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

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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!

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!

##### 1 Answer

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.

As a forenote, anything in

**What do we know?**

- the

- the

- the

Having said this, we can sketch a graph.

Let

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

**For the cat**

We can use the equation

**For the mouse**

Again,

Since at

Solving simultaneously:

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

Let

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

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!