Constant acceleration problem: Dog chasing a surprised cat

[abdol83]

Homework Statement

A cat is sleeping on the floor in the middle of a 3.0-m-wide room when a barking dog enters with a speed of 1.50 m/s. As the dog enters, the cat (as only cats can do) immediately accelerates
at 0.85 m/s2 toward an open window on the opposite side of the room. The dog (all bark and no bite) is a bit startled by the cat and begins to slow down at 0.10 m/s2 as soon as it enters the
room. Does the dog catch the cat before the cat is able to leap through the window?

Relevant Equations

Cat: x - x0 = 0.5 * a * (t - t0) ^ 2
Dog: x - x0 = v0 * (t-t0) - 0.5 * a * (t - t0) ^ 2

Here I'm not worried about the solution as I got it. There are two ways I could come up with:
Either finding the time it takes the cat to leap through the window, use that time to find what distance does the dog cross through the room
Or: finding both times, for the dog and the cat to cross the room, make a comparison.

The problem is when trying to find the time for the dog, I end up with quadratic equation with two positive times, which is illogical as, how can an object travel the same distance with two times (given everything else is the same)?

 

[kuruman]

How abut posting how you got the time for the dog so that we can look at it? Remember that the equations that you wrote express position as a function of time not distance traveled. Because the one-dimensional motion is parabolic in time, an object can be at the same position at two separate times. When you throw a ball straight up in the air, it is at the same position at an earlier time on its way up and at a later time on its way down. Both times are positive roots of the quadratic equation.

 

[erobz]

Homework Statement: A cat is sleeping on the floor in the middle of a 3.0-m-wide room when a barking dog enters with a speed of 1.50 m/s. As the dog enters, the cat (as only cats can do) immediately accelerates
at 0.85 m/s2 toward an open window on the opposite side of the room. The dog (all bark and no bite) is a bit startled by the cat and begins to slow down at 0.10 m/s2 as soon as it enters the
room. Does the dog catch the cat before the cat is able to leap through the window?
Relevant Equations: Cat: x - x0 = 0.5 * a * (t - t0) ^ 2
Dog: x - x0 = v0 * (t-t0) - 0.5 * a * (t - t0) ^ 2

Here I'm not worried about the solution as I got it. There are two ways I could come up with:
Either finding the time it takes the cat to leap through the window, use that time to find what distance does the dog cross through the room
Or: finding both times, for the dog and the cat to cross the room, make a comparison.

The problem is when trying to find the time for the dog, I end up with quadratic equation with two positive times, which is illogical as, how can an object travel the same distance with two times (given everything else is the same)?

I think you are making an error. Make a diagram, show the initial position of the cat at time $t=0$ when the dog enters the room at initial speed $v_o$ label the accelerations of the cat and dog as $a_c$ and $a_d$ respectively.

Show the quadratic you arrive at, and how you arrive at the equations. In the solution, pay close attention to the radical.

Please present your solution using the mathematical format supported by the site. See LaTeX Guide

 

[haruspex]

finding the time it takes the cat to leap through the window, use that time to find what distance does the dog cross through the room
Or: finding both times, for the dog and the cat to cross the room, make a comparison.

Neither approach will necessarily work. In one case, you are finding both positions at a particular time, in the other finding both times for a particular distance. Both overlook the possibility that the dog reaches the cat at an earlier time and distance but falls behind later. You should be looking for the existence of an unknown time and distance combination.

The problem is when trying to find the time for the dog, I end up with quadratic equation with two positive times, which is illogical as, how can an object travel the same distance with two times (given everything else is the same)?

It often happens that an equation only correctly represents the scenario between certain constraints. In this case, your equation for the dog's motion is only valid from the point where the dog enters the room to (whichever comes first of) the dog's reaching the window or coming to a halt.
The equation may generate spurious solutions outside that range.

 

[PeroK]

The problem is when trying to find the time for the dog, I end up with quadratic equation with two positive times, which is illogical as, how can an object travel the same distance with two times (given everything else is the same)?

What goes up, must come down! An object thrown upwards passes each vertical point on its trajectory twice (except the highest point). Note that the SUVAT equations involve displacement, not distance.

And, in this case, what goes out the window, must come back in through the window! Or, at least, if the dog keeps accelerating at the same rate, it will eventually stop and start to retreat with ever increasing speed.

 

[abdol83]

This is exactly what I was looking for!
Thank you very much indeed!

 

[erobz]

This is exactly what I was looking for!
Thank you very much indeed!

So, (regardless of the window) does the dog ever catch the cat?

 

[kuruman]

Both overlook the possibility that the dog reaches the cat at an earlier time and distance but falls behind later.

I think that the best approach is, as others have already suggested, to make a plot. The one on the right shows the above possibility with the sleeping cat at 0.5 m from the entrance, the other parameters being the same.

