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The problem:

Two particles A and B start from rest and move for equal time on a straight line. The particle A has an acceleration a for the first half of the total time and 2a for the second half. The particle B has an acceleration 2a for the first half and a for the second half. Which particle has covered larger distance?

Using Newton's 2nd equation, I found out the distance travelled by particle A when it was moving with an acceleration a and 2a separately and added both which came out to be 5/8 at^2. In In the second case of motion of particle B since the acceleration is changing from 2a to a I took acceleration negative. Similarly for B I got the total distance 3/8 at^2. Thus, this means particle A has covered larger distance.

Why does it happen that even though both the particles travelled with acceleration a and 2a for same time, still, particle A travelled larger distance? Please give me an answer based on intuition.

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Let's draw a graph of velocity against time for the two particles $A$ and $B$. For convenience I've made the total time $2t$:

Velocity time graph

The red line shows the velocity for particle $A$ while the green line shows the velocity for particle $B$.

When we draw a velocity:time graph the distance travelled is the area under the line. More precisely it is the integral of the velocity wih respect to time, but this is just equal to the area under the line. The area under the green line is obviously greater than the area under the red line, so we can immediately see that particle $B$ travelled farther.

Incidentally, you have misinterpreted the question. Particle $B$ accelerates at $2a$ for the first half of the time then accelerates at $a$ for the second half. So its speed increases throughout. In fact $B$ travels a distance of $3\tfrac{1}{2}at^2$ while particle $A$ travels a distance of $2\tfrac{1}{2}at^2$. You don't even need to use the SUVAT equation as you can read these figures directly off the graph.

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    $\begingroup$ I love it when a simple picture conveys the principle so clearly. Excellent! $\endgroup$ – Floris Jul 25 '15 at 13:40
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I find that sometimes intuition works better with more extreme examples. Let's change the problem up a bit. Instead of some itty-bitty difference in acceleration (a vs 2a), lets choose a big acceleration for a short period of time.

In this modified example, both A and B are going to be fired out of a cannon. The firing is going to take just 0.1s, but with a very high acceleration (i.e. the result of gunpowder). Compare this to your original example where the two objects spend equal time accelerating at a and 2a. In this new example, the object spends a tiny fraction of their time accelerating really hard, and then the coast for the rest of the experiment (let's only pay attention to horizontal distance... that way we can ignore gravity for now.)

Lets say you're going to start the experiment at t=0s, and measure their position at t=10s. Cannonball A(corresponding to your particle A, accelerating at a then 2a) begins accelerating at t=9.9s - it hardly accelerates at first, then it accelerates really fast. Meanwhile, Cannnonball B(corresponding to your particle B, accelerating at 2a then a) begins accelerating at t=0.0.

It should be trivial to see that the cannon ball which was fired at t=0s will travel further by t=10s than the cannon ball which was fired at t=9.9s will have traveled by t=10s. This is exactly the same answer as you get with your problem, only taken to an extreme to make it easier to get an intuitive grasp.

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    $\begingroup$ Right - the sooner you get your speed the longer you can take advantage of it. Using the extremes is a good way to show this. $\endgroup$ – Floris Jul 25 '15 at 13:42
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The same reasoning that goes for racing cars: Acceleration is important if you are slow, because then the relative change in velocity over time will be high

edit: (also the velocity is the accumulated acceleration, meaning that the earlier you have high accelaration the better because you will "keep" that to the end)

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  • $\begingroup$ And distance is accumulated velocity. That is the crux here. They end up with the same velocity but B spent longer time going fast. $\endgroup$ – Floris Jul 25 '15 at 13:43
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I think your answer is incorrect. First the intuitive argument:

Assume you had stopped both particles after half of the time. It is obvious that in this case they had covered the same distance. At first B is first, but in the second half A would catch up. If you don't stop the particles before they continue with a different acceleration also the velocity they obtained by that time will contribute to the covered distance. Because particle B has accelerated more in the first half it will be start faster in the second half. Thus, intuitively, B should be the faster particle.

This is also true if you do the calculation. Unfortunately you didn't show yours, but I guess your mistake is that you took the second acceleration of B negative. There is no reason for doing this. B will continue becoming faster and faster, however at a slower rate.

[EDIT] Here are some more hints:

After time $t/2$ the particles have traveled the distance $\frac 1 2 a \left(\frac t 2 \right)^2$ and $\frac 1 2 \cdot 2 a \left(\frac t 2 \right)^2$ for particles A and B, respectively. But keep in mind that even if they stopped accelerating both would continue with velocity $a \frac t 2$ (for A) and $2 \cdot a \frac t 2$ (for B).

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I'm assuming you used the following kinematic equation in your solution: $$d = x_0 + v_{0}t + \frac{1}{2}a_{0}t^{2}$$ Remember that after the initial acceleration, the particles will have covered some distance and gained some velocity. Hence, in the second step where you have a new acceleration, you no longer have a zero initial position and velocity.

I got $\; d_{A} = \frac{7}{8}at^2$ and $d_{B} = \frac{11}{8}at^{2}$ for the distances covered by particle A and B, respectively (although I didn't double check). Which agrees with the intuition given by the previous answers.

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