# Insanely difficult mechanics question!!! Watch

Announcements

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."

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

0

reply

Report

#2

(Original post by

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!

**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!

0

reply

Report

#3

(Original post by

Please don’t post full solutions on the maths forum, it is against the rules. Only hints should be offered

**Y12_FurtherMaths**)Please don’t post full solutions on the maths forum, it is against the rules. Only hints should be offered

0

reply

Report

#4

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.

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.

0

reply

Report

#5

(Original post by

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.

**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.

0

reply

(Original post by

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.

**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.

0

reply

Report

#7

(Original post by

Why would they be at rest?

**bingbong654**)Why would they be at rest?

0

reply

Report

#8

Not sure how to do b) but I got a) as 1.24m/s

s=13900m

u=0

a=?

t=150s

s=ut+at

2S/t

a=1.24ms

^{2}s=13900m

u=0

a=?

t=150s

s=ut+at

^{2}/22S/t

^{2}=aa=1.24ms

^{-2}(3sf)
0

reply

Report

#9

(Original post by

are you in yr13 maths?

**Rohit94994**)are you in yr13 maths?

0

reply

Report

#10

(Original post by

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.

**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.

0

reply

Report

#11

(Original post by

Not sure how to do b) but I got a) as 1.24m/s

s=13900m

u=0

a=?

t=150s

s=ut+at

2S/t

a=1.24ms

**for_real_though**)Not sure how to do b) but I got a) as 1.24m/s

^{2}s=13900m

u=0

a=?

t=150s

s=ut+at

^{2}/22S/t

^{2}=aa=1.24ms

^{-2}(3sf)
0

reply

**for_real_though**)

Not sure how to do b) but I got a) as 1.24m/s

^{2}

s=13900m

u=0

a=?

t=150s

s=ut+at

^{2}/2

2S/t

^{2}=a

a=1.24ms

^{-2}(3sf)

0

reply

Report

#13

(Original post by

Theres 3600 seconds in an hour though?

**ZombieTheWolf**)Theres 3600 seconds in an hour though?

0

reply

Report

#14

(Original post by

Oh my...I converted it to minutes...

**for_real_though**)Oh my...I converted it to minutes...

0

reply

Report

#16

(Original post by

That's wrong

**Rohit94994**)That's wrong

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.

0

reply

(Original post by

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.

**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.

0

reply

Report

#19

**for_real_though**)

Not sure how to do b) but I got a) as 1.24m/s

^{2}

s=13900m

u=0

a=?

t=150s

s=ut+at

^{2}/2

2S/t

^{2}=a

a=1.24ms

^{-2}(3sf)

0

reply

Report

#20

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).

0

reply

X

### Quick Reply

Back

to top

to top