mechanics question Suvat equations

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lavely
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For the fourth marking point of part b, the mark scheme used speed=distance/time rather than Suvat. I used the Suvat equation s=1/2(u+v)t and got the wrong answer because of this. I.e. s=1/2(1.83+0)t and then rearranged it to get t2 as 1.2. Can someone explain why they have done so and why I am wrong? thank you
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mqb2766
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(Original post by lavely)
For the fourth marking point of part b, the mark scheme used speed=distance/time rather than Suvat. I used the Suvat equation s=1/2(u+v)t and got the wrong answer because of this. I.e. s=1/2(1.83+0)t and then rearranged it to get t2 as 1.2. Can someone explain why they have done so and why I am wrong? thank you
For that "suvat" phase, acceleration is zero so u=v.
Q has hit the floor, P is travelling with constant velocity.
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lavely
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this makes so much sense, thank you! So I could have used the equation I used but make the U and V both 1.83, which would give the same results as using speed= distance / time
(Original post by mqb2766)
For that "suvat" phase, acceleration is zero so u=v.
Q has hit the floor, P is travelling with constant velocity.
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mqb2766
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(Original post by lavely)
this makes so much sense, thank you! So I could have used the equation I used but make the U and V both 1.83, which would give the same results as using speed= distance / time
Of course, the average of the 1.83 and 1.83 is indeed 1.83. Forgetting about the signs (scalar or vector) the y5? speed-distance-time triangle is just suvat with a=0 and all the 5 equations are still valid with this assumptions.
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lavely
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thank you I really appreciate your help
(Original post by mqb2766)
Of course, the average of the 1.83 and 1.83 is indeed 1.83. Forgetting about the signs (scalar or vector) the y5? speed-distance-time triangle is just suvat with a=0 and all the 5 equations are still valid with this assumptions.
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