Wqyds
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Question 1:
Two light gates are arranged vertically, 0.8m apart. M ball is dropped some height above the gates, and takes 0.21s to pass between the gates. Calculate the speed at the upper and lower gate.

Question 2:
A car is travelling at a constant velocity of 15m/s. It passes a second car, and the second car sets of from rest with a constant acceleration of 3m/s2.
How long does it take for the second car to catch up to the first car, and at what speed it the second car travelling at at that point?

Any help would be appreciated
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hmmdali
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For question 1 you can take acceleration = 9.8m/s^2 and then use s=ut+(at^2)/2 and s=vt-(at^2)/2 to calculate the velocity at the upper and lower gates with u being the upper and v, the lower.
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hmmdali
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Question 2: I labelled the car moving at 15m/s as c1 and the other car as c2. I then equated the time at which the displacement between the 2 cars would be equal. Using the time I worked out I used v=u+at to find the speed the car would be travelling at that point.
Last edited by hmmdali; 9 months ago
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Wqyds
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(Original post by hmmdali)
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I labelled the car moving at 15m/s as c1 and the other car as c2. I then equated the time at which the displacement between the 2 cars would be equal. Using the time I worked out I used v=u+at to find the speed the car would be travelling at that point.
Thank you so much. Q1 was a lot simpler than I thought, as I was trying to calculate the initial drop height. And Q2 makes a lot of sense now. Thanks for the help.
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hmmdali
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(Original post by Anon13231)
I still don't understand question 1
(Original post by hmmdali)
For question 1 you can take acceleration = 9.8m/s^2 and then use s=ut+(at^2)/2 and s=vt-(at^2)/2 to calculate the velocity at the upper and lower gates with u being the upper and v, the lower.
You just substitute the known values into each of the equations. s=0.8 a=9.8 and t=0.21. Then rearrange to find u and v.
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Muttley79
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(Original post by hmmdali)
Question 2: I labelled the car moving at 15m/s as c1 and the other car as c2.
PLEASE note it is against the rules to post solutions
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