# physis

A motorist traveling at 18 m/s encounters a deer in the road 42m ahead.
If the maximum acceleration the vehicle's brakes are capable of is -6 m/s^, what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer?

If his or her reaction time is 1.1333s, how fast will (s) he be traveling when(s) he reaches the deer? Answer in units of m/s.

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1. look at his stopping distance>

0=18^2 -12d

d= 18^2/12=18(3/2)=27m
so he can stop WITH NO REACTION TIME.
Now, then he can have a reaction time equivalent to 42-27m=15m
time=15/18 sec

Now if his reactiontime is 1.1333 sec he travels..
18*1.13333=20.4 m, so he has 42-20.4=21.6 m left, so he cant stop..

Vf^2=42^2-12 *21.6
solve for vf at the deer.

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2. Here is an answer I found. The numbers are different, but the process is correct.

Let's first find the stopping distance once deceleration begins. This can be found using one of the kinematic equations of motion. We have the following information:

u = initial velocity = 16 m/s

a = acceleration = -5 m/s²

v = final velocity = 0 m/s

d = stopping distance = ?

v² = u² + 2ad. Since v = 0, we must have

d = -u²/2d = (16 m/s)²/2(5 m/s²) = 25.6 m.

So it takes her 25.6 m once she starts braking. Her maximum possible reaction time must be the time it takes for her to travel 41 m - 25.6 m = 15.4 m. As that's before she begins braking, her velocity would be constant over that period, and therefore the reaction time is just distance divided by the speed

= (15.4 m)/(16 m/s) = 0.9625 s.

B: During that time she will travel a distance of (1.1925 s)(16 m/s) = 19.08 m, meaning she decelerates for 41 m - 19.08 m = 21.92 m before hitting the deer.

Here we use v² = u² + 2ad again but this time v is our unknown that we're finding instead of d.

v² = (16 m/s)² + 2(-5 m/s²)(21.92 m)

= 256 - 219.2

= 36.8

and taking the square root gives a speed of 6.1 m/s²

020

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