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A level Maths help!?

Author :

Submitted : 2018-06-14 07:40:24    Popularity:     

Tags: level  Maths  

I have tried to do this question multiple times but keep getting the answer as 13s: ‘the maximum acceleration for a lift is 1ms^-2 and the maximum decceleration js 4ms^-2. Use a speed-time graph to find the minimum time taken for the lift to go from groun

Answers:

a) For the acceleration v^2 = 2as so 25 = 2s_1 where s = distance
s_1 = 12.5 m
time = 5 seconds
For the deceleration 25 = 8s_2
s_2 = 25/8 m
time = 5/4 seconds
Time during acceleration and deceleration = 25/4 seconds
= 6.25 seconds
Distance during acceleration and deceleration = 125/8 m = 15.625 m
Distance at 5 m/s = 40-15.625 = 24.375 m
Time at 5 m/s = 4.875 seconds
Total time = 4.875+6.25 = 11.125 seconds
b) Time to accelerate to v m/s = v/1 = v seconds
Time to decelerate = v/4 seconds
Total time = 5v/4 seconds
Distance to accelerate to v m/s = s_1
2(s_1) = v^2
Distance to decelerate from v m/s = s_2
8s_2 = v^2
∴ s_1 = 4s_2
s_1+s_2 = 40
5(s_2) = 40
s_2 = 8m
s_1 = 32 m
v = 8 m/s
Total time = 5v/4 = 10 seconds

I suppose the vertical displacement is 40 meters, although that is not actually possible for a typical elevator in a 40 meter building.

To accelerate from 0 to 5 ms⁻¹ requires 5 s, at average speed 2.5ms⁻¹.
To decelerate from 5 to 0 ms⁻¹ requires 1.25 s, at average speed 2.5ms⁻¹.
That accounts for 2.5ms⁻¹ for 6.25 s, which comes to 15.625 m.

The remaining 25.375 m would be covered at 5 ms⁻¹, which takes 4.875 s.

6.25 s + 4.875 s = 11.125 s

For part (b), let v be the maximum speed.
acceleration time = v/1 = v
deceleration time = v/4 = 0.25v
total time = 1.25v

average speed = 0.5v
total time = 40/(0.5v) = 80/v

1.25v = 80/v
v² = 64
v = 8 ms⁻¹

time = 1.25(8) = 10 s

With no speed limit the minimum time required would be 10 seconds.

Time to accelerate to maximum speed:
v = at
5 = 1t
t = 5 seconds

The distance traveled in this time:
x = 1/2 a t^2
x = 1/2 (1) (5)^2
x = 12.5 meters

Time to decelerate to stop:
v = at
-5 = -4t
t = 1.25 seconds

The distance traveled in this time:
x = 1/2 a t^2
x = 1/2 (4) (1.25)^2
x = 3.125 meters

The total distance is 40m, so the distance traveled at constant speed is:
40 - 12.5 - 3.125 = 24.375 meters

The time traveled at constant speed is:
x = vt
24.375 = 5t
t = 4.875 seconds

The total time is therefore:
5 + 1.25 + 4.875 = 11.125 seconds

You accidentally used 3.125 meters instead of 1.25 seconds.


If there is no speed restriction, the lift accelerates to a maximum speed v in time t₁.
v = at
v = t₁

The distance traveled is:
x = 1/2 a t^2
x₁ = 1/2 t₁^2

The lift then decelerates from this speed.
v = at
v = 4t₂

The distance traveled is:
x = 1/2 at^2
x₂ = 2t₂^2

v = v
t₁ = 4t₂

x₁ + x₂ = 40
1/2 t₁^2 + 2t₂^2 = 40
1/2 (4t₂)^2 + 2t₂^2 = 40
8t₂^2 + 2t₂^2 = 40
t₂^2 = 4
t₂ = 2 seconds

t₁ = 8 seconds

t = t₁ + t₂
t = 10 seconds

Here's how I'd do it.
Distance traveled is the area under a speed-time graph.

If the elevator constantly accelerates until it's time to decelerate, then the speed-time graph is a line going up with a slope of 1 m/s^2 and then down with a slope of 4 m/s^2. In both cases the lines go to 0 m/s since it starts from rest and ends up at rest.

So those two lines form a triangle. The base of that triangle is the total time t. Let's say the maximum velocity, the height of the triangle, is vmax. The time to accelerate to vmax at 1 m/s^2 is vmax/1 = vmax seconds. The time to decelerate from vmax to 0 at 4 m/s^2 is vmax/4 seconds.

Total time: t = vmax + (vmax/4) = (5/4) vmax seconds.
Total distance traveled = area of triangle = (1/2) * (5/4) vmax * vmax = (5/8)vmax^2 = 40 m.

So vmax = sqrt(40*8/5) = 8 m/s. And the total time we found was (5/4) vmax seconds = (5/4) * 8 = 10 s.

This is the answer to part (b). If there's no limit, the shape is a triangle, the maximum velocity is 8 m/s and we can do it in 10 s.

For part (a), the lines can only rise as high as 5 m/s. So the shape is a trapezoid. First it rises to 5 m/s at 1 m/s^2, which takes 5 seconds during which it travels (1/2) * 5 * 5 = 12.5 m. (I'm doing (1/2)bh for the triangles formed by these lines).
At the end it decelerates from 5 m to 0 at 4 m/s^2, which takes 1.25 s, during which it travels (1/2) * 1.25 * 5 = 3.125 m.
That leaves 40 - 12.5 - 3.125 = 24.375 m to cover in the middle, which it does at 5 m/s, which takes 4.875 s.

Total time = 5 s + 1.25 s + 4.875 s = 11.125 s.

"I have tried to do this question multiple times but keep getting the answer as 13s"
If you had posted HOW you arrived at this answer, I could check your reasoning and tell you what assumptions you might be making that are wrong. The answer tells me no information, and so there's no way for you to learn from it.

With a limit of 5 m/s:
Calculate the time needed to accelerate to 5 m/s and the time needed to decelerate from 5 m/s to a stop.
Calculate the distance traveled during those two stages.
Add the additional time needed to traverse the rest of the 40m at 5 m/s.

With no speed limit:
There are only two stages. The lift will accelerate until it needs to start decelerating.
Call the maximum speed X. Repeat your calculations as in (a) for time and distance in terms of X.
Set the distance equal to 40m and solve for time.



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