# How to solve a related rates problem?

An object is thrown from the ground with initial velocity 15 meters per second. It moves vertically according to $s = 15t – 4.9t_2$ where $s$ is in meters and $t$ in seconds, and $v$ is velocity and $a$ is acceleration, find:

a) $v$ and $a$ when $t = 4$ seconds. b) The maximum height. c) When its height will be 25 meters. d) When it hits the ground.

What would be the steps for solving that? I think these problems involve differentiating and substituting, but I'm not sure how to go about doing that.

The general equation for the motion is $s(t)=s_0+v_0t+\frac{1}{2}at^2$.

So in your case $s_0=0$, $v_0=15\,\mathrm{ms^{-1}}$ and $a=9.8\,\mathrm{ms^{-2}}$ (with the negative sign because falling). Thus $v(t)=v_0=15\,\mathrm{ms^{-1}}$ and $a(t)=a=9.8\,\mathrm{ms^{-2}}$. The trajectory $s(t)$ is a parabola and recalling that the vertex of a parabola $$f(x)=ax^2+bx+c=a\left(x+\frac{b}{2a}\right)+\frac{4ac-b^2}{4a}$$ is at point $\left(-\frac{b}{2a},\,\frac{4ac-b^2}{4a}\right)$, we have that the maximum height is at $t_\max=\frac{15}{9.8}\approx 1.53\,\mathrm{s}$ and is $$s(t_\max)=s_\max=-\frac{15^2}{4(-4.9)}=\frac{225}{19.6}\approx 11.48\,\mathrm{m}$$

If $s_0=25\,\mathrm{m}$ the revised height is $$s(t)=s_0+v_0t+\frac{1}{2}at^2=25+15t-4.9t^2$$ which has zeros at $t\approx -1.20\mathrm s$ and $t\approx 4.26 \mathrm s$. We ignore the negative value and claim that the object will hit the ground after $t\approx 4.26 \mathrm s$.

I assume you mean $t^2$ instead of $t_2$.

You have given the position $s$ as a function $s(t)$. Remember that the velocity is the derivative of the position: $v(t) = s'(t)$ and that the acceleration is the derivative of the velocity: $a(t) = v'(t) = s''(t)$.

A (differentiable) function $s(t)$ can only have a maximum at $t=t^*$, where its first derivative vanishes ($s'(t^*)=0$). If the second derivative is negative at that position ($s''(t^*) < 0$) it is actually a maximum.

The object will be 25 meters high, when you have $s(t_{25}) = 25$. You will have to solve for $t_{25}$.

The object is at the ground, when its height is zero: $s(t_0)=0$. Again, solve for $t_0$ this time.