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# Hard OCR M1 SUVAT question... (Rep+) watch

1. Hi guys,

I am in real need of some help of this question!

A particle travels in straight line with uniform acceleration. The particle passes through 3 points, A, B and C lying in that order at times t=0, t=2s, t=5s. If BC=30meters, and the speed of the particle when at B is 7ms^-1 find the acceleration of the particle and its speed when at A.

Okay so I am guessing it is a simultaneous equation right?

So yeah I am pretty much stuck aha.

From AB; S=x
U=u
V=doesnt matter
A=a
t=2s

From AC S=30+x
U=7ms^-1
V= doesnt matter
A=a
T= 7s

I really have no clue what to do from here now as I can't solve it!
2. (Original post by J DOT A)
Hi guys,

I am in real need of some help of this question!

A particle travels in straight line with uniform acceleration. The particle passes through 3 points, A, B and C lying in that order at times t=0, t=2s, t=5s. If BC=30meters, and the speed of the particle when at B is 7ms^-1 find the acceleration of the particle and its speed when at A.

Okay so I am guessing it is a simultaneous equation right?

So yeah I am pretty much stuck aha.

From AB; S=x
U=u
V=doesnt matter
A=a
t=2s

From AC S=30+x
U=7ms^-1
V= doesnt matter
A=a
T= 7s

I really have no clue what to do from here now as I can't solve it!
Use s=ut+1/2at^2 for BC to get the acceleration of the particle and then use v=u+at for AB to find the initial speed (u).
3. (Original post by Gemini92)
Use s=ut+1/2at^2 for BC to get the acceleration of the particle and then use v=u+at for AB to find the initial speed (u).
Oh right so is the acceleration constant throughout process? If I use from BC I get acceleration as -3/5 ms^2. The answer is 2ms^2?
4. (Original post by J DOT A)
Oh right so is the acceleration constant throughout process? If I use from BC I get acceleration as -3/5 ms^2. The answer is 2ms^2?
Yes, the acceleration is constant, that's what's meant by uniform acceleration. I got the acceleration to be 2ms^-2. What values are you using to work out the acceleration?
5. (Original post by Gemini92)
Yes, the acceleration is constant, that's what's meant by uniform acceleration. I got the acceleration to be 2ms^-2. What values are you using to work out the acceleration?

7. U(a)=x (unknown)
V(a)=U(b)=7
BC=30
from v=u+at
7=x+2a

then

s=U(b)t+0.5at^2
30=21+4.5a
a=2

then U(a)=7-4
U(a)=3 m/s

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