distance you still have:
distancebeforepressingbrake=42- Vi*.5
after pressing break, distance to stop:
vf^2=vi^2-2ad
0=vi^2-2(10)d but the max d is above, or
0=vi^2-20(42-vi*.50)
vi^2+10vi -840=0
v=(-10+-sqrt(100 +3440)/2= -5+-1/2 (59.5)= about 25 m/s
check my math.
You're driving down the highway late one night at 18 m/s when a deer steps onto the road 42 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 10 m/s2 .
A.)How much distance is between you and the deer when you come to a stop?
I got 17m.
B.)What is the maximum speed you could have and still not hit the deer?
I really don't know how to get Vmax.
Someone please help me. Thank You!
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