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annacEstepha
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You're driving down the highway late one night at 16m/s when a deer steps onto the road 44m in front of you. Your reaction time before stepping on the brakes is 0.50s , and the maximum deceleration of your car is 10m/s2 .


How much distance is between you and the deer? What is the maximum speed you could have and still not hit the deer?

Verified Answer :

Have you considered hitting it one piece at a time?
 
Reaction Distance: Speed 0.50 s
 
Initial Problem: 16 m/s 0.50 s = 8 m
 
Braking Acceleration: a(t) = -10 m/s^2 --

This is constant.


Braking Velocity: s(t) = -10 m/s^2 (t) + "Initial Velocity"


Initial Problem: s(t) = -10 m/s^2 (t) + 16 m/s
 
Braking Location: x(t) = -5 m/s^2 (t^2) + 16 m/s (t) - "Initial Location"
 
Initial Problem: x(t) = -5 m/s^2 (t^2) + 16 m/s (t) - (44 m - 8 m)

 
Having said all that, the first question makes no sense. Does it mean the distance when the vehicle stops?
 
s(t) = -10 m/s^2 (t) + 16 m/s = 0 ==> t = 16/10 s = 8/5 s
 
That's how long it takes to stop.
 
Where are we when that happens?
x(8/5) = -5 m/s^2 ((8/5 s)^2) + 16 m/s (8/5 s) - (44 m - 8 m) = -23.2 m
 
This seems to be a little different from your response.

Now, the challenge is to do this all over again, but missing the initial velocity.
 
Initial Velocity: V -- Let's just call it this so we can talk about it.
 
Initial Distance is the same: 44 m -- No change.


Reaction Distance: V/2


Braking Acceleration: a(t) = -10 m/s^2 --

No Change
Braking Velocity: s(t) = -10 m/s^2 (t) + V -- Missing V!
 
Braking Location: x(t) = -5 m/s^2 (t^2) + V (t) - (44 m - V/2 m)
 
When do we stop? s(t) = -10 m/s^2 (t) + V = 0 ==> t = V/10 s


Where do we stop? x(t) = -5 m/s^2 ((V/10 s)^2) + V (V/10 s) - (44 m - V/2 m) = 0 -- We might just brush the poor, frightened deer.



One piece at a time! Slowly. Methodically.

Answer:

Part a)

d = 23.2 m

Part b)

v = 25.1 m/s

Explanation:

Part a)

As we know that the speed of the car is given as

\(v = 16 m/s\)

reaction time of the driver = 0.50 s

now the distance moved by the car during reaction time

\(d = velocity \times time\)

\(d = 16(0.50) = 8m\)

After this car is decelerated due to brakes with deceleration

\(a = -10 m/s^2\)

so the distance moved by the car till it will stop

\(v_f^2 - v_i^2 = 2ad\)

\(0 - 16^2 = 2(-10)d\)

\(d = 12.8 m\)

so total distance moved by the car

\(d = 12.8 + 8 = 20.8 m\)

Now the distance between the car and deer is given as

\(x = 44 - 20.8\)

\(x = 23.2 m\)

Part b)

Let the driver is moving with speed "v"

now the distance moved in reaction time

\(d_1 = v(0.50)\)

now after applying brakes the distance moved by it

\(v_f^2 - v_i^2 = 2 ad\)

so we have

\(0 - v^2 = 2(-10)d_2\)

so total distance covered is given as

\(d = d_1 + d_2\)

\(d = \frac{v^2}{20} + 0.50v\)

for maximum speed v the distance moved is

\(d = 44 m\)

so we have

\(44 = \frac{v^2}{20} + 0.50v\)

by solving above equation we have

\(v = 25.1 m/s\)