# Will The Cars Stop In Time? - Check My Work Please :)

• valleyldp
valleyldp
Homework Statement
STORY PROBLEM:
Sarah is speeding down the highway westward at 65 mph. She notices ahead that an accident has resulted in a pile-up in the middle of the road. She slams on her brakes 35 meters from the accident, causing her car to slow at a rate of 9.1 m/s^2. Andrew is traveling the other way on the same highway at 45 m/s eastward. He also notices the accident and begins to brake 20 meters from the accident. His car decelerates at 8.8 m/s^2. Assuming a damaged car will keep them home from work, which of the two people will be able to make it to work the next day?

STEPS TO SOLVE:
1. Draw a diagram to model the situation with relevant items labeled. Make sure you indicate which direction is east or west, and which of those you chose to be positive.

2. What are the signs for the following information as defined by your model? (Correct or not).

3. Construct a velocity versus time graph of both Sarah's and Andrew's motion in the same reference frame.

4. Write out the stopping time needed for each person indicated by the graph. (Correct sig figs and units required)

5. Use a kinematic equation to determine the distance needed for each person to stop. (Correct sig figs and units required)

6. Which personal will make it to work the next day?
Relevant Equations
Δx=v_iΔt+½a(Δt)^2
1. Model attached below.
1. Changed Sarah's 65 mph to m/s and got 29 m/s.
2. Sarah's initial position is positive. Her initial velocity is negative. Her acceleration is positive; Andrew's initial position is negative. His initial velocity is positive. His acceleration is negative.
3. Graph is attached below.
4. Graph indicates a stopping time of 3.2 seconds for Sarah and 5.1 seconds for Andrew. (Two sig figs because measurements in initial problem were two sig figs.)
5. Worked Out Equations (Δx=v_iΔt+½a(Δt)^2)
1. Sarah
1. Δx = unknown
2. a = -9.1 m/s^2
3. Initial Velocity = 29 m/s
4. Δt = 3.2 seconds
5. Answer: 46 meters
2. Andrew
1. Δx = unknown
2. a = -8.8 m/s^2
3. Initial Velocity = 45 m/s
4. Δt = 5.1 seconds
5. Answer: 120 meters
6. Neither person will make it to work the next day because the required stopping distance is more than the given distances.

#### Attachments

• Screenshot 2023-09-15 5.34.34 PM.png
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Last edited by a moderator:
I agree with your numbers and conclusions. However your plot is not worth much without labels explaining what is what.

MatinSAR
Labels for the lines? I'm a bit confused on what to label.

Label the axes with what variables including units and the lines to distinguish which of the two cars are what color. If you have plotted velocity on the vertical axis and time on the horizontal axis, then you need to explain the negative time ##t=-3.187## s which, by the way, has too many sig figs as does the other time.

It seems that you have problems with what happens when ##t =0##. Conventionally, ##t=0## is when each car starts braking. Look at the blue line. It shows that as time increases, the speed increases because the velocity is positive and moves away from zero. That cannot be because supposedly the car is braking. Both lines should start on the vertical axis and there shouod be nothing to the left of it.

By the way, the stopping distance is quickly found using the kinematic equation $$2ax=v_{\!f}^2-v_{\!0}^2.$$

MatinSAR
Given the speeds involved and the short distance to the accident before they noticed it, both drivers were guilty of careless driving!

MatinSAR
A few additional thoughts...

valleyldp said:
Sarah
1. Δx = unknown
2. a = -9.1 m/s^2
3. Initial Velocity = 29 m/s
4. Δt = 3.2 seconds
5. Answer: 46 meters

valleyldp said:
1. Andrew
1. Δx = unknown
2. a = -8.8 m/s^2
3. Initial Velocity = 45 m/s
4. Δt = 5.1 seconds
5. Answer: 120 meters
You have given both accelerations as negative. That conflicts with what you said ealier.

You have given both initial velocities as positive.. That conflicts with what you said ealier.

valleyldp said:
Your drawing shows a large gap between the 35m and 20m lengths. There should be no gap Also, the drawing doesn't clearly show the directions for east and west vectors. Add arrows,

But the main fault is with the graph, as already been explained by @kuruman in Post #4.

MatinSAR
I have fixed all the things that have been suggested. My new graph is posted below. Does it look better?

Note - I couldn't figure out how to get desmos to round the x-intercept, so I just recalculated the line equations to cross the y-axis at exactly 3.2 and 5.1. Does this work?

valleyldp said:
I have fixed all the things that have been suggested. My new graph is posted below. Does it look better?

Note - I couldn't figure out how to get desmos to round the x-intercept, so I just recalculated the line equations to cross the y-axis at exactly 3.2 and 5.1. Does this work?

I also added the constant velocity that happened before the acceleration started.

valleyldp said:
View attachment 332025

I also added the constant velocity that happened before the acceleration started.
One initial velocity is positive and the other one is negative (because the directions are opposite). But your graph shows both as positive.

One acceleration is positive and the other one is negative (do you see why?). But your graph shows both as negative. (The gradient of a v-t graph equals acceleration; the gradients of both of your lines (for t>0) are negative).

MatinSAR and Lnewqban
In my model one acceleration is positive and the other is negative, but I couldn't figure out how to make a graph that would properly include both of these, so for simplicities sake I made both of them have negative acceleration and positive velocity. Do you have any idea how I could do it the correct way?

I got it I think....would it look like this?

#### Attachments

• Screenshot 2023-09-16 12.47.52 PM.png
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Steve4Physics and Lnewqban
valleyldp said:
I got it I think....would it look like this?
Exactly. Good job!

MatinSAR
Awesome! Thank you all for your help!

Steve4Physics and Lnewqban
I have added the times (gray bars) when each driver collides with the pileup. Sarah will hit the pile up at about 30 mph and Andrew at about 90 mph. It looks like Andrew will take an early retirement from his job ##~\dots~## in the morgue.

Steve4Physics, MatinSAR and valleyldp

## What factors determine whether a car will stop in time?

The ability of a car to stop in time depends on several factors including the speed of the car, the condition of the brakes, the reaction time of the driver, the type of road surface, and weather conditions. Each of these factors can significantly affect stopping distance.

## How do you calculate the stopping distance of a car?

Stopping distance is typically calculated by adding the reaction distance (the distance a car travels during the driver's reaction time) and the braking distance (the distance it takes for the car to come to a complete stop once the brakes are applied). The formula often used is: Stopping Distance = Reaction Distance + Braking Distance.

## What is the role of reaction time in stopping a car?

Reaction time is the period between the driver perceiving a need to stop and actually applying the brakes. The average reaction time is about 1.5 seconds, and during this time, the car continues to travel at its initial speed. Faster reaction times can significantly reduce the overall stopping distance.

## How does speed affect stopping distance?

Speed has a quadratic effect on braking distance, meaning that if you double the speed of the car, the braking distance increases by a factor of four. This is because kinetic energy, which needs to be dissipated for the car to stop, is proportional to the square of the speed.

## Why is road surface condition important for stopping distance?

The condition of the road surface greatly affects the friction between the tires and the road. Wet, icy, or gravel-covered roads provide less friction, increasing the braking distance. Conversely, dry, well-maintained roads provide better friction and shorter braking distances.

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