Insanely difficult mechanics question!!! Watch

Rohit94994
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Here's the question; "In an orienteering competition, a competitor moves in a straight line past three checkpoints. P, Q and R, where PQ=2.4km and QR=11.5km. The competitor is modelled as a particle moving with constant acceleration. She takes 1 hour 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.



Another question "Two particles P and Q have masses 0.5kg and 0.4kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed above a horizontal floor. Both particles are held, with string taut, at a height of 2m above the floor. The particles are released from rest and in subsequent ,onion Q does not reach the pulley. When particles have been moving for 0.2s, the string breaks. Find the further time elapses until Q hits the floor."
Last edited by Rohit94994; 1 year ago
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Y12_FurtherMaths
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(Original post by ZombieTheWolf)
What I have so far for a)
Convert everything to metres and seconds.

From P to Q = 2400m/3600 seconds = 0.67m/s
From Q to R = 11500/5400 = 2.13m/s

We are not told she started from rest at any of the points, and by how the question is set out I assume the order of points is P Q R. If she is moving at a speed of 0.67m/s bettwen P and Q I assume she started with this speed. And I assume she finishes with 2.13m/s at R.
If P is her inital velocity, and R is her final and it takes her 2.5 hours then do v-u/t. 2.13-0.67/9000 = 1.6×10^-4 m/s^2


Not sure about B.
Sorry if I got it wrong!
Please don’t post full solutions on the maths forum, it is against the rules. Only hints should be offered
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ZombieTheWolf
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(Original post by Y12_FurtherMaths)
Please don’t post full solutions on the maths forum, it is against the rules. Only hints should be offered
I won't next time I had no idea you couldn't do this as it doesn't say anywhere but thank you for letting me know.
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ZombieTheWolf
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Convert everything to metres and seconds.

If they are orienteering, they will be at rest at some of the checkpoints.

Find their intial and final velocities between each of the points.
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bingbong654
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(Original post by ZombieTheWolf)
Convert everything to metres and seconds.

If they are orienteering, they will be at rest at some of the checkpoints.

Find their intial and final velocities between each of the points.
Why would they be at rest?
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Rohit94994
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(Original post by ZombieTheWolf)
I won't next time I had no idea you couldn't do this as it doesn't say anywhere but thank you for letting me know.
are you in yr13 maths?
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ZombieTheWolf
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(Original post by bingbong654)
Why would they be at rest?
If P Q and R are checkpoints, in orienteering this usually means there are signs you have to find and tick off. Either way, you have to stop and look for the sign. From personal experience, I've never witnessed anyone running past an orienteering checkpoimt without stopping at all but I maybe completely wrong.
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for_real_though
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Not sure how to do b) but I got a) as 1.24m/s2
s=13900m
u=0
a=?
t=150s

s=ut+at2/2
2S/t2=a
a=1.24ms-2 (3sf)
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ZombieTheWolf
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(Original post by Rohit94994)
are you in yr13 maths?
Year 13 physics, but I just recently had to relearn Year 12 mechanics because of personal circumstances.
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bingbong654
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(Original post by ZombieTheWolf)
If P Q and R are checkpoints, in orienteering this usually means there are signs you have to find and tick off. Either way, you have to stop and look for the sign. From personal experience, I've never witnessed anyone running past an orienteering checkpoimt without stopping at all but I maybe completely wrong.
Sure in real life you would have to stop, but this is a particle moving with constant acceleration so she can't stop even if she wanted to!
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ZombieTheWolf
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(Original post by for_real_though)
Not sure how to do b) but I got a) as 1.24m/s2
s=13900m
u=0
a=?
t=150s

s=ut+at2/2
2S/t2=a
a=1.24ms-2 (3sf)
Theres 3600 seconds in an hour though?
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Rohit94994
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(Original post by for_real_though)
Not sure how to do b) but I got a) as 1.24m/s2
s=13900m
u=0
a=?
t=150s

s=ut+at2/2
2S/t2=a
a=1.24ms-2 (3sf)
That's wrong
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for_real_though
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(Original post by ZombieTheWolf)
Theres 3600 seconds in an hour though?
Oh my...:facepalm::facepalm:I converted it to minutes...
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ZombieTheWolf
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(Original post by for_real_though)
Oh my...:facepalm::facepalm:I converted it to minutes...
No worries!! I always makes the stupidest mistakes honestly 😂
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bingbong654
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a=2s/t^2
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Justvisited
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(Original post by Rohit94994)
That's wrong
If you know the answers already, why are you asking?

Zombie Wolf got it wrong by getting confused/misunderstanding the question when they say "If she is moving at a speed of 0.67m/s bettwen P and Q I assume she started with this speed." The whole point is that the runner is always accelerating, so the 0.67 is the average speed over PQ, i.e. the speed at P is less than this.

We have two unknown to solve for, a and u (the speed at P), so we need two simultaneous equations with them in.

We can get this from the suvat formula s = ut + 1/2 at² for PQ and for PR, in metres and seconds, thus:

(1) 2400 = 3600u + 1/2 a 3600² and

(2) 11500 = 9000u + 1/2 a 9000².

To first get a, multiply (1) by 2.5 and then take it from (2), which eliminates u and you can then solve for a. Then you can solve for u as well.
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bingbong654
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then v = sqrt(2as)
v(p) = 2s/t if you sub in a
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Rohit94994
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(Original post by Justvisited)
If you know the answers already, why are you asking?

Zombie Wolf got it wrong by getting confused/misunderstanding the question when they say "If she is moving at a speed of 0.67m/s bettwen P and Q I assume she started with this speed." The whole point is that the runner is always accelerating, so the 0.67 is the average speed over PQ, i.e. the speed at P is less than this.

We have two unknown to solve for, a and u (the speed at P), so we need two simultaneous equations with them in.

We can get this from the suvat formula s = ut + 1/2 at² for PQ and for PR, in metres and seconds, thus:

(1) 2400 = 3600u + 1/2 a 3600² and

(2) 11500 = 9000u + 1/2 a 9000².

To first get a, multiply (1) by 2.5 and then take it from (2), which eliminates u and you can then solve for a. Then you can solve for u as well.
It's because I can't get to the answer myself.
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lconlon13
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(Original post by for_real_though)
Not sure how to do b) but I got a) as 1.24m/s2
s=13900m
u=0
a=?
t=150s

s=ut+at2/2
2S/t2=a
a=1.24ms-2 (3sf)
u isn't 0 since part b is calculate u, you have to eliminate u
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username2337287
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Work out equations for P-Q and PR, since the initial velocity would be the same. You would have a pair of simultaneous equations, and therefore from this, you could respectively work out the acceleration of the particle and the initial velocity, (speed at which it passes P).
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