You are driving down the highway and you brake to avoid hitting a deer

[skysunsand]

Homework Statement

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s^2.

How much distance is between you and the deer when you come to a stop?
and
What is the maximum speed you could have and not hit the deer?

Homework Equations

I know we have to use both of these equations-

x= x0 + v0t+ 1/2at^2
and
v=v0+at

The Attempt at a Solution

Reaction time eats up .5 of your seconds, so in the other half of a second, you are now 25 m from the deer.
So for you, x=20t+1/2at^2

And then your velocity-

v= 20+at

Then for the deer, since he's doing nothing, he would be

x=25? Because he's not doing anything else but standing 25 feet away?

And then his velocity wouldn't exist?

So then what would you do to solve this?

 

[cepheid]

Yes, the deer is stationary. You don't really need to worry about its equations of motion.

Solve for the value of x (the position of the car) at the time when v = 0. This tells you how far in front of the deer you are when you come to a stop.

I would set x to be equal to your distance from the deer (who is at x = 0, the origin), so that x₀ (the initial position, at t = 0) is +25 m. Don't forget that the formula is then:

x = x₀ + v₀t + (1/2)at²

If you take "forward" to be the negative direction (since your position is getting closer to 0 as you move forward), then your initial velocity is negative and your acceleration is positive (since it is rearward).

You're free, of course, to choose the exact opposite sign convention (in which case your initial position is -25 m, your initial velocity is positive, and your acceleration is negative). Either way will work.

I put my tip on how to actually solve the problem in italics. The rest is just helpful additional information.

 

[PeterO]

Homework Statement

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s^2.

How much distance is between you and the deer when you come to a stop?
and
What is the maximum speed you could have and not hit the deer?

Homework Equations

I know we have to use both of these equations-

x= x0 + v0t+ 1/2at^2
and
v=v0+at

The Attempt at a Solution

Reaction time eats up .5 of your seconds, so in the other half of a second, you are now 25 m from the deer.
So for you, x=20t+1/2at^2

And then your velocity-

v= 20+at

Then for the deer, since he's doing nothing, he would be

x=25? Because he's not doing anything else but standing 25 feet away?

And then his velocity wouldn't exist?

So then what would you do to solve this?

I copy here a line from your solution:

"Reaction time eats up .5 of your seconds, so in the other half of a second, you are now 25 m from the deer."

Now the question

"How much distance is between you and the deer when you come to a stop?
and
What is the maximum speed you could have and not hit the deer?"

Notice how the question was concerned about distances and speeds, while your first thought was about times??

I think you should be using formulae which yield displacement and velocity rather than time.

 

[Lobezno]

You're right, reaction time does mean constant velocity, which as you say, means that effectively deceleration doesn't start until 25m away. Then the problem starts:
Distance to Deer: 25m
Deceleration: -10 m s^-2
Initial Velocity: 20 m s

Using those equations, you're going to need to find how much time it will take to reach 0 velocity. So, equal the first equation to zero and you'll get a time value:
v=u+at
v=0=20+(-10t)

Insert that value into the first equation and you'll get how far the car has traveled in t seconds. If you use the sign convention suggested by the user above, I would get slightly confused, but it's completely up to you. Just note that if you take the signs to be the ones I said, the distance traveled will be something positive, so you'll have to subtract it from 25 metres, because that's how far away the deer is. You may well find, that if the distance is more than 25, you get a negative answer. Not to worry, that just means the imaginary deer is dead, because you drove straight past it!

(Actually, the equations would change once you hit the deer, since your deceleration will be larger, but not to worry about that! If you ignore the force exerted by hitting the deer, I get that you've gone 5 metres beyond the deer.)

 

[rwisz]

I would approach this problem by first defining the variables and units involved. In this case, we have distance (x) in meters, initial velocity (v0) in meters per second, acceleration (a) in meters per second squared, and time (t) in seconds. I would also make sure to convert all units to a consistent system, such as SI units.

Next, I would analyze the situation and identify the relevant equations that can help us solve the problem. In this case, we can use the equations of motion:

x = x0 + v0t + 1/2at^2
v = v0 + at

Where x0 is the initial position, which we can assume to be 0 in this case since we are starting from a stop, and v is the final velocity, which we can assume to be 0 since we are coming to a stop.

We are given the initial velocity (v0 = 20 m/s), the maximum deceleration (a = -10 m/s^2), and the reaction time (t = 0.50 s). We can use these values to solve for the distance (x) between the car and the deer when the car comes to a stop:

x = 0 + (20 m/s)(0.50 s) + 1/2(-10 m/s^2)(0.50 s)^2
x = 10 m + (-1.25 m)
x = 8.75 m

Therefore, when the car comes to a stop, it will be 8.75 meters away from the deer.

To answer the second question, we can use the same equation for distance and solve for the initial velocity (v0) that would result in the car coming to a stop exactly at the position of the deer (x = 35 m):

35 m = 0 + v0t + 1/2(-10 m/s^2)t^2
35 m = v0(0.50 s) + (-2.50 m/s^2)(0.50 s)^2
35 m = v0(0.50 s) + (-1.25 m)
v0 = 35 m/0.50 s + 1.25 m
v0 = 71.25 m/s

Therefore, the maximum speed the car could have and not hit the deer is 71.25 m/s