# "How far did the car move while braking?"

• Austin Gibson
In summary, a car traveling down a neighborhood street must suddenly apply its brakes to avoid hitting a child who runs into the path of the car to retrieve a ball. The car slows down from 14.5î + 4.50 m/s to 8.25î + 3.50 m/s in 1.06 seconds. To find the distance the car traveled while braking, we can use the equation (v_i + v_f)/2 = D/t. After solving for the displacement in each direction separately, we can combine them to find the final distance traveled, which is 12.78 meters in vector form (12.0575i + 4.24k). It is important to note that the question
Austin Gibson

## Homework Statement

:[/B]

A ball rolls onto the path of your car as you drive down a quiet neighborhood street. To avoid hitting the child that runs to retrieve the ball, you apply your brakes for 1.06 s. The car slows down from 14.5î + 4.50
m/s to 8.25î + 3.50
m/s.

How far did the car move while braking?

Picture for verification: https://gyazo.com/d3851efe6c4dc5eb7164ef5e72afaf0f

NOTE: I already solved the first question (a) in the picture. I'm stuck on (b).

:
[/B]
(vi +vf)/2 = D/t

## The Attempt at a Solution

[([ (14.5i + 8.25k) + (4.50i + 3.50k) ] * 1.06)] / 2 = 12.0575i + 4.24k.
After applying the Pythagorean theorem, my result was 12.78 meters. That answer was rejected.

#### Attachments

6.5 KB · Views: 417
Last edited by a moderator:
I can't see the connection between your "Relevant equations" and your "attempt at a solution". In particular I can't see how all the numbers in your solution are connected to symbols in the equation.

Can you show symbolically what you did with ##(v_i + v_f)/2 = D/t## before putting in the numbers?

You need to find how far the car travels in each direction separately and then combine the displacements to find the distance.

On edit: I got the same distance, 12.78 m, but note that the expected answer is a vector ##\vec d## which you have to enter in unit vector notation.

Last edited:
Austin Gibson
kuruman said:
You need to find how far the car travels in each direction separately and then combine the displacements to find the distance.

[(14.5i + 8.25i) * 1.06] / 2 = 12.0575
[(4.5k + 3.5k) * 1.06] / 2 = 4.24
12.0575 + 4.24 = 16.2975 m <------ ?

Please see the edit in my previous post. You need to enter the displacement as a vector.

Austin Gibson
kuruman said:
You need to find how far the car travels in each direction separately and then combine the displacements to find the distance.

I wondered about that, but I think it works as a vector equation. Let's see...

$$x_f = x_i + v_{ix} t + 0.5 a_x t^2 \\ y_f = y_i + v_{iy} t + 0.5 a_y t^2$$
becomes
$$\vec D = \vec d_f - \vec d_i \\ = \vec v_i t + 0.5 \vec a t^2 \\ = \vec v_i t + 0.5 \frac {\vec v_f - \vec v_i}{t} t^2 \\ = \vec v_i t + 0.5 \left(\vec v_f - \vec v_i \right) t = 0.5\left(\vec v_f + \vec v_i \right) t$$

So actually you can interpret the displacement equation as a vector equation.

Austin Gibson
RPinPA said:
So actually you can interpret the displacement equation as a vector equation.
Yes you can.

Austin Gibson
kuruman said:
Please see the edit in my previous post. You need to enter the displacement as a vector.
Therefore, 16.2975 m in vector form is 12.0575i + 4.24k?

Austin Gibson said:
Therefore, 16.2975 m in vector form is 12.0575i + 4.24k?
Yes, enter that and see what happens. And it is not 16.2975 m, it is 12.78 m as entered before (not that it matters).

Austin Gibson
kuruman said:
You need to find how far the car travels in each direction separately and then combine the displacements to find the distance.

