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?
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.
Thanks sooo much for your help in advance!