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- anonymous

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 40 m in front of you. Your reaction time before stepping on the brakes is
0.50 s , and the maximum deceleration of your car is 12 m/2^2
How much distance is between you and the deer when you come to a stop?
What is the maximum speed you could have and still not hit the deer?

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- anonymous

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- anonymous

I solved the problem, but I am not so sure about my answer.

- anonymous

cam i see wat you did?

- anonymous

yeah sure!

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- anonymous

I first found the time needed for the car to stop, that's when the final velocity is zero. Apply it to the formula:
\[v_f=v_i+at \implies 0=20-12t \implies t=20/12=1.67\]
(since it's decelerating, I took a to be negative)

- anonymous

but we know that the reaction was 0.5 s late so the total time needed is t=1.67+0.5=2.17

- anonymous

are you following so far?

- anonymous

yes so far so good

- anonymous

ok I will just write the rest.
now we will have to find the distance he is going to drive during this period of time, which is:
\[x={v_i}t+0.5at^2\]
just direct substitution in the formula, you will get:
\[x=20 (2.17)+0.5(-12)(2.17)^2=15.15 m\]
so the distance he traveles before the car stops is 15.15 meter

- anonymous

now it's easy to find the distance between the car and the deer, which is
40-15.15=24.85 meter

- anonymous

I dunno if you have the final answers?

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