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Cat and mouse question

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Homework Statement



A mouse is resting in a grass field. At t=0 s, the mouse spots a cat that is 10.0m away and is racing towards him a at a constant speed of 5.0 m/s. After taking 0.5 s to react, the mouse starts accelerating from rest at 1.5 m/s^2 away from the cat. How far away from the cat's initial position is it when the cat and mouse first meet?

Homework Equations



x = vt + (1/2)at^2

The Attempt at a Solution



Okay so we start off by saying for the cat:

x = vt, because acceleration is zero

then for the mouse:

x - 10 = (1/2)a(t + 0.5)^2 because the mouse has initial velocity of zero

then, we combine these two to get:

(v cat)t - 10 = (1/2)a(t + 0.5)^2

(v cat)t - 10 = (1/2)at^2 + (1/2)at + (1/8)a <<< after expanding

0 = .75t^2 - 4.25t + 10.1875

using a quadratic solver, t does not exist (that is, it is not a real number).

Where did I go wrong? When I had this marked (above), I actually accidentally used 0 = .75t^2 - 4.25t - 10.1875 (note the MINUS sign before the 10.1875 instead of a plus from above), and i got time = 7.5 seconds. I used this time to figure out the distance, and I got 37 meters, and i got 9/10 for my solution.

After looking back i realized there is no solution, because my algebra was wrong.

What happened?

Thanks sooo much for your help in advance!
 

Answers and Replies

  • #2
ehild
Homework Helper
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Think it over: When the mouse starts to accelerate the cat is already closer than 10 m to it.
The reaction time of the mouse means that it starts to move later. If you add the reaction time, it means that the mouse has a displacement already at t=0 before it started to move.

Place the origin somewhere and decide when you start to measure time. I suggest to start time when the mouse is set to move.
Write the coordinate of the mouse and the coordinate of the cat: they must be equal.

ehild
 
  • #3
265
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Ahh very true...so I should instead subtract 0.5 from t for the mouse...i will try this and see what i get!

Thank you!
 

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