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Velocity time graph -M1
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Qer
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#1
how to draw velocity-time graph for a question like that
A small ball is projected vertically upwards from ground level with speed u m s–1. The ball
takes 4 s to return to ground level.
(a) Draw, in the space below, a velocity-time graph to represent the motion of the ball
during the first 4 s
A small ball is projected vertically upwards from ground level with speed u m s–1. The ball
takes 4 s to return to ground level.
(a) Draw, in the space below, a velocity-time graph to represent the motion of the ball
during the first 4 s
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Plagioclase
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#2
(Original post by Qer)
how to draw velocity-time graph for a question like that
A small ball is projected vertically upwards from ground level with speed u m s–1. The ball
takes 4 s to return to ground level.
(a) Draw, in the space below, a velocity-time graph to represent the motion of the ball
during the first 4 s
how to draw velocity-time graph for a question like that
A small ball is projected vertically upwards from ground level with speed u m s–1. The ball
takes 4 s to return to ground level.
(a) Draw, in the space below, a velocity-time graph to represent the motion of the ball
during the first 4 s
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Qer
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#3
(Original post by Plagioclase)
Think about what forces are going to affect the ball once it's in the air (clue - using the assumptions generally made in M1, there's only one). Then think about how that force will affect the velocity.
Think about what forces are going to affect the ball once it's in the air (clue - using the assumptions generally made in M1, there's only one). Then think about how that force will affect the velocity.
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Plagioclase
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#4
(Original post by Qer)
Its velocity becomes zero due to air resistance/
Its velocity becomes zero due to air resistance/
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Qer
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#5
(Original post by Plagioclase)
Generally in M1, you ignore air resistance unless it specifically tells you to take it into account. The only force you need to consider here is gravity.
Generally in M1, you ignore air resistance unless it specifically tells you to take it into account. The only force you need to consider here is gravity.
so here is graph
its start at u ,I don't understand that why it goes to -u?
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Plagioclase
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#6
. Through 
, this means that the ball is undergoing uniform acceleration (in the 
direction, which on a velocity-time graph is a straight line with gradient 
). You know that the end velocity is because, assuming no air resistance, no energy is lost from the system so when the ball is at ground level, the kinetic energy (and hence the speed) is equal at the start and at the end.
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