kellejalALII
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You're driving down the highway late one night at 18m/s when a deer steps onto the road 39m in front of you. Your reaction time before stepping on the brakes is 0.50s , and the maximum deceleration of your car is 12m/s2 .

How much distance is between you and the deer


when you come to a stop?

=17m


What is the maximum speed you could have and still not hit the deer?

= ?

Verified Answer :

olemakpadu
AMind you before your reaction time, you had be going at a uniform speed 18m/s, so for the reaction time of 0.5 seconds, you had covered a distance of:

18m/s0.5s = 9 m

For the second part which involved deceleration, using: 

v = u - at,  Noting that there is deceleration. 

u = 18m/s, v = final velocity = 0, a = -12m/s². 

Let us solve for the time.

v = u + at

0 = 18 - 12t

12t = 18

t = 18/12 = 1.5 seconds.

Let us compute for the distance covered during the 1.5s

s = ut + 1/2at²,       a = -12 m/s²

s = 181.5 -0.512*1.5² = 13.5m


So the total distance covered = Distance covered from reaction time + Distance covered from deceleration
              
                                              = 9m + 13.5m = 22.5m

So you have covered 22.5m out of the initial 39m.

Distance between you and the dear:   39 - 22.5  = 6.5m

So you have 6.5m between you and the deer. So you did not hit the deer.

b) Maximum speed you still have:

Well through trial and error, if you maintain the same values of deceleration, reaction time, distance between the car and the deer, you could have a speed of 25 m/s and still not hit the deer. Once it is higher than that by a significant amount you would hit the deer.

Answer:

Part a)

\(x = 16.5 m\)

Part b)

\(T = 0.5 + 1.5 = 2 s\)

Part c)

\(v_i = 26.8 m/s\)

Explanation:

As we know that initial speed of the car is

\(v_i = 18 m/s\)

Now reaction time is given as

t =0.50 s

so the distance moved by the car is given as

\(d = v_i t\)

\(d = (18)(0.5) = 9m\)

now the deceleration of the car is given as

\(a = -12 m/s^2\)

so the distance after which it is stopped is given as

\(v_f^2 - v_i^2 = 2 a d\)

\(0 - 18^2 = 2(-12)d\)

\(d = 13.5 m\)

so distance between car and deer is given as

\(x = 39 - 9 - 13.5\)

\(x = 16.5 m\)

Part b)

time taken to stop the car is given as

\(v_f = v_i + at\)

\(0 = 18 - 12 t\)

\(t = 1.5 s\)

so total time to stop is given as

\(T = 0.5 + 1.5 = 2 s\)

Part c)

maximum stopping distance could be

\(d = 39 - 9 = 30 m\)

so here we can have

\(v_f^2 - v_i^2 = 2 a d\)

\(0 - v_i^2 = 2(-12)(30)\)

\(v_i = 26.8 m/s\)

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