy=0.584$$

Answer 2

As the graph is $y = -\sqrt x$, the slope is $\frac{dy}{dx} = \frac{-1}{2\sqrt x} = \frac{1}{2y}$.

Now, the slope at a point is $\tan \theta = \frac{dy}{dx}$

Also, due to the normal force by the surface $y = -\sqrt x$, the net acceleration the particle experiences is in the tangential direction. That is, $\vec a = g\sin\theta (\cos\theta\,\hat x-\sin\theta\,\hat y)$.

If we just consider the $y$ direction, then we have \begin{align} \frac{d^2y}{dt^2} &= -g \sin^2\theta \\ \implies v_y\frac{dv_y}{dy}&= -g\frac{\tan^2\theta}{1+\tan^2\theta} \\ &= \frac{-g}{1+4y^2} \\ \end{align}

Integrating, as $v(0,0)) = 0$ and $y=0$ at $t=0$, we get - \begin{align} v_y^2 &= -g\tan^{-1}(2y) \\ \implies \frac{dy}{dt}&= \sqrt{-g\tan^{-1}(2y)} \\ \implies t &= \int_0^{-1} \frac{dy}{\sqrt{-g\tan^{-1}(2y)}} \\ \end{align}

Unfortunately, the last integral does not have a closed form solution, so must be done numerically.