One can also find the minimum safe distance of the cat from the entrance by finding the time at which the two parabolas have equal slopes and substituting that time in one of the position equations.

FinalIy, cannot resist noting that a passerby below the window might be astonished to observe that it's literally raining cats and dogs.

 

[haruspex]

I think that the best approach is, as others have already suggested, to make a plot.

Sure, but you had already made that point. I was trying, as I generally do, to nail the specific error in the OP's approach.

 

[abdol83]

So, (regardless of the window) does the dog ever catch the cat?

No, luckily the cat escapes!

 

[abdol83]

View attachment 352771I think that the best approach is, as others have already suggested, to make a plot. The one on the right shows the above possibility with the sleeping cat at 0.5 m from the entrance, the other parameters being the same.

One can also find the minimum safe distance of the cat from the entrance by finding the time at which the two parabolas have equal slopes and substituting that time in one of the position equations.

FinalIy, cannot resist noting that a passerby below the window might be astonished to observe that it's literally raining cats and dogs.

I like your graphical representation, thank you indeed. The cat actually starts from the middle of the room, which is at position 1.5 meters, but that's a technical detail, your graphs truly depict the opposing signs of accelerations. Thanks again!

 

[PeroK]

No, luckily the cat escapes!

Just looking at the numbers, I would put a completely different interpretation on the animals' behaviour. Note that $1.5 m/s$ is walking speed and $0.85m/s^2$ is exceptionally sluggish for a cat. I'd imagine a cat could generate 5-10 times that acceleration.

A cat is sleeping on the floor in the middle of a 3.0-m-wide room when a barking dog wanders in. The cat reluctantly gives up its sleeping berth, moves slowly to the open window and jumps out. The dog moves slowly to the window and looks out to make sure the cat has gone, then settles down for a snooze in the middle of the room.

 

[kuruman]

I like your graphical representation, thank you indeed. The cat actually starts from the middle of the room, which is at position 1.5 meters, but that's a technical detail, your graphs truly depict the opposing signs of accelerations. Thanks again!

Yes, I used a different starting position $x_0$ for the cat to illustrate the catch-up condition graphically relative to the window. Now that I think about I think that the best way to approach this problem is to calculate minimum the "safe distance" $x_{\text{min}}$ of the cat from the entrance and compare with $x_0$.

This can be done by setting the equations of the animals' parabolas equal to each other. The roots of the resulting quadratic equation will give the times at which the two animals are at the same position. However, there is no need to find these times. We just have to find the discriminant of the equation. There will be two roots if the discriminant is positive, one root if it zero and no roots if it is negative. We are interested in a negative discriminant because that is the threshold for a possible catch-up. Thus, $$\begin{align}
&x_0+ \frac{1}{2}a_ct^2=v_0t-\frac {1}{2}a_d t^2 \nonumber \\
& (a_c+a_d)t^2-2v_0t+2x_0=0 \nonumber \\
& \text{Disc.}=v_0^2-2x_0(a_c+a_d) <0 \implies x_0>\frac{v_0^2}{2(a_c+a_d)}.\nonumber
\end{align}$$ Here, the critical value for $x_0$ above which the cat is guaranteed to escape is $$x_{\text{crit.}}=\frac{(1.5~\text{m/s})^2}{2(0.85+0.1)(\text{m/s}^2)}=1.18~\text{m}.$$

The graph for this problem is shown on the right with the critical value for $x_0$ added.
Note that

• Since the cat starts above the critical value, it escapes and there is no possibility that it will be caught no matter where the window is.
• If the cat starts at the critical value, the red line showing its position is tangent to both the green critical line and the blue line for the dog. The cat may or may not be caught depending on where the window is.
• If the cat starts below the critical value, it may still escape if the window is close enough to it. In the limit that the window is at the cat's starting position, the cat will always escape. Perhaps that is why cats like to sleep on window sills.

 

[kuruman]

A cat is sleeping on the floor in the middle of a 3.0-m-wide room when a barking dog wanders in. The cat reluctantly gives up its sleeping berth, moves slowly to the open window and jumps out. The dog moves slowly to the window and looks out to make sure the cat has gone, then settles down for a snooze in the middle of the room.

I should point out that this scenario will not work in the real world under the assumption that both animals are undergoing constant acceleration. Under that assumption, the dog will come to instantaneous rest when he is 8.25 m horizontally away from the window but the pull of gravity will displace him vertically down.

That said, the scenario is possible in the cartoon world where the Cartoon Laws of Physics are in effect. In that case, said animals are cartoon characters, Tom, the cat, and Spike (sometimes Butch or Killer), the dog. According to Cartoon Law I, "Any body suspended in space will remain in space until made aware of its situation." Since Spike is portrayed as irreparably witless, he will have plenty of time to go horizontally out, verify that Tom is gone and make it back into the room under constant horizontal acceleration.