On edit: I got the same distance, but note that the expected answer is a vector ##\vec d## which you have to enter in unit vector notation.
Isn't the question flawed? The car has changed direction. If it is skidding the direction will not have changed, or at least not in a predictable way. If it is not skidding then the applied force has changed direction, so we do not know the path taken.
Even if there were no change of direction, we are not told the applied force is constant. If it decelerates very quickly at first and more slowly later it will cover less distance than in the converse case.
Taking the applied force to be constant despite the change of direction would be quite unrealistic.

kuruman said:
Yes, enter that and see what happens. And it is not 16.2975 m, it 12.78 m as entered before (not that it matters).

Your guidance is appreciated. I calculated a magnitude of 12.78 yesterday (it rejected that), but, as you mentioned, the question demands it in vector notation. If a line is above a variable (in this case "d"), is that to signify vector notation? If that's true, I was oblivious of that until now because the website would prompt me to type the answer in vector form. Therefore, I solved these in the past without recognizing that.

#### Attachments

• b3310a95a41c89057debb8fa50979669.png
1.7 KB · Views: 301
The first answer was correct (12.0575i + 4.24k), but I was entering the magnitude rather than the answer in vector notation assuming that's what they were requesting.

Austin Gibson said:
If a line is above a variable (in this case "d"), is that to signify vector notation?
Yes, it means a vector is expected. Also, it's not a line, it's supposed to be an arrow indicating a vector. However, @haruspex raised a point that I have to think about.

Austin Gibson
kuruman said:
Yes, it means a vector is expected. Also, it's not a line, it's supposed to be an arrow indicating a vector. However, @haruspex raised a point that I have to think about.
The clarification is appreciated.

haruspex said:
Isn't the question flawed? The car has changed direction. If it is skidding the direction will not have changed, or at least not in a predictable way. If it is not skidding then the applied force has changed direction, so we do not know the path taken.
Even if there were no change of direction, we are not told the applied force is constant. If it decelerates very quickly at first and more slowly later it will cover less distance than in the converse case.
Taking the applied force to be constant despite the change of direction would be quite unrealistic.
I agree. The problem cannot be solved unless one assumes that the force and hence the acceleration are constant. OP got the "right" answer under this assumption. However, this results in acceleration ##\vec a=(-5.90~\hat i-0.943~\hat k)m/s^2##. It can be used to find the angle between the initial velocity and the acceleration as 172o and the angle between the final velocity and the acceleration as 166o. One may consider that this happens because (a) the centripetal component is changing; (b) the tangential component is changing; (c) both components are changing. The choices follow from the assumption that the acceleration is constant but all of them are inconsistent with that assumption. Therefore the assumption is incorrect.

## 1. How is the distance the car moves while braking calculated?

The distance the car moves while braking is calculated by using the formula d = vi * t + (1/2) * a * t^2, where d is the distance, vi is the initial velocity, a is the deceleration rate, and t is the time taken to stop.

## 2. How does the weight of the car affect the distance it moves while braking?

The weight of the car can affect the distance it moves while braking, as a heavier car will require more force to stop and therefore may take longer to come to a complete stop.

## 3. What factors other than weight can affect the distance the car moves while braking?

Other factors that can affect the distance the car moves while braking include the type and condition of the brakes, the road surface and conditions, and the speed of the car.

## 4. Can the distance a car moves while braking be shortened?

Yes, the distance a car moves while braking can be shortened by maintaining proper brake maintenance, driving at a safe speed, and leaving enough space between vehicles to allow for safe stopping distances.

## 5. Is it possible for a car to move farther while braking than while driving?

No, it is not possible for a car to move farther while braking than while driving. The car is actively being decelerated while braking, so it will always travel a shorter distance than when it is accelerating.

Replies
9
Views
1K
Replies
29
Views
3K
Replies
1
Views
5K
Replies
1
Views
10K
Replies
7
Views
1K
Replies
1
Views
1K
Replies
20
Views
11K
Replies
9
Views
2K
Replies
4
Views
10K
Replies
5
Views
15K