Motion problem -- The plug and chug at the end

[opus]

Homework Statement

A car and motorcycle are racing along a runway. The motorcycle first takes the lead because it's constant acceleration $a_m=8.40~\left(\frac{m}{s}\right)^2$ is greater than the car's constant acceleration $a_c=5.60~\left(\frac{m}{s}\right)^2$.
But it soon loses to the car because it reaches its greatest speed $v_m=58.8~\frac{m}{s}$ before the car reaches its greatest speed $v_c=106~\frac{m}{s}$
How long does it take for the car to reach the motorcycle?

Homework Equations

$v=v_0+at$
$x-x_0=v_0t+\frac{1}{2}at^2$

The Attempt at a Solution

This is a problem out of the text with step by step directions. What was done was form an equation that relates the positions of the car and motorcycle: $x_c=x_{m1}+x_{m2}$ where the m1 is when the motorcycle is accelerating, and m2 is once it has reached its max speed and the acceleration is zero.

The equations were derived as follows:

$x_c=\frac{1}{2}a_ct^2$
$x_{m1}=\frac{1}{2}\frac{v_m^2}{a_m}$
$x_{m2}=v_m\left(t-7.00sec\right)$

Thus we have
$$x_c=x_{m1}+x_{m2}$$
$$\frac{1}{2}a_ct^2=\frac{1}{2}\frac{v_m^2}{a_m}+v_m\left(t-7.00sec\right)$$

Now the text says that this is a quadratic, and obviously we need to solve for $t$.

Now if you'll look at my attached image, you can see what a disaster I am making of this. We want $t$, but there are also seconds in the equations. I'd imagine the meters should cancel.
Please help

 

[TSny]

The $a_m$ in the denominator of the first term on the right should not be squared in your hand-written notes.

I'm not sure what difficulty you are having with the units of the last term on the right. It looks good. $(58.8 \rm m)t$ has units of meters multiplied by seconds. Also, the $411.6 \, \rm {m \cdot s}$ has units of meters multiplied by seconds. Both of these quantities are divided by seconds, so overall they yield meters. This is as it should be since the terms in the starting equation represent distances.

 

[opus]

(58.8m)t(58.8 \rm m)t has units of meters multiplied by seconds

I guess this part is the root of my troubles. Yes, $t$ is measured in seconds, but it's still the variable $t$ which I'm solving for.

Also, the 411.6m⋅s411.6 \, \rm {m \cdot s} has units of meters multiplied by seconds.

This one is what makes more sense and I can clearly see the seconds cancel, but the seconds is a unit, and the $t$ in the other term is a variable.

 

[TSny]

Take a very simple example. Suppose a car moves at a constant speed v = 2 m/s. To find the time it takes the car to travel 10 m, you could set up the equation x = vt as

$10 \, m = (2 \, m/s) t$

Do the units bother you here? Yes, t is a variable. But it is a variable that has units of seconds.

If you solve the above equation for the time t, you get

$t = \frac {10\, m}{ 2 \, m/s}$

You can see that the units in the answer for t reduce to seconds. m cancels and the s comes up to the numerator.

 

[opus]

So then for $\frac{58.8mt}{s}$, if I know that $t$ is in seconds, I can cancel the $t$ and $s$?

 

[TSny]

So then for $\frac{58.8mt}{s}$, if I know that $t$ is in seconds, I can cancel the $t$ and $s$?

If you are determining the overall units of $\frac{58.8m \,t}{s}$, then you must consider the units for t. Then, yes, the seconds cancel and the overall unit is meters.

However, if you are solving for t in an equation, then you would not cancel the seconds in t with any other seconds. You can see how this works out in post #13. When you solve for t, you will get a number and a unit of seconds.

 

[opus]

Ok, I'm almost entirely on board with you and I'm almost there. I'm having troubles actually isolating the $t$. Can you spot any errors?

 

[TSny]

Ok, I'm almost entirely on board with you and I'm almost there. I'm having troubles actually isolating the $t$. Can you spot any errors?

In going from (f) to (g) you dropped $s^2$ in the term on the right side. So (h) is missing units of $s^2$ in the term on the right. Move this term to the left so that you have zero on the right. You can then solve for t using the quadratic formula. I would express the term with the 58.8 as (58.8 s) t.

