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Mehcanics Question

https://www.quora.com/profile/Bravewarrior/p-169718943
Here is the question. I'm stuck on part b so some help would be great!
Reply 1
Original post by pigeonwarrior
https://www.quora.com/profile/Bravewarrior/p-169718943
Here is the question. I'm stuck on part b so some help would be great!

Youve got the answer, so find the time when they have travelled the same distance then sub that into the speed expression. What dont you understand.
Original post by mqb2766
Youve got the answer, so find the time when they have travelled the same distance then sub that into the speed expression. What dont you understand.

When they are working out the distance the motorbike has travelled in the mark scheme, they did 1.5(T-15). I'm a bit confused as to how they got that
Reply 3
Original post by pigeonwarrior
When they are working out the distance the motorbike has travelled in the mark scheme, they did 1.5(T-15). I'm a bit confused as to how they got that

s = 1/2 a t^2
a = 3/2
t = (T-15)
Original post by mqb2766
s = 1/2 a t^2
a = 3/2
t = (T-15)

Thank you! ๐Ÿ™‚
Reply 5
Original post by pigeonwarrior
Thank you! ๐Ÿ™‚

Theyve written it the way they have as part a) is about drawing a velocity - time graph and the distance is the area under that. For this suvat section, it would be a triangle with base (T-15) and height 3/2(T-15) and the area is the half the product of the two. So either think of it as the usual suvat or the area of a triangle.
Original post by mqb2766
Theyve written it the way they have as part a) is about drawing a velocity - time graph and the distance is the area under that. For this suvat section, it would be a triangle with base (T-15) and height 3/2(T-15) and the area is the half the product of the two. So either think of it as the usual suvat or the area of a triangle.

That makes more sense, thank you! ๐Ÿ™‚

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