You\'re driving down the highway late one night at 16.0 m/s when a deer steps on

Question

You're driving down the highway late one night at 16.0 m/s when a deer steps onto the road 39.0 M in front of you. You reaction time before stepping on the breaks is 0.50 s, and the maximum deceleration of your car is 12.0 m/s^2? How much distance is in between you and the deer when you come to a stop? What is the maximum speed you could have and still not hit the dear? Make sure you follow the steps below: Read the whole problem and make sure you understand it. Then read it again. Decide on the objects under study and what the time interval is. Choose coordinate axis and draw a diagram. Write down the known (given) quantities, and then the unknown ones that you need to find. What physics applies here? Plan an approach to a solution. Which equations relate the known and unknown quantities? Are they valid in this situation? Solve algebraically for the unknown quantities, and check that your result is sensible (correct dimensions).

Explanation / Answer

It takes 0.5s to apply his brakes

So, he moves at 16 m/s for 0.5 s

So, distance moved in this time = velocity*time = 16*0.5 = 8 m

Initial velocity , u = 16 m/s

acceleration = -12 m/s2 <----- negative sign for deceleration

Using the equation of motion :

v^2 = u^2 + 2as

here, v = final velocity = 0 <--- it comes to rest

s = displacement before coming to rest

So, 0^2 = 16^2 + 2*(-12)*s

So, s = 10.67 m

So, total distance moved before he comes to rest = 10.67 + 8 = 18.67 m

So, the remaining distance between you and the deer = 39 - 18.67 = 20.33 m <------- answer

Let the maximum speed so that you dont hit be u

So, distance moved in initial 0.5s = 0.5*u

So, for distance moved while decelerating:

v^2 = u^2 + 2as

So, 0^2 = u^2 + 2*(-12)*(39 - 0.5*u)

So, u = 25.2 m/s <-------answer

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