 

[opus]

Drats. Thanks! Ok and one final question (apologies if these are off the mark. I haven't done this kind of stuff yet).
With the a,b,c for the quadratic formula,
would $b=-58.8ts$ or would it be $b=58.8t$? Similarly, $c=205.8$ or $c=205.8s^2$? A dumb question I know, but I don't want to start throwing units out if it's not appropriate and then be left with bad problem solving habits.

 

[opus]

To be more clear,
$x=\frac{{58.8s+-\sqrt\{(58.8s)^2-4(2.8)(205.8s^2)}}{2(2.8)}$ ??

 

[TSny]

To be more clear,
$x=\frac{{58.8s+-\sqrt\{(58.8s)^2-4(2.8)(205.8s^2)}}{2(2.8)}$ ??

Yes. Looks good. Of course the x on the left is actually t.

 

[PeroK]

To be more clear,
$x=\frac{{58.8s+-\sqrt\{(58.8s)^2-4(2.8)(205.8s^2)}}{2(2.8)}$ ??

A couple of observations.

First, in the problem it's not clear whether the car catches the motorcycle before it reaches its maximum speed. If it does, then the equation you used in the original post is correct. But, if not, then you will get a solution here that relies on the car reaching a higher speed. So, I suggest you need to check this.

Second, your equation still has $\pm$ in it. Which solution is it?

Third, do you really want to plug and chug these questions? Surely an algebraic approach is better. For example, you could shoot for the following formula (assuming the car catches the motorcycle before it reaches its maximum speed):

$t = \frac{v_m}{a_c}(1 + \sqrt{1 - \frac{a_c}{a_m}})$

And, if you note that $\frac{a_c}{a_m} = \frac23$, then you don't have any complicated numerical expressions to deal with.

 

[jbriggs444]

First, in the problem it's not clear whether the car catches the motorcycle before it reaches its maximum speed.

The motorcycle accelerates more rapidly than the car. It is pulling away, not being caught. That said, it is prudent to think about the possibility.

 

[PeroK]

The motorcycle accelerates more rapidly than the car. It is pulling away, not being caught. That said, it is prudent to think about the possibility.

The problem with pronouns:

.. it's not clear whether the car catches the motorcycle before it (the car) reaches its maximum speed.

 

[opus]

Thanks for the responses all.
Yes, the car never reaches max speed in the problem, and passes the motorcycle after the motorcycle has reached max speed.

\pm in it. Which solution is it?

So after the plug an chug, I got $t=16.6,4.4$ seconds. It is stated in the problem that the car passes the motorcycle after the motorcycle has reached max speed (which is at 7s). Therefore, 4.4s is a mathematically correct solution, but not the solution to the physical problem.

Third, do you really want to plug and chug these questions? Surely an algebraic approach is better. For example, you could shoot for the following formula (assuming the car catches the motorcycle before it reaches its maximum speed):

t=vmac(1+√1−acam)t = \frac{v_m}{a_c}(1 + \sqrt{1 - \frac{a_c}{a_m}})

And, if you note that acam=23\frac{a_c}{a_m} = \frac23, then you don't have any complicated numerical expressions to deal with.

I think it was good to see the long strung out version but I definitely would not like to do all of that every time.
How did you derive what you have there, and how did you notice that $\frac{5.60}{8.40}=\frac{2}{3}$?

 

[opus]

Let me ask another question as well.

For $x_{m1}$:
We use $v=v_0+at$
Then $t_m=\frac{v_m}{a_m}=\frac{58.8m/s}{8.40m/s^2}=7.00m/s$
This is the time that it took the motorcycle to reach maximum speed.

Now from $x-x_0=v_0t+\frac{1}{2}at^2$,
$x_{m1}=0+\frac{1}{2}a_mt_m^2$
$=\frac{1}{2}a_mt_m^2$
$=\frac{1}{2}a_m\left(\frac{v_m}{a_m}\right)^2$
$=\frac{1}{2}\frac{v_m^2}{a_m}$

Why can't I just plug in 7.00s for $t$ to find the position that the motorcycle covered in $x_{m1}$? Is it because the velocity is not constant, and would need an integral to find the new position with just this info?

