2
$\begingroup$

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.

  • 0
    Oh, I was just working through my calculus homework and this came up.2011-10-02
  • 0
    HINT:Instantaneous velocity is actually derivative of distance with respect to time.2011-10-02
  • 0
    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 0.2011-10-02
  • 2
    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 http://physics.stackexchange.com, though.2011-10-02

4 Answers 4

0

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?

  • 0
    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.2011-10-02
  • 0
    I have no idea how to solve for t in that equation.2011-10-02
  • 0
    @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.2011-10-02
  • 0
    @jordan As for solving for t in my hint, moving everything to the right hand side and it's a basic quadratic equation...2011-10-02
  • 0
    But that is the same as the derivative when the height is any number so how does that help me?2011-10-02
  • 0
    @Jordan _Precisely_. Well, almost precisely. $15-2(1.86t) = v(t)$ is the velocity (derivative) at any _time_ $t$. But, we have height $h$ as a function of time $t$ ($h = 15t-1.86t^2$) So, if we can just find $t$ when $h$ = 25, you can plug that t into the equation for velocity and you're home free. For a general quadratic equation $ax^2 + bx + c = 0$ the solutions are given by the equation: $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ Here, your equation is: $$25 = 15t-1.86t^2$$ $$-1.86t^2 + 15t - 25 = 0$$ Given that, just plug the values into the general equation and you'll have solutions.2011-10-02
  • 0
    So basically what is happening is that I take the derivative of the right side and then subtract 25? I don't quite understand why this is being done, it feels like I am just loosely using math rules to manipulate the problem so it gives me the right answer. it just doesn't feel intuitive unless I know what the answer is I guess. To me if I want to derivative of $25=15t−1.86t2$ I just get 0=15=2(1.86t)2011-10-02
  • 0
    @jordan you can't plug in the h=25 before you take the derivative. As didier, Hennring, and many other members have told you, math is _not_ a set of strict rules. You have to think about the problems and apply the _tools_ you have to solve them, sometimes in new and different ways. If you wanted to know the slope of $f(x)=x^2$ at x=10, you would _not_ plug 10 in then differentiate. You would find $f'(x) = 2x$ and the calculate $f'(10)$. We're doing the same thing here, finding $t_{1}$ such that $s(t_{1}) = 25$ _and_ $s'(t)$ then substituting $t_{1}$ into $s'(t)$2011-10-02
  • 0
    @jordan remember, _tools_, not _rules_.2011-10-02
  • 0
    I don';t understand this...so there is no way to really do math properly, I just do it until I have the correct answer? This feels like guessing to me.2011-10-02
  • 0
    @jordan Ideally, you don't know the correct answer at the outset, only the validity and use of your tools (derivatives, algebra, etc). Then, by careful application either by the predefined 'rules' as you call them (recipes as didier or someone called them) or novel 'recipes' that nevertheless use the tools correctly, you can know completely and undoubtedly that the answer is right. To clarify: there is a way to do math properly, and it is by using the tools you have carefully so that you know every step is valid, and thus the entire result you derive is valid.2011-10-02
  • 0
    But I don't understand how I should know that I can't just take the derivative of 25=15t-1.86t^2 what tells me that I can't do that?2011-10-02
  • 0
    Honestly I am just getting too frustrated to even do this, spent all day doing homework and I got nothing done.2011-10-02
  • 0
    @jordan Math can be frustrating, it's complicated stuff. Some of us here definitely make it look easy, but that's the product of decades of constant work. If it hasn't been your focus (which isn't bad in the slightest) it's just going to be a bit harder for you. The best advice I can give you is what everyone else has said: don't think about rules, tricks, etc. Instead, try to conceptualize everything you run. Here, the function s(t) = h is good example. What does it look like? What does it tell you about the point in describes? What follows from it?2011-10-03
  • 0
    @jordan For me, the answers are: It's a point moving along a line, s(t) shows me where that point is at anytime time t. Knowing the position at every point in time means pretty much everything (velocity, acceleration, location, distance traveled, etc) follows from that. Moreover that function exists inside the logical framework of mathematics, within which I expect everything to be consistent. If I follow the rules anything that begins corresponding to reality should end corresponding to reality. Hence, you shouldn't 'know' _ex ante_ you can't plug in h=25 before you differentiate2011-10-03
  • 0
    @jordan but you should be able to see ex post when your result is that the velocity is 0 at all time t that something is wrong. Thus, you take a step back, assume that to be false, and try something else. Hopefully some of this advice helps. The last thing I'll add: stick with it. It seems hard because it is, but it's hard because it's worth it.2011-10-03
  • 0
    I don't know, I just find this humiliating that I am struggling so hard with math that 16 and 17 year olds breeze through. I don't think this is worth it.2011-10-03
3

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.

