# Velocity time graph -M1

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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
(a) Draw, in the space below, a velocity-time graph to represent the motion of the ball
during the first 4 s
0
4 years ago
#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
(a) Draw, in the space below, a velocity-time graph to represent the motion of the ball
during the first 4 s
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.
0
#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.
Its velocity becomes zero due to air resistance/
0
4 years ago
#4
(Original post by Qer)
Its velocity becomes zero due to air resistance/
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.
0
#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.

so here is graph

its start at u ,I don't understand that why it goes to -u?
0
4 years ago
#6
(Original post by Qer)

so here is graph

its start at u ,I don't understand that why it goes to -u?
Under gravity, there's only one force affecting the ball, . 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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