# Maths/Physics Help!

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#1
H

Guys I’m being really dumb today and I can’t figure this question out, help me please
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#2
For distance I put 32m initially but that was wrong, and I put -2 and -8 for the displacement but they weren’t right
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4 weeks ago
#3
What’s your working for those questions?
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#4
(Original post by 0ptics)
What’s your working for those questions?
To find the distance, I tried to find the area, so I did 8x8/2 and I got 32m,

For the displacement I have no working out tbh
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4 weeks ago
#5
Distance is always the area under a velocity time graph. So distance= sum of the area of the two triangles = (6x6)/2 + 2x2/2 = which is 8m

Displacement is essentially distance but with a direction, so since it is moving backwards (negative velocity) from 6-8 seconds, then it will be the area of the big triangle, minus the area of the small triangle (from 6-8s). This would be 6- (2x2)/2= 4m
Last edited by chinjyanson; 4 weeks ago
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#6
(Original post by chinjyanson)
Distance is always the area under a velocity time graph. So distance= sum of the area of the two triangles = (6x6)/2 + 2x2/2 = which is 8m

Displacement is essentially distance but with a direction, so since it is moving backwards (negative velocity) from 6-8 seconds, then it will be the area of the big triangle, minus the area of the small triangle (from 6-8s). This would be 6- (2x2)/2= 4m
Thank you so much, I understand all of this, I just can’t see the two triangles, I can only see one big one
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#7
(Original post by chinjyanson)
Distance is always the area under a velocity time graph. So distance= sum of the area of the two triangles = (6x6)/2 + 2x2/2 = which is 8m

Displacement is essentially distance but with a direction, so since it is moving backwards (negative velocity) from 6-8 seconds, then it will be the area of the big triangle, minus the area of the small triangle (from 6-8s). This would be 6- (2x2)/2= 4m
Also how is the distance 8m because 6x6/2 is equal to 18 and 2x2/2 is 2m so wouldn’t it be 20m
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#8
(Original post by chinjyanson)
Distance is always the area under a velocity time graph. So distance= sum of the area of the two triangles = (6x6)/2 + 2x2/2 = which is 8m

Displacement is essentially distance but with a direction, so since it is moving backwards (negative velocity) from 6-8 seconds, then it will be the area of the big triangle, minus the area of the small triangle (from 6-8s). This would be 6- (2x2)/2= 4m
Omg I see the two triangles now tysm
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4 weeks ago
#9
(Original post by zainabxxx)
Also how is the distance 8m because 6x6/2 is equal to 18 and 2x2/2 is 2m so wouldn’t it be 20m
Oh yes my bad it is.
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#10
(Original post by chinjyanson)
Oh yes my bad it is.
The displacement is not 4m
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4 weeks ago
#11
(Original post by zainabxxx)
H

Guys I’m being really dumb today and I can’t figure this question out, help me please
What is the average speed of the particle?
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4 weeks ago
#12
(Original post by chinjyanson)
So distance= sum of the area of the two triangles = (6x6)/2 + 2x2/2 = which is 8m
18+2 isn't 8.
(Original post by chinjyanson)
Displacement is essentially distance but with a direction, so since it is moving backwards (negative velocity) from 6-8 seconds, then it will be the area of the big triangle, minus the area of the small triangle (from 6-8s). This would be 6- (2x2)/2= 4m
I agree with the logic, but the equation is wrong.
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4 weeks ago
#13
(Original post by zainabxxx)
Also how is the distance 8m because 6x6/2 is equal to 18 and 2x2/2 is 2m so wouldn’t it be 20m
Yes.

For the displacement, an easy way is average speed * time.
Last edited by RogerOxon; 4 weeks ago
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#14
(Original post by RogerOxon)
Yes.

For the displacement, an easy way is average speed * time.
That would give you the distance, I think you mean average velocity!
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4 weeks ago
#15
(Original post by chinjyanson)
Distance is always the area under a velocity time graph.
In future please read the forum rules before posting - we do not give solutions
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4 weeks ago
#16
(Original post by zainabxxx)
That would give you the distance, I think you mean average velocity!
Perhaps, depending on how you define terms.

Both average speed and velocity would be useful here.
Last edited by RogerOxon; 4 weeks ago
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4 weeks ago
#17
(Original post by zainabxxx)
That would give you the distance, I think you mean average velocity!
No, velocity is a vector quantity meaning that it has a direction. This means that velocity can have a negative direction. Apply this to the formula and you get a negative distance, which is wrong as distance is scalar, hence it can’t have a negative value.

Average speed is the correct term here.
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