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Mechanics Help please

Hi,

I have two questions that i need help with, My teacher isn't really great so I'm having problems in Mechanics, I really could do with some help. :frown:

The questions are:

1. A girl standing on a bridge throws a stone vertically upwards at 6 metres per second. It hits the water below the bridge after 2 seconds. Find the speed at which the stone hits the water and the initial height of the stone above the water.

2. A train starts from rest and after 20 seconds it is travelling at 30 metres per second. It travels at this speed for 1 minute. Find the acceleration in the first 20 seconds and the total distance travelled. What assumptions are inherent in this question?


Thank you so much in advance...!!! :biggrin:
Reply 1
Original post by SalmaT
1. A girl standing on a bridge throws a stone vertically upwards at 6 metres per second. It hits the water below the bridge after 2 seconds. Find the speed at which the stone hits the water and the initial height of the stone above the water.

2. A train starts from rest and after 20 seconds it is travelling at 30 metres per second. It travels at this speed for 1 minute. Find the acceleration in the first 20 seconds and the total distance travelled. What assumptions are inherent in this question?


For the first one, you should use SUVAT.

You've been given 3 things *alarm bells should be ringing*.

[br]S=?[br][br]U=6m/s[br][br]V=?[br][br]A=9.81m/s2[br]S=?[br][br]U=6m/s[br][br]V=?[br][br]A=-9.81m/s^2
(acceleration due to gravity).
T=2T=2

Now look through the SUVAT equations and find a suitable equation that will give you the answer.

For the second one,

use the formula a=ΔVelocity÷Timetakena=\Delta Velocity\div Time taken (also SUVAT, but this is a general formula you should know).

to find the total distance travelled/displacement you could plot a velocity-time graph and work out the area under the graph.

by looking at the (vt) graph you can explain one of the assumptions.
(edited 12 years ago)
Reply 2
I'm very much confused about what you're trying to say about the second one. =/
Reply 3
Original post by SalmaT
I'm very much confused about what you're trying to say about the second one. =/


The triangle represents 'Delta', a Greek letter that represents the 'change/difference' in something. In other words, Final Velocity-Starting Velocity. I'm sorry for using that but I think it's very useful to get to grips with the Greek letters as they crop up a lot in Physics/Maths.

To answer the SUVAT equations you should be able to just mention the definition of SUVAT and what things you aren't including in your calculations (other additional forces).
(edited 12 years ago)
Reply 4
Oh ok...I don't know what delta is...never heard of it in maths...
But...Thank you... :smile:
Reply 5
Original post by SalmaT
Hi,

I have two questions that i need help with, My teacher isn't really great so I'm having problems in Mechanics, I really could do with some help. :frown:

The questions are:

1. A girl standing on a bridge throws a stone vertically upwards at 6 metres per second. It hits the water below the bridge after 2 seconds. Find the speed at which the stone hits the water and the initial height of the stone above the water.

2. A train starts from rest and after 20 seconds it is travelling at 30 metres per second. It travels at this speed for 1 minute. Find the acceleration in the first 20 seconds and the total distance travelled. What assumptions are inherent in this question?


Thank you so much in advance...!!! :biggrin:


Below are the solutions:

1) Taking upwards direction as positive,
v = u + at
v = 6 - 9.8*2 = -13.6m/s
The velocity is 13.6m/s downwards (the negative sign shows that velocity is acting opposite to the direction we have chosen).

2) Acceleration = (v-u)/t = (30-0)/20 = 1.5m/s^2
Distance travelled between 0 to 30s,
v^2 = u^2 + 2as
30^2 = 0^2 + 2*1.5*s
900/3=s=300m
Distance travelled between 30 to 90s,
s = vt
s = 30 * 60 = 1800m

Total distance travelled = 300+1800 = 2100m
Reply 6
Thank you...!!! :smile:
2100m seems a bit too far to throw a stone. I used the suvat equation s=ut 1/2at^2 and i got the answer 31.62m which sounds more accurate. i may be wrong.
Original post by Not-so-smart
2100m seems a bit too far to throw a stone. I used the suvat equation s=ut 1/2at^2 and i got the answer 31.62m which sounds more accurate. i may be wrong.


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