# Velocity word problem

I know how to find velocity but I just can't make sense of this problem. If a rock is thrown verically upward from the surface of mars with velocity 15m/s, its height after t seconds is $h=15t - 1.86t^2$

What is the velocity of the rock when its height is 25 m on its way up and its way down.

So when I think I need to do is set it equal to 25 so $25=15t-1.86t^2$ and then from that I should get a positive and a negative answer maybe? I am not sure what to do from here. I know I will need to find the derivative but not sure how.

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Oh, I was just working through my calculus homework and this came up. – user138246 Oct 2 '11 at 15:44
HINT:Instantaneous velocity is actually derivative of distance with respect to time. – VelvetThunder Oct 2 '11 at 15:47
I know that but if I take the derivative of $25=15t−1.86t2$ I just get 0=15-2(1.86t) which I know doesn't help because that is the same as the derivative when it is at 0. – user138246 Oct 2 '11 at 15:49
Sorry, I'd misread the question and thought it was just a matter of solving the given equation. I still think it would be better placed at physics.stackexchange.com, though. – joriki Oct 2 '11 at 15:51

Added: The rock leaves the surface of mars with maximum speed, moves up rectilinearly with decreasing speed until it reaches the top, where the speed is zero, and then moves down with increasing speed until it reaches the surface of mars at the point of departure with a speed that is equal (in absolute value) to the starting speed. This can be justified by finding the equation of the velocity of the rock (equation $(2)$ below).

I need to do is set it equal to 25 so $25=15t−1.86t^2$.

Yes. Note that this equation has two solutions I compute below.

I know I will need to find the derivative but not sure how.

$$$$25=15t-1.86t^{2},\tag{1}$$$$

which has two solutions, let's call them $t_{1}$ and $t_{2}$, the equation of the velocity of the rock as a function of $t$ and evaluate it at those two instants $t_{1}$ and $t_{2}$. Since the velocity is the derivative of the height $h(t)$ the equation is$^1$

$$$$v=h^{\prime }(t)=\frac{d}{dt}\left( 15t-1.86t^{2}\right) =15-2\times 1.86t.$$\tag{2}$$

I decided to insert here the following picture which represents the motion of the rock as a function of the time $t$, i.e. the graph of $h(t)=15t-1.86t^{2}$ (the distance to the surface of mars), to which I added the line $h=25$ m, though it is not necessary to find the analytical solution of the problem. This line intersects the graph at instants I compute below $t_1, t_2$, with $t_1<t_2$.

The solutions of $(1)$ are (the approximated values are not necessary to be computed, you just need to use the exact values):

$$\begin{eqnarray*}t_{1} &=&\frac{15-\sqrt{15^{2}-4\times 25\times 1.86}}{2\times 1.86}\approx 2.3535\text{ h,} \\t_{2} &=&\frac{15+\sqrt{15^{2}-4\times 25\times 1.86}}{2\times 1.86}\approx 5.711\text{ h.} \end{eqnarray*}\tag{1'}$$

Added: Here is a picture of the graph of $h'(t)$ (green) together with the graph of $h(t)$ (black)

By $(2)$ the velocity at the instant $t_{1}$ is $$\begin{eqnarray} v_{1}=v(t_1) &=&15-2\times 1.86\left( \frac{15-\sqrt{15^{2}-4\times 25\times 1.86}}{2\times 1.86}\right) \\ &=&15-\left( 15-\sqrt{39}\right) \\ &=&\sqrt{39}\approx 6.245\text{ m/s} \\ &&\text{(the rock is moving upward),}\tag{3} \end{eqnarray}$$

while the velocity at the instant $t_{2}$ is $$\begin{eqnarray} v_{2}=v(t_2)&=&15-2\times 1.86\left( \frac{15+\sqrt{15^{2}-4\times 25\times 1.86}}{2\times 1.86}\right) \\ &=&15-\left( 15+\sqrt{39}\right) \nonumber \\ &=&-\sqrt{39}\approx -6.245\text{ m/s} \\ &&\text{(the rock is moving downward).} \tag{4} \end{eqnarray}$$

Note that as one should have expected the velocities are symmetric.

Comment: The velocity is a vector pointing up ($v>0$) or down ($v<0$) in the present case, because the motion is rectilinear. Its absolute value is the speed. At $h=25$ we have $v(t_1)>0$, $v(t_2)<0$ and $|v(t_1)|=|v(t_2)|$.

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$^1$ The derivative $h^{\prime }(t)$ can be computed in detail as follows: $$\begin{eqnarray*} h^{\prime }(t) &=&\frac{d}{dt}\left( 15t-1.86t^{2}\right) \\ &=&\frac{d}{dt}\left( 15t\right) -\frac{d}{dt}\left( 1.86t^{2}\right) \qquad \text{sum rule} \\ &=&15\frac{dt}{dt}-1.86\frac{d}{dt}\left( t^{2}\right) \qquad \text{product rule} \\ &=&15\times 1-1.86\times 2t\qquad \text{power rule} \\ &=&15-1.86\times 2t\text{.} \end{eqnarray*}$$

