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?
Answer in units of s.
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.
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
2ad = -u², and therefore
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
Why did the deer cross the road? To make the motorist ask these complicated physics questions! But don't worry, I'll help you out.
To avoid hitting the deer, the motorist needs to come to a stop before reaching it. Let's assume the motorist takes time 't' to react and applies maximum braking acceleration 'a' to stop the car.
The distance covered during the reaction time 't' can be calculated using the equation:
distance = initial velocity * time + 0.5 * acceleration * time^2.
In this case, the initial velocity is 18 m/s, the distance is 42 m, and the acceleration is -6 m/s^2 (negative because we're slowing down). So, we can rearrange the equation and solve for time 't':
42 m = 18 m/s * t + 0.5 * (-6 m/s^2) * t^2.
Solving this quadratic equation will give us two possible answers, but we discard the negative value since time cannot be negative. So, the maximum reaction time 't' is approximately 1.1333 s.
Now, to find the final velocity 'v' when the motorist reaches the deer, we can use the equation:
final velocity = initial velocity + acceleration * time.
Plugging the values in:
final velocity = 18 m/s + (-6 m/s^2) * 1.1333 s.
Calculating the expression gives us the final velocity of approximately 11.6 m/s.
So, the motorist will be traveling at approximately 11.6 m/s when they reach the deer. Let's hope they have some quick reflexes and give the deer a frightener!
To determine the maximum reaction time of the motorist that will allow them to avoid hitting the deer, we can use the equations of motion.
Let's break down the problem step by step:
Step 1: Calculate the time it would take for the motorist to reach the deer without any reaction time.
Using the equation of motion:
distance = initial velocity * time + 0.5 * acceleration * (time^2)
Given:
Initial velocity (v0) = 18 m/s
Distance (d) = 42 m
Acceleration (a) = 0 (since the motorist is assuming no reaction time)
Substituting the values:
42 = 18t + 0.5 * 0 * (t^2)
42 = 18t
Solving for t:
t = 42 / 18
t = 2.33 s
Step 2: Determine the maximum reaction time.
The maximum reaction time (t_reaction) is the time it takes for the vehicle to come to a complete stop from the initial velocity (v0) using the maximum acceleration (a).
Given:
Acceleration (a) = -6 m/s^2
Using the equation of motion:
v = v0 + at
As the vehicle comes to a stop, the final velocity (v) will be 0 m/s. Therefore:
0 = 18 + (-6)t_reaction
Solving for t_reaction:
-6t_reaction = -18
t_reaction = 18 / 6
t_reaction = 3 s
The maximum reaction time of the motorist that will allow them to avoid hitting the deer is 3 seconds.
To find the final velocity when the motorist reaches the deer after the reaction time, we can use the same equation of motion as in Step 2.
Given:
Initial velocity (v0) = 18 m/s
Acceleration (a) = -6 m/s^2
Reaction time (t_reaction) = 1.1333 s (rounded)
Using the equation of motion:
v = v0 + at
Substituting the values:
v = 18 + (-6) * 1.1333
v = 18 - 6.8
v = 11.2 m/s
Therefore, if the motorist has a reaction time of 1.1333 seconds, their speed when reaching the deer will be approximately 11.2 m/s.
look at his stopping distance>
vf^2=vi^2+2ad
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=vi^2-2ad
Vf^2=42^2-12 *21.6
solve for vf at the deer.