Question

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?

Answer 1

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

Answer 2

cam i see wat you did?

Answer 3

yeah sure!

Answer 4

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)

Answer 5

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

Answer 6

are you following so far?

Answer 7

yes so far so good

Answer 8

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

Answer 9

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

Answer 10

I dunno if you have the final answers?