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Ok...I haven't quite worked out everything yet but I am already confused by the answer, how can there be two velocities? Isn't it distance over time? I mean, I guess we are taking distance from a fixed point then? Is this always assumed? – user138246 Oct 2 '11 at 21:43
@Jordan Carlyon: There are two velocities because there are two directions, upwards and downwards. When the rock is going up the velocity is positive (the derivative $h'(x)>0$), when it is coming down the velocity is negative (the derivative $h'(x)<0$). When the rock gets the top the velocity is zero. – Américo Tavares Oct 2 '11 at 21:50
@Jordan Carlyon: the first velocity is attained at $t_1$, the second at $t_2$. – Américo Tavares Oct 2 '11 at 21:52
@Jordan Carlyon: Yes, it is the distance to surface of mars, i.e the hight. – Américo Tavares Oct 2 '11 at 21:53
@Jordan Carlyon: You see that because from the given formula for $h$, at $t=0$ $h=0$ (surface of mars). – Américo Tavares Oct 2 '11 at 21:57

The question asks us to find the velocity. The velocity at any time $t$ is equal to $\frac{dh}{dt}$. We have in general $$\frac{dh}{dt}=15-3.72t.\qquad\text{(Equation 1)}$$

If we had been asked for the velocity at time $t=3$, for example, life would be easy, we would just substitute $3$ for $t$ in the above equation.

Unfortunately, we have been asked something more complicated, namely the two possible velocities when the height of the rock is $25$.

The height $h$ of the rock is $25$ when $$25=15t-1.86t^2. \qquad\text{(Equation 2)}$$
From this equation, we should be able to find the two times $t$ when $h=25$. And once we know these two times $t$, simple substitution in Equation 1 will give us the velocities.

Equation 2 is equivalent to $1.86t^2-15t+25=0$. This is a standard quadratic equation, with slightly messy coefficients. To solve for $t$, we use the Quadratic Formula. We get $$t=\frac{15\pm\sqrt{15^2-(4)(1.86)(25)}}{3.72}.\qquad\text{(Equation 3)}$$

Finally, we use the calculator to find good approximations to the two times $t$ when the height of the rock is $25$, and substitute in Equation 1.

Comment: We suggested using the calculator to find the two values of $t$ at which the height is $25$. But there is a better way. The velocity at time $t$ is $15+3.72t$. Substitute the values of $t$ obtained in Equation 3. There is very nice simplification, and we get that the two velocities are $\pm\sqrt{15^2-(4)(1.86)(25)}=\pm\sqrt{39}$. Thus it turns out (by symmetry, this is no big surprise!) that the speed of the rock is the same when it reaches height $25$ on its way up as when it reaches height $25$ again on its way down. But the velocities (positive for up, negative for down) are different. The great simplification that we got by not using the calculator immediately hints that there may be a simpler approach to the problem. And indeed there is, but the approach that we described in the main part of the answer is the most straightforward one. However, it can be in general quite useful not to bring out the calculator too early.

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What is 3.72 and why is it involved in this problem? This is so frustrating for me, I have spent all morning on two problems and I still have over 14 hours of homework to do. – user138246 Oct 2 '11 at 18:30
I could have given more detail. You know the Quadratic Formula for the solutions of $ax^2+bx+c=0$. It is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. In our case, $a=1.86$, so $2a=3.72$. – André Nicolas Oct 2 '11 at 18:35

You presumably know some formulae which look something like $s = ut +\frac{1}{2}at^2$ and $v=u+at$, where $s$ is distance at time $t$, $u$ is the initial velocity, $v$ is the velocity at time $t$, and $a$ is the constant acceleration.

So if $s=h$, you can find what the initial velocity $u$ and acceleration $a$ are.

You could solve $25 = 15t - 1.86 t^2$ for $t$ to get two solutions, and then use them to find the two values of $v$.

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I have no idea what ut+1/2at^2 is. – user138246 Oct 2 '11 at 17:39
@Jordan: they are equations of motion under constant acceleration. Basic physics/mechanics. – Henry Oct 2 '11 at 20:47
I don't think I am suppose to know that for this problem. – user138246 Oct 2 '11 at 21:18

You have something of the right start with plugging in 25 for h, but you've taken the derivative too early. Remember that the position (height) equation is given in terms of t. Thus its derivative, whose output will give you velocity, is also in terms of t. You found the correct derivative $\frac{dh}{dt} = 15-2(1.86t)$. So, all you need now, is the values of t such that h is 25.

HINT: t will be the solutions of the equation you gave above: $25=15t-1.86t^2$

Can you go from there?

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I don't understand what you mean by I found the derivative, all I did is set the equation equal to 25 but I don't know how to find the derivative. – user138246 Oct 2 '11 at 17:31
I have no idea how to solve for t in that equation. – user138246 Oct 2 '11 at 17:40
@jordan "I know that but if I take the derivative of 25=15t−1.86t2 I just get 0=15-2(1.86t) which I know doesn't help because that is the same as the derivative when it is at 0." That was your comment two hours ago. What you did taking the equation from 25=15t+1.86t2 to 0=15+2(1.86t) was finding the derivative. The only modification I made in my answer was to replace h=25 with just h. Then, $\frac{d}{dt}(h) = \frac{dh}{dt} = velocity = v$. The derivative ($\frac{d}{dt}$) of the RHS is simply what you stated in your comment above. – Drew Christianson Oct 2 '11 at 18:15
@jordan As for solving for t in my hint, moving everything to the right hand side and it's a basic quadratic equation... – Drew Christianson Oct 2 '11 at 18:17
But that is the same as the derivative when the height is any number so how does that help me? – user138246 Oct 2 '11 at 18:19