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M1 - Average velocity doubt

1. A car is moving along a straight road with uniform acceleration. The car passes a check-point
A with speed 12 m/s and another check-point C with speed 32 m/s. The distance between
A and C is 1100 m.
a Find the time taken by the car to move from A to C.
Given that B is the mid-point of AC,
b find the speed with which the car passes B. I have already solved both parts but my question is concerned about part b; why cant we say (v+u)/2 will give the speed at the midpoint ?
2. (Original post by tryingtomaths)
A car is moving along a straight road with uniform acceleration. The car passes a check-point
A with speed 12 m/s and another check-point C with speed 32 m/s. The distance between
A and C is 1100 m.
a Find the time taken by the car to move from A to C.
Given that B is the mid-point of AC,
b find the speed with which the car passes B. I have already solved both parts but my question is concerned about part b; why cant we say (v+u)/2 will give the speed at the midpoint ?
Not sure if there's a better way to explain this, but let me give you an example.

Imagine there's no acceleration per se but a car travels at 12 m/s and every 5 seconds on the dot, the speed jumps by +5 m/s, so at t=0 the car is going at 12 m/s, at t=3 it's still 12 m/s.. t = 4.9999 is 12 m/s.. then t= 5.0, it goes at 17 m/s... same till t = 10 where it goes to 22 m/s.. and ends when it reaches 32 m/s.

In the first 5 seconds it travels 5 * 12m = 60m. The next 5 seconds, it travels 5 * 17m = 85m. The next 5 at 22 m/s, 110m. Then 135m, then what happens after doesn't matter.

As you can see, the total distance travelled is 5 * (12 + 17 + 22 + 27)) = 390 m. But by the time it reaches 22 m/s (the halfway point in terms of speed) it has travelled less than half of the distance (145m instead of 195m), this is because at higher speeds, in the same space of time (5 seconds) the car travels more, so the majority of the distance is covered at a higher speed - not equally as you might think. This is why at the halfway point, the speed is not usually the average - it is usually higher than the average.

I appreciate that may be difficult to get your head around, and that's just taking out the acceleration but if you put it in the effect is the same.

So remember, half the distance does not necessarily correlate to the speed being the midpoint between the max and min. It is usually over the midpoint (unless it's decelerating, in which case it's the other way round) because greater velocity means that more distance is covered, that probably makes no sense.. but the distance covered from going from 22 m/s to 32 m/s is much more than the distance from 12 m/s to 22 m/s so for the distance to be equal, there has to be more balance (eg it might be something like 12m/s to 24 m/s is half then 24 to 32 m/s is the other half)
3. Consider this:
From A to C, displacement = s
Using SUVAT:

Final Velocity (at C) = u^2 + 2as
Initial Velocity (at C) = u
Average velocity = (u^2 + 2as + u)/2

From A to B, displacement = s/2
Velocity at B = u^2 + 2a x (s/2) = u^2 + as

(u^2 + 2as + u)/2 is NOT always equal to u^2 + as,

Further calculations, assuming they are equal:
(u^2 + 2as + u)/2 = u^2 + as
(u^2)/2 + u/2 + as = u^2 + as
Cancel as out.
(u^2)/2 + u/2 = u^2
This is NOT always true

Method 2:
Acceleration, a, is a given constant. Therefore 2a is a constant. Let 2a be constant m.
Initial velocity, u, is a given constant, therefore u^2 must be constant. Let u^2 be constant c.

Now we are investigating the relationship of velocity at a point, v, and the displacement, s.
Let v^2 = y and let s = x

v^2 = u^2 + 2as
=> y = c + mx
in other words this is
y = mx + c
The relationship between v^2 and s is directly proportional, so if displacement is double v^2 is double.
But if v^2 is double, v is NOT double.

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Updated: October 4, 2016
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