# Maths/Physics Help!

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Guys I’m being really dumb today and I can’t figure this question out, help me please

Guys I’m being really dumb today and I can’t figure this question out, help me please

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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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(Original post by

What’s your working for those questions?

**0ptics**)What’s your working for those questions?

For the displacement I have no working out tbh

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#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

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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(Original post by

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

**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

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**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

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**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

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#9

(Original post by

Also how is the distance 8m because 6x6/2 is equal to 18 and 2x2/2 is 2m so wouldn’t it be 20m

**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

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(Original post by

Oh yes my bad it is.

**chinjyanson**)Oh yes my bad it is.

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#11

(Original post by

H

Guys I’m being really dumb today and I can’t figure this question out, help me please

**zainabxxx**)H

Guys I’m being really dumb today and I can’t figure this question out, help me please

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#12

(Original post by

So distance= sum of the area of the two triangles = (6x6)/2 + 2x2/2 = which is 8m

**chinjyanson**)So distance= sum of the area of the two triangles = (6x6)/2 + 2x2/2 = which is 8m

(Original post by

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

**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

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#13

**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

For the displacement, an easy way is average speed * time.

Last edited by RogerOxon; 4 weeks ago

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#15

(Original post by

Distance is always the area under a velocity time graph.

**chinjyanson**)Distance is always the area under a velocity time graph.

**do not**give solutions

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#16

(Original post by

That would give you the distance, I think you mean average velocity!

**zainabxxx**)That would give you the distance, I think you mean average velocity!

Both average speed and velocity would be useful here.

Last edited by RogerOxon; 4 weeks ago

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#17

(Original post by

That would give you the distance, I think you mean average velocity!

**zainabxxx**)That would give you the distance, I think you mean average velocity!

Average speed is the correct term here.

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