At what time will bike and car be side by side again - Mechanics

[chwala]

Homework Statement

See attached

Relevant Equations

Kinematics

Find the question below;

For part (a), i used the graph to find $t=22$

For part (b), i considered the points;

$(8,20)$, $(13.333,20)$ and $(0,0)$

it follows that,
Area=$\sqrt {25.454(25.454-21.54)(25.454-24.036)(25.454-5.333)}$
$\sqrt {2842.58}$=$53.31$

There may be other better methods, my textbook only provides the final solution with no working...these are how the solutions look like (from textbook).

 

[mjc123]

For part (b), i considered the points;
(8,20), (13.333,20) and (0,0) it follows that,
Area=√25.454(25.454−21.54)(25.454−24.036)(25.454−5.333)
√2842.58=53.31

What is your reasoning, and where do you get those numbers from?

 

[chwala]

What is your reasoning, and where do you get those numbers from?

The greatest distance between the bike and the car is bound by the given coordinates...and one can find area of the region bound by the three co ordinates...one can get the co ordinates directly from the graph...let me know if this is still not clear...

 

[mjc123]

I still don't see where you get the numbers 25.454, 21.54 etc., or the big square root expression. The area of the triangle is simply 1/2*base* height = 1/2*5.333*20 = 53.33.

 

[chwala]

I still don't see where you get the numbers 25.454, 21.54 etc., or the big square root expression. The area of the triangle is simply 1/2*base* height = 1/2*5.333*20 = 53.33.

OK, i made use of
1. L= $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$
2. A=$\sqrt {s(s-a)(s-b)(s-c)}$

 

[chwala]

I still don't see where you get the numbers 25.454, 21.54 etc., or the big square root expression. The area of the triangle is simply 1/2*base* height = 1/2*5.333*20 = 53.33.

We are looking at the region where the motorbike is in front of the car...

 

[mjc123]

OK, just seems a rather complicated way of doing the calculation...

 

[Merlin3189]

We are looking at the region where the motorbike is in front of the car...

Since you have a v-t graph, I don't see how you can tell by inspection the region where this is true.

I think you mean, your coordinates span the time when the bike is going faster than the car. The bike remains in front for a further 8.667 seconds, while the now faster car is catching up.

Relevant Equations:: Kinematics

Might help if you mentioned the slightly unusual Heron's formula.
Especially when most people would use A= bh/2 for such a simple triangle with b and h known, so no need to use Pythagoras to calculate the other two sides of the triangle.

The greatest distance between the bike and the car is bound by the given coordinates...

(8,20), (13.333,20) and (0,0)

True, but if you're asking people to check your work, IMO some explanation would be sensible. Until I drew the graph myself, it was not even clear to me where your coordinates came from: the coordinates on the graph were not in their correct positions and the vertical axis had no scale. (I did wonder why your vertical axis went above 30m/sec, why the horizontal axis went so high, why both axes went negative, what the point of labeling (12,30) and (22,33) was? These all seemed to obscure rather than help.)

I'm not wanting to pick holes in your work just for the sake of it. If you want others to tell you if there is a better method, they need to know what method you used! You say you already have the answer, so just telling you that it is correct is no help. But largely, all YOU gave was your answer, not your method/ thinking. (And I am thinking of your other, even more mystifying thread, as well.)

 

[chwala]

Since you have a v-t graph, I don't see how you can tell by inspection the region where this is true.

I think you mean, your coordinates span the time when the bike is going faster than the car. The bike remains in front for a further 8.667 seconds, while the now faster car is catching up.Might help if you mentioned the slightly unusual Heron's formula.
Especially when most people would use A= bh/2 for such a simple triangle with b and h known, so no need to use Pythagoras to calculate the other two sides of the triangle.True, but if you're asking people to check your work, IMO some explanation would be sensible. Until I drew the graph myself, it was not even clear to me where your coordinates came from: the coordinates on the graph were not in their correct positions and the vertical axis had no scale. (I did wonder why your vertical axis went above 30m/sec, why the horizontal axis went so high, why both axes went negative, what the point of labeling (12,30) and (22,33) was? These all seemed to obscure rather than help.)

I'm not wanting to pick holes in your work just for the sake of it. If you want others to tell you if there is a better method, they need to know what method you used! You say you already have the answer, so just telling you that it is correct is no help. But largely, all YOU gave was your answer, not your method/ thinking. (And I am thinking of your other, even more mystifying thread, as well.)

I have noted your remarks, i assumed that the reader would probably be conversant with the math concepts being used, i will be more clearer in the future and also make sure to state which concept or math rule is used on a given problem. Cheers...

 

[chwala]

Since you have a v-t graph, I don't see how you can tell by inspection the region where this is true.

I think you mean, your coordinates span the time when the bike is going faster than the car. The bike remains in front for a further 8.667 seconds, while the now faster car is catching up.Might help if you mentioned the slightly unusual Heron's formula.
Especially when most people would use A= bh/2 for such a simple triangle with b and h known, so no need to use Pythagoras to calculate the other two sides of the triangle.True, but if you're asking people to check your work, IMO some explanation would be sensible. Until I drew the graph myself, it was not even clear to me where your coordinates came from: the coordinates on the graph were not in their correct positions and the vertical axis had no scale. (I did wonder why your vertical axis went above 30m/sec, why the horizontal axis went so high, why both axes went negative, what the point of labeling (12,30) and (22,33) was? These all seemed to obscure rather than help.)

I'm not wanting to pick holes in your work just for the sake of it. If you want others to tell you if there is a better method, they need to know what method you used! You say you already have the answer, so just telling you that it is correct is no help. But largely, all YOU gave was your answer, not your method/ thinking. (And I am thinking of your other, even more mystifying thread, as well.)

"Since you have a v-t graph, I don't see how you can tell by inspection the region where this is true."

Using the three co ordinates $(8,20)$, $(13.333,20)$ and $(0,0)$ would do, i guess this should be clear enough..

Might help if you mentioned the slightly unusual Heron's formula.
Especially when most people would use 'A= bh/2 for such a simple triangle with b and h known', so no need to use Pythagoras to calculate the other two sides of the triangle.

Unless i am missing something, i do not think one can use the right angle triangle directly to find the required distance. Not unless, you want to find area of rectangle...then subtract area of two triangles...

 

[mjc123]

You do know that A = bh/2 applies to all triangles, not just right-angle ones, don't you?

For an arbitrary triangle, the vertical height may not be obvious, but in this case it is (and you need to use it to calculate two of the sides by Pythagoras).