prajeetboi
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Last Question of my HW and im stuck pls help asap its due tomorrow

In an orienteering competition, a competitor moves in a straight line past three checkpoints, P, Q and R, where PQ = 2.4 KM and QR = 11.5 KM. The competitor is modelled as a particle moving with constant acceleration. She takes 1 hour to travel from P to Q and 1.5 hours to travel from Q to R. Find:

a) the acceleration of the competitor

b) her speed at the instant she passes P
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mqb2766
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(Original post by MikeOxlong69)
Last Question of my HW and im stuck pls help asap its due tomorrow

In an orienteering competition, a competitor moves in a straight line past three checkpoints, P, Q and R, where PQ = 2.4 KM and QR = 11.5 KM. The competitor is modelled as a particle moving with constant acceleration. She takes 1 hour to travel from P to Q and 1.5 hours to travel from Q to R. Find:

a) the acceleration of the competitor

b) her speed at the instant she passes
Which suvat equation(s) might be useful? What do you know/don't know?
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prajeetboi
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i know 5 suvat equation which are as follows:

u = v - at
2s/(u+v) = t
v^2 = U^2 + 2as
s = Vt - (1/2)at^2
S = ut + (1/2)at^2

i cant figure out which one to use. i can only derive 2 pieces of information, and i can't figure out the third to solve for acceleration.
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mqb2766
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(Original post by MikeOxlong69)
i know 5 suvat equation which are as follows:

u = v - at
2s/(u+v) = t
v^2 = U^2 + 2as
s = Vt - (1/2)at^2
S = ut + (1/2)at^2

i cant figure out which one to use. i can only derive 2 pieces of information, and i can't figure out the third to solve for acceleration.
Agreed you have two bits of information, s and t and you want to solve for a.
However, you have two sets of such information. Can you think how you can use two (possible the same equation with different s and t) to eliminate the need for the third bit of information.

Id concentrate on the last equation and think about P->Q and P->R.
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boulderingislife
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(Original post by MikeOxlong69)
Last Question of my HW and im stuck pls help asap its due tomorrow

In an orienteering competition, a competitor moves in a straight line past three checkpoints, P, Q and R, where PQ = 2.4 KM and QR = 11.5 KM. The competitor is modelled as a particle moving with constant acceleration. She takes 1 hour to travel from P to Q and 1.5 hours to travel from Q to R. Find:

a) the acceleration of the competitor

b) her speed at the instant she passes B
Looks pretty simple.
S = 2.4+11.5
T=2.5hrs
A=?
U=0
What eqtn can you use?
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mqb2766
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(Original post by boulderingislife)
Looks pretty simple.
S = 2.4+11.5
T=2.5hrs
A=?
U=0
What eqtn can you use?
It doesn't say u=0 anywhere?
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boulderingislife
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(Original post by mqb2766)
It doesn't say u=0 anywhere?
You’re right. I just (probably falsely) assumed the runner started from rest.
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prajeetboi
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oh i see now - its simultaneous equations
i got these equations:

2.4 = u+ (1/2)a
13.9 = 2.5u + 3.125a

solved this for:
u = 0.293 ms^-1
a = 4.2 ms^-2

and then for b:
isnt it just 0.293 ms^-1??
because we worked that out in part a no??

let me know if this is right
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DFranklin
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(Original post by MikeOxlong69)
and then for b:
isnt it just 0.293 ms^-1??
because we worked that out in part a no??

let me know if this is right
We don't know what point is actually meant for (b): you wrote B but that point is never mentioned or defined in the rest of the question.
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boulderingislife
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(Original post by MikeOxlong69)
oh i see now - its simultaneous equations
i got these equations:

2.4 = u+ (1/2)a
13.9 = 2.5u + 3.125a

solved this for:
u = 0.293 ms^-1
a = 4.2 ms^-2

and then for b:
isnt it just 0.293 ms^-1??
because we worked that out in part a no??

let me know if this is right
Units need converting.
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prajeetboi
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(Original post by DFranklin)
We don't know what point is actually meant fthor (b): you wrote B but that point is never mentioned or defined in the rest of the question.
thanks for spotting that, i'll change the original post now

its meant to say P
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mqb2766
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(Original post by MikeOxlong69)
oh i see now - its simultaneous equations
i got these equations:

2.4 = u+ (1/2)a
13.9 = 2.5u + 3.125a

solved this for:
u = 0.293 ms^-1
a = 4.2 ms^-2

and then for b:
isnt it just 0.293 ms^-1??
because we worked that out in part a no??

let me know if this is right
Not checked the vslues , but right approach. Sub back in if you want to check.
In a) you could just eliminate u and solve for a.
Then in b), you could find u, assuming that's the question asked - passing through P?
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boulderingislife
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(Original post by MikeOxlong69)
thanks for spotting that, i'll change the original post now

its meant to say P
You are right with the simultaneous eqtns but use km/hr or m/s, not both. Your answer looks like it should be in kh/h?
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prajeetboi
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(Original post by boulderingislife)
Units need converting.
omg im so dumb lol
when u solve the equations u get the following:

u = 0.293 kmh^-1
a = 4.2 kmh^-2

which when converted to SI units gives:

u = 0.293*1000*(3600)^-1 = 0.0814 ms^-1
a = 4.2*1000*(3600)^-2 = 3.24*10^-4 ms^-2

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prajeetboi
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(Original post by MikeOxlong69)
omg im so dumb lol
when u solve the equations u get the following:

u = 0.293 kmh^-1
a = 4.2 kmh^-2

which when converted to SI units gives:

u = 0.293*1000*(3600)^-1 = 0.0814 ms^-1
a = 4.2*1000*(3600)^-2 = 3.24*10^-4 ms^-2

and then i guess for part b the answer is 0.0814 ms^-1
right????

boulderingislife
mqb2766
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boulderingislife
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(Original post by MikeOxlong69)
oh i see now - its simultaneous equations
i got these equations:

2.4 = u+ (1/2)a
13.9 = 2.5u + 3.125a

solved this for:
u = 0.293 ms^-1
a = 4.2 ms^-2

and then for b:
isnt it just 0.293 ms^-1??
because we worked that out in part a no??

let me know if this is right
Above is m/s. Km/hr is fine as long as you use the correct units 😜
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