  • 0
    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.2011-10-02
  • 0
    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 $2a=3.72$.2011-10-02
2

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.

You just have to add to your equation

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

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$^1$

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

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

enter image description here

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)

enter image description here

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 $|v(t_1)|=|v(t_2)|$.

--

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

  • 0
    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?2011-10-02
  • 0
    @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.2011-10-02
  • 0
    @Jordan Carlyon: the first velocity is attained at $t_1$, the second at $t_2$.2011-10-02
  • 0
    @Jordan Carlyon: Yes, it is the distance to surface of mars, i.e the hight.2011-10-02
  • 0
    @Jordan Carlyon: You see that because from the given formula for $h$, at $t=0$ $h=0$ (surface of mars).2011-10-02
  • 0
    Ok I am a little confused as to what actually happened "The velocity at t1 is" I don't really see how any of that was composed or where these numbers are coming from. $\begin{equation} v=h^{\prime }(t)=\frac{d}{dt}\left( 15t-1.86t^{2}\right) =15-2\times 1.86t. \end{equation}\tag{2}$ Doesnt quite make sense to me either, the derivative of h with respect to t is the same as the derivative of t times 15t-1.86 squared? Which is then equal to something else I am unfamiliar with.2011-10-02
  • 0
    @Jordan Carlyon: The derivative $h^{\prime }(t)$ can be computed in detail as follows: $$\begin{eqnarray*} h^{\prime }(t) &=&\frac{d}{dt}\left( 15t-1.86t^{2}\right) \\ &=&\frac{d}{dt}\left( 15t\right) -\frac{d}{dt}\left( 1.86t^{2}\right) \qquad \text{sum rule} \\ &=&15\frac{dt}{dt}-1.86\frac{d}{dt}\left( t^{2}\right) \qquad \text{product rule} \\ &=&15\times 1-1.86\times 2t\qquad \text{power rule} \\ &=&15-1.86\times 2t\text{.} \end{eqnarray*}$$2011-10-02
  • 0
    Ok that makes more sense to me I guess.2011-10-02
  • 0
    @Jordan Carlyon: Do you still have doubts?2011-10-04
  • 0
    Honestly I completely forgot how to do this problem and will need to spend an hour or so on it today.2011-10-04
  • 0
    @Jordan Carlyon: OK, take your time.2011-10-04
  • 0
    I don't really follow what is happening I guess. First of all what is the approximated value and how is that calculated? $\begin{eqnarray*}t_{1} &=&\frac{15-\sqrt{15^{2}-4\times 25\times 1.86}}{2\times 1.86}\approx 2.3535\text{ h,} \\t_{2} &=&\frac{15+\sqrt{15^{2}-4\times 25\times 1.86}}{2\times 1.86}\approx 5.711\text{ h.} \end{eqnarray*}\tag{1'}$ And then in the actual solution for velocity what are the numbers in the parentheses?2011-10-04
  • 0
    @JordanCarlyon: Equation $(1)$ is a quadratic equation $$\begin{equation} 25=15t-1.86t^{2}\Leftrightarrow 1.86t^{2}-15t+25=0\tag{1}\end{equation}.$$ I added the approximated values of its roots $t_1,t_2$ so that you could recognize them in the graphs. I computed them with a calculator. In $(3)$ and $(4)$ I used algebra ($2\times 1.86$ gets cancelled) to find the exact numeric values $-\sqrt{39}$ and $\sqrt{39}$.2011-10-04
1

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 $v$.

  • 0
    I have no idea what ut+1/2at^2 is.2011-10-02
  • 0
    @Jordan: they are equations of motion under constant acceleration. Basic physics/mechanics.2011-10-02
  • 0
    I don't think I am suppose to know that for this problem.2011-10-02