- #1
ujellytek
- 35
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Homework Statement
There is a speedy car chase! A thief is getting away from a police officer. The distance between them at the start is 100m. The thief starts to accelerate at 5.0m/s^2 [F] from rest. Meanwhile the police is chasing him at its max velocity of 30m/s (sad... I know XD). Show that the police officer NEVER CATCHES UP to the thief.
Please take a gander at section 3 and answer this:
i) First off, is my solution correct?
ii) Ok... this seems to be the correct solution because they are NOT supposed to meet-up/collide during the car chase at all, because subbing in 6s into both equations gives 90m and 80m, but how can the quad formula say that at 6s they do meet (why do we get this mathematical inconsistency?)?
iii)Or did I just do It wrong? Please explain this in more depth if you want, or just show me how to solve it, this is kind of a bonus question and I need ALL the marks I can to get into software engineering, please help!
~Thanks
P.S. I will watch this thread inthe next little while, and in 10 hours plus a constant span of 75min (my period 1 spare) [i'm also interested in question 2)ii) quite a bit].
Homework Equations
The Attempt at a Solution
So since the officer stars 100 meters behind I got: d=t*30-100
and for the thief I got: d= v*t+0.5*a*t^2 which becomes d=2.5*t^2 because thief starts at rest
Then I make these two equations equal to each other: t*30-100=2.5*t^2 and I make it equal to 0;
thus 0=2.5*t^2-30t+100 and when I apply the quadratics formula I get i2 seconds and 6 seconds.
Now when I sub 2s into the officers displacement equation I get: d(6s)=(6)*30-100= 80m
and, for the thief I get: d(6s)=2.5(6)^2= 90m
I get a mathematical inconsistency and that satisfies the question.
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