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# Mechanics - Motion with uniform acceleration watch

1. A lift ascends from rest with an acceleration of 0.5 m s^-2 before slowing with an acceleration of -0.75 m s^-2 for the next stop.
If the total journey time is 10 seconds, what is the distance between the two stops?

I know how to answer all other questions the section, but I can't seem to get this.

Can someone just point me in the right direction as in which formula would best be used for it.

I thought of using s = ut + 1/2at^2 but I don't think it was right :s as there's two acceleratiosn in the question

Thanks
2. acceleration = ms^-2 and not m/s.....that is for velocity.
3. (Original post by boromir9111)
acceleration = ms^-2 and not m/s.....that is for velocity.
Yeah, forgot about those. Was gonna copy and paste the small -2 from word, then forgot about it
4. minus one from the other?
5. (Original post by boromir9111)
minus one from the other?
0.5 - -0.75 = 1.25 m s^-2 acceleration?

I put that into the s = ut + 1/2at^2 and it doesn't give me the correct answer
6. A lift would normally have a period of constant velocity, I'd have thought? Have you got all the information from the question there? If so I'm afraid I don't know...
7. I will help with this in a minute. Just drawing a diagram for you right now
8. I got the answer as 15m. I'll do a solution in a min...
9. (Original post by Iepnauy)
...
If you've not done so already, draw a velocity time graph, and it might make things a lot clearer.
10. Velocity-time graph divided it into two triangles

we know that a/b=0.5 (b/a=2 use this later)
and -a/c=-0.75
to find the ratio b/c : -(b/a)*(-a/c)=-2*-3/4=3/2
if b+c=10, and b/c=3/2 then b is 6 seconds and c is 4 seconds.
so a is 3m/s.

the area under the triangle is then 3*10/2=15m
this is the distance between the two stops
11. Are you familliar with distance-time graphs? This is the one for the lift:

Where V is the maximum velocity of the lift.

Looking at the graph you can see there are two right angled triangles:

The heights of both of these triangles is V
so: V=0.5x=0.75y
x=1.5y (divide by 0.5) This will be important later

You also know that the total time of the journey is 10 seconds so:
x + y = 10
1.5y + y = 10
2.5y=10
y=4

To work out the distance travelled you must work out the are underneath thie graph. Which is the area of this triangle:

Use V=0.75y to work out V and remember that the area of a triangle is 0.5 x height x length.

Hope I helped but if you need me to go more in depth just shout
12. (Original post by BenOliver21)
Velocity-time graph divided it into two triangles

we know that a/b=0.5 (b/a=2 use this later)
and -a/c=-0.75
to find the ratio b/c : -(b/a)*(-a/c)=-2*-3/4=3/2
if b+c=10, and b/c=3/2 then b is 6 seconds and c is 4 seconds.
so a is 3m/s.

the area under the triangle is then 3*10/2=15m
this is the distance between the two stops
Oh I see! That explains it . I should've drawn a diagram to help me :/

Thanks again
13. (Original post by ElMoro)
Are you familliar with distance-time graphs? This is the one for the lift:

Where V is the maximum velocity of the lift.

Looking at the graph you can see there are two right angled triangles:

The heights of both of these triangles is V
so: V=0.5x=0.75y
x=1.5y (divide by 0.5) This will be important later

You also know that the total time of the journey is 10 seconds so:
x + y = 10
1.5y + y = 10
2.5y=10
y=4

To work out the distance travelled you must work out the are underneath thie graph. Which is the area of this triangle:

Use V=0.75y to work out V and remember that the area of a triangle is 0.5 x height x length.

Hope I helped but if you need me to go more in depth just shout
Much clearer diagram + explanation I think I've got it now
14. You're welcome

I also wanted to point out that it's actually a velocity-time graph not a displacement time graph. I made a mistake. But the rest of the solution is fine.

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Updated: September 27, 2010
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