EK Phys CH1 Q10

[ridethecliche]

Alright, for some reason I keep getting the wrong answer. Can someone explain how they did this because I feel like I'm missing something really basic here...

I just take the area under the triangle because it's asking for distance, so that's:

.5x40x10=200.

The answer is 150. The question asks for distance, do they mean to ask for displacement?

Thanks.

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[ridethecliche]

This is for the in class questions and not the 30 min exam passages.

Thanks!

 

[ilovemcat]

Alright, for some reason I keep getting the wrong answer. Can someone explain how they did this because I feel like I'm missing something really basic here...

I just take the area under the triangle because it's asking for distance, so that's:

.5x40x10=200.

The answer is 150. The question asks for distance, do they mean to ask for displacement?

Thanks.

I'm not sure I really understand where you got your numbers from but the question is asking for total distance (not total displacement) from 0 seconds to 10 seconds:

Therefore we just add up the individual areas under each triangle using this equation: (1/2)base x height

(1/2)(5)(20) + (1/2)(5)(20) = Total Area
100 meters = Total Area

 

[ilovemcat]

Alright, for some reason I keep getting the wrong answer. Can someone explain how they did this because I feel like I'm missing something really basic here...

I just take the area under the triangle because it's asking for distance, so that's:

.5x40x10=200.

The answer is 150. The question asks for distance, do they mean to ask for displacement?

Thanks.

I think I understand what you did. You took the total time as the base, and the total height difference as the height, then plugged it into the equation for area under a triangle. If that's what you did - you should see why you're wrong. What you basically did was find the TOTAL distance from 0 to 15 seconds, which is not what the question asked for. Had they asked that, then the answer would of been 150 meters. Either way, I'd be very careful about approaching the problem that way. You should take it step-by-step and find the area for each individual triangle.

 

[ridethecliche]

I think I understand what you did. You took the total time as the base, and the total height difference as the height, then plugged it into the equation for area under a triangle. If that's what you did - you should see why you're wrong. What you basically did was find the TOTAL distance from 0 to 15 seconds, which is not what the question asked for. Had they asked that, then the answer would of been 150 meters. Either way, I'd be very careful about approaching the problem that way. You should take it step-by-step and find the area for each individual triangle.

I did it from 0-10 seconds. The base is 10 and the height is 40 (-20 to 20 v).

.5 (bh)=.5 (10x40)= 200.

I didn't do t=15, I did b=10 (0-10 sec). So basically i did the area under the curve from 0-10 seconds.

Why is this wrong? I thought it was the derivative (area under curve).

If I broke it into two chunks, 0-5 and then 5-10 and did it the ek way using the averages, then I would get.

.5|10x5| + .5 |-10x5| which is correct (100), but I'm not sure why area under the curve doesn't work.

 

[ilovemcat]

I did it from 0-10 seconds. The base is 10 and the height is 40 (-20 to 20 v).

.5 (bh)=.5 (10x40)= 200.

I didn't do t=15, I did b=10 (0-10 sec). So basically i did the area under the curve from 0-10 seconds.

Why is this wrong? I thought it was the derivative (area under curve).

If I broke it into two chunks, 0-5 and then 5-10 and did it the ek way using the averages, then I would get.

.5|10x5| + .5 |-10x5| which is correct (100), but I'm not sure why area under the curve doesn't work.

The distance (area) under a velocity vs. time graph IS the derivative. You're just using the incorrect values. Look again at the graph between time 0 to time 5. Everything below that isn't part of the data collected. It's just empty unplotted space. And basically, by using 40 for the height, you're including this unplotted space into the total area which is why your answer is incorrect and larger than its suppose to be. You need to find the areas under each triangle individually (break them up if you have to), then add them together to find the total distance. If the area is a rectange, then just multiple the length times the width. Just make sure you're taking the area under the plotted data and not what's around it.

 

[ridethecliche]

The distance (area) under a velocity vs. time graph IS the derivative. You're just using the incorrect values. Look again at the graph between time 0 to time 5. Everything below that isn't part of the data collected. It's just empty unplotted space. And basically, by using 40 for the height, you're including this unplotted space into the total area which is why your answer is incorrect and larger than its suppose to be. You need to find the areas under each triangle individually (break them up if you have to), then add them together to find the total distance. If the area is a rectange, then just multiple the length times the width. Just make sure you're taking the area under the plotted data and not what's around it.

Haha thanks, I realized that last night when I was looking at it googly eyed.

That square chunk of 'nothingness' was screwing me up.

Thanks