"How far did the car move while braking?"

[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).

Homework Equations

:
[/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.

 

[RPinPA]

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?

 

[kuruman]

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.

 

[Austin Gibson]

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 <------ ?

 

[kuruman]

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

 

[RPinPA]

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.

 

[kuruman]

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

Yes you can.

 

[Austin Gibson]

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?

 

[kuruman]

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).

 

[haruspex]

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.

 

[Austin Gibson]

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.

 

[Austin Gibson]

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.

 

[kuruman]

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]

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.

 

[kuruman]

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 172^o and the angle between the final velocity and the acceleration as 166^o. 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.