Mechanics/ Maths Question

dmg24

Labman wrote: »

Same question is here:
-
http://assets.cambridge.org/052154/9000/sample/0521549000ws.pdf

...bad news is answer is missing...good news is that it (hopefully!) explains how to work it out!

Found that too ... but the notes didn't help (I think I am beyond help on this subject!). I'm definitely going to Google difficult questions in future ... would it be ok to let my pupils do the same?!! x

Labman_2

dmg24 wrote: »

Found that too ... but the notes didn't help (I think I am beyond help on this subject!). I'm definitely going to Google difficult questions in future ... would it be ok to let my pupils do the same?!! x

It's been a long time since I did physics/maths/mechanics but I'd have thought answer should be in m/s/s

Winnifred worked in a bank & should be good with numbers...might be worth a PM?

dmg24

Labman wrote: »

It's been a long time since I did physics/maths/mechanics but I'd have thought answer should be in m/s/s

Winnifred worked in a bank & should be good with numbers...might be worth a PM?

Yep, answer is in ms^-2 ... I didn't do mechanics at school, far too like physics and far too male for me ... (yes, I got called a 'he' again tonight!).

As for your second suggestion ... he might swing his 'thingy' at me! x

Labman_2

dmg24 wrote: »

Yep, answer is in ms^-2 ... I didn't do mechanics at school, far too like physics and far too male for me ... (yes, I got called a 'he' again tonight!).

As for your second suggestion ... he might swing his 'thingy' at me! x

....I can hear my mechanics teacher Mr Campbell saying 1/36 of what boy.....'elephants/camels/tigers'! Also are you not mixing 'imperial' & 'metric' units?

Saw you way were called a 'he' earlier, in amongst the poor english/spelling & grammar...was going to mention it... but decided not to start to correct!

I'm sure if you asked Winnie he'd slow his 'swinging' so you had time to measure how fast he was moving!

dmg24

I'll be a much more gentle teacher, no raised voices in my class (I save them for the weekend)!

Have started on the next chapter, Force and Motion (ooer missus!), and am lost already ... this is not maths, it's physics!

Right I'm off to bed with my textbook ... got to be up before 9am to get my Westlife tickets (Cliff is Wednesday) ...

If you can lend me a magnifying glass I'll look into the little swinger and report back! x

Labman_2

dmg24 wrote: »

......If you can lend me a magnifying glass I'll look into the little swinger and report back! x

I can do better than that....being a 'Labman' I can supply a microscope & a pair of tweezers! Goodnight GIRL!

ffeindadifyr

I thnik the solution ias much simpler than this and relies on the assumption that the acceleration is constant.

If so - the average velocity between the first two Sticks (0km and 1km) is 1/100 km/s. And the average velocity between sticks 2 and 3 (1km and 2km) is 1/80 km/s. Now as these are average velocities where the increase in velocity is uniform (Constant accel) then we can assume that the velocity at the mid points - 0.5km and 1.5km are the average values bewteen 0km-1km and 1km-2km respectively.

So the mechanics is now much simpler - we use the equation a = (v-u)/t = (0.0125-0.01)/80 = 0.0000313 km/s^2 = 0.0313 m/s^2

I hope this is correct - feel free to correct me!

dmg24

ffeindadifyr wrote: »

I thnik the solution ias much simpler than this and relies on the assumption that the acceleration is constant.

If so - the average velocity between the first two Sticks (0km and 1km) is 1/100 km/s. And the average velocity between sticks 2 and 3 (1km and 2km) is 1/80 km/s. Now as these are average velocities where the increase in velocity is uniform (Constant accel) then we can assume that the velocity at the mid points - 0.5km and 1.5km are the average values bewteen 0km-1km and 1km-2km respectively.

So the mechanics is now much simpler - we use the equation a = (v-u)/t = (0.0125-0.01)/80 = 0.0000313 km/s^2 = 0.0313 m/s^2

I hope this is correct - feel free to correct me!

I just don't know! This was one of the options that I tried, but it seemed too simple?

I'll update when I get some feedback ... and later on today I'll post the next point I'm stuck on! x

m277

For the first kilometre you have the following info for "suvat":
s= 1
u= unknown and don't care
v= same as the starting speed for the next km
a= want to find out
t= 100

For the second kilometre you have:
s= 1
u= same as the finishing speed in the first km
v= unknown and don't care
a= want to find out
t= 80

We have a link between the two kilometres. This is the speed as the bike passes the middle post. For clarity in the calculations lets call this speed w.

So for the first km we use the following suvat equation:
s=vt - (a*t^2)/2
Remembering v = w in this case, we have:
1 = 100w - (a* 100^2)/2
Call the equation above number (1)

For the second km we use:
s= ut + (a*t^2)/2
In this case you have u=w, hence:
1 = 80w + (a* 80^2)/2
Call the equation above number (2)

Now you have simultaneous equations with a and w as the unknowns. But you don't want to find out w so eliminate that and then find a. This should be the easier bit as its GCSE stuff.