What’s your working for those questions?
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
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
(edited 2 years ago)
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
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
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
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
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
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
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.
Original post by chinjyanson
Oh yes my bad it is.
The displacement is not 4m
Original post by zainabxxx
H
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
What is the average speed of the particle?
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.
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.
(edited 2 years ago)
Original post by RogerOxon
Yes.
For the displacement, an easy way is average speed * time.
For the displacement, an easy way is average speed * time.
That would give you the distance, I think you mean average velocity!
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
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.
(edited 2 years ago)
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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