 

[PeroK]

Let me ask another question as well.

For $x_{m1}$:
We use $v=v_0+at$
Then $t_m=\frac{v_m}{a_m}=\frac{58.8m/s}{8.40m/s^2}=7.00m/s$
This is the time that it took the motorcycle to reach maximum speed.

Now from $x-x_0=v_0t+\frac{1}{2}at^2$,
$x_{m1}=0+\frac{1}{2}a_mt_m^2$
$=\frac{1}{2}a_mt_m^2$
$=\frac{1}{2}a_m\left(\frac{v_m}{a_m}\right)^2$
$=\frac{1}{2}\frac{v_m^2}{a_m}$

Why can't I just plug in 7.00s for $t$ to find the position that the motorcycle covered in $x_{m1}$? Is it because the velocity is not constant, and would need an integral to find the new position with just this info?

I don't follow what you are doing there. In general, if you have constant acceleration $a$ up to a maximum speed $v$ and then constant speed of $v$, then the distance covered at time $t$ is:

$x = \frac12 at_0^2 + v(t-t_0)$

Where $t_0 = \frac{v}{a}$ is the time it takes to reach maximim speed. Now, if you substitute this expression for $t_0$ into the above expression you get:

$x = vt - \frac12at_0^2 = vt - \frac12 \frac{v^2}{a}$

The last equation is useful, because you've eliminated $t_0$.

 

[opus]

Ok I see. The work in my post is from the text, where they are (I think) showing that you can use the two constant acceleration equations simultaneously.
So then, is it possible, without an integral, to find out the distance covered by the motorcycle in the first stage (where velocity is not constant, and acceleration is constant)?

 

[PeroK]

Ok I see. The work in my post is from the text, where they are (I think) showing that you can use the two constant acceleration equations simultaneously.
So then, is it possible, without an integral, to find out the distance covered by the motorcycle in the first stage (where velocity is not constant, and acceleration is constant)?

Yes, it's the area of a triangle. That's where $\frac12 at^2$ comes from. The triangle has base $t$ and height $at$. The equations I posted above are essentially just the area of a triangle plus a rectangle. You don't need integration for that either!

 

[opus]

Super cool. So why then, don't we just use that and 7.00s to find the position that the motorcycle is in when it hits max speed?
That is,
If $x=\frac{1}{2}at^2$ and the time to max speed is 7 seconds, then
Distance covered when the motorcycle hits max speed = x = $\frac{1}{2}\frac{8.40m}{s^2}{(7.00s)^2}$? When x is the position at this time, but since we are starting at 0, then it's the displacement as well.

 

[PeroK]

Super cool. So why then, don't we just use that and 7.00s to find the position that the motorcycle is in when it hits max speed?
That is,
If $x=\frac{1}{2}at^2$ and the time to max speed is 7 seconds, then
Distance covered when the motorcycle hits max speed = x = $\frac{1}{2}\frac{8.40m}{s^2}{(7.00s)^2}$? When x is the position at this time, but since we are starting at 0, then it's the displacement as well.

That's not very useful by itself. Knowing where the bike is when it reaches max speed is only one variable in the equation.

In some problems it would be a good idea to calculate that, but in this problem it doesn't help much.

 

[opus]

But isn't it equal to $m_1$? which is an entire term in the original equation? It seems it would greatly simplify the calculation at the end.

 

[PeroK]

But isn't it equal to $m_1$? which is an entire term in the original equation? It seems it would greatly simplify the calculation at the end.

If you can find a simpler solution using this, then fine.

 

[opus]

I guess I don't understand why we say that $x_{m1}=\frac{1}{2}a_mt_m^2=\frac{1}{2}\frac{v_m^2}{a_m}$
Rather than $x_{m1}=\frac{1}{2}\frac{8.40m}{s^2}{7.00s^2}$ The text went directly out of it's way to not use the 7 seconds that we found for $x_{m1}$

 

[opus]

So that we’re reading from the same sheet of music.

 

[PeroK]

So that we’re reading from the same sheet of music.

Although it deals with cars and motorcycles, that's a fairly pedestrian approach if you ask me!

 

[opus]

Although it deals with cars and motorcycles, that's a fairly pedestrian approach if you ask me!

LOL.