Question

Constant speed vs constant acceleration - When will B catch up A?

A cat is waiting on the sidewalk. He sees a mouse running at a constant speed of 1.3 km/h. Precisely when the mouse is in front of the cat, the cat starts running after it. The cat's constant acceleration is 0.45 m/s^2.

A) How long will it take for the cat to catch the mouse?

B) What will be the distance the mouse traveled when the cat catches it?

I'm not sure what kind of function I would need to create in order to get the answer to those questions.

Thank you for your help!

Answer 1

A) 1.61 seconds
B) 0.580 metres

Explanation:

Perhaps the best way to visualise this is on a speed-time graph. To do this, we need to pick apart the question.

`color(red)"When I (Steve J.) edited this answer to correct an error"`
`color(red)"I did not study or make changes to the graph. Caution."`

As a forenote, anything in `color(red)"red"` relates to the motion of the `color(red)"cat"` and anything in `color(green)"green"` relates to the `color(green)"mouse"`.

What do we know?
- the `color(green)"mouse"` has an initial speed of `color(green)(1.3kmh^-1)` and has this speed throughout. This is the same as `color(green)(13/36 ms^-1)`
- the `color(red)"cat"` has an itial speed of `color(red)0` and accelerates at `color(red)(0.45ms^-2)`
- the `color(red)"cat"` starts moving immediately, at `t=0`. Yes, the question does say "precisely after", but this is so we don't say that cat eats the mouse at t=0 so we're done.

Having said this, we can sketch a graph.

Let `color(blue)T` be the time taken for the `color(red)"cat"` to catch the `color(green)"mouse"`

We can use equations of motion to find equations of motion for both the cat and the mouse.

For the cat

`s=s_(cat)`
`u=0`
`v="/"`
`a=0.45`
`t=color(blue)T`

We can use the equation `s=ut+1/2at^2` to form the equation:

`color(red)(s_(cat))=0.0color(blue)T+1/2xx0.45color(blue)(T^2)`

`color(red)(s_(cat))=0.0color(blue)T+9/40color(blue)(T^2)`

For the mouse

`s=s_(mouse)`
`u=13/36`
`v=13/36`
`a=0`
`t=color(blue)T`

Again, `s=ut+1/2at^2`. Since `a=0`, this simplifies to `s=ut`

`color(green)(s_(mouse))=13/36color(blue)T`

Since at `t=color(blue)T`, the cat and the mouse have the same displacement:

`color(green)(s_(mouse)=color(red)(s_(cat))`

Solving simultaneously:

`13/36color(blue)T=0.0color(blue)T+9/40color(blue)T^2`
`9/40color(blue)T^2-13/36color(blue)T=0`
`81color(blue)T^2-130color(blue)T=0`
`color(blue)T(81color(blue)T-130)=0`
`color(blue)T=130/81 (color(blue)T!=0)`

So it takes the cat 1.61 seconds to catch up with the mouse (3sf).

For part B, we can substitute into either equation, since the distance `s` at that point will be the same for either. I will substitute into the `color(green)("mouse")` equation.

Let `color(blue)T=130/81`

`s=13/36xx130/81`
`s=1690/2916`

So the mouse (and also the cat) has travelled 0.580m (3sf).

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These numbers personally seem a bit too small, so I'm not sure on my arithmetic, although I think my logic is correct. Hope this helps!