Question

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?
= ?

Answer 1

a) Mind 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/s*0.5s = 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 - 12*t

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 = 18*1.5 -0.5*12*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 2

Part a)

`x = 16.5 m`

Part b)

`T = 0.5 + 1.5 = 2 s`

Part c)

`v_i = 26.8 m/s`

Step-by-step 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`