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Acceleration in Classical Mechanics

One might wonder as to whether it is really necessary to write an entire article based only on the concept of acceleration and that too in the context of Classical Mechanics (rather than in Relativistic Mechanics). I am sure you would agree that even a primary school student of physics knows what acceleration means. So, does this article explore something new about acceleration?

No, it just is an effort to make anybody who loves to solve problems (in classical mechanics) involving acceleration understand the real logic of acceleration.

Acceleration as we understand is defined as the 'rate of change of velocity'. Consider a car moving with a velocity of 100 m/s. Let the velocity of the car increase to 120 m/s in the next two seconds. Now, the change in velocity for two seconds is 20 m/s. Hence, the acceleration is (20/2) m/s²,i.e 10 m/s², which simply means that the change in velocity of the car was 10 m/s.

[NOTE: The change in velocity might be either an increase in velocity or decrease in velocity, and the acceleration is positive or negative accordingly. Negative acceleration is also called retardation.
Also note that since velocity is a vector, change in direction of a moving body means a change in its velocity, which in turn means acceleration. Hence we can have acceleration even when the speed remains unchanged.]

Now we focus our attention on solving problems involving acceleration in Classical Mechanics.

Consider a car at rest. Now if the car has to move then it MUST accelerate because, for the car to move it must have a velocity greater than zero and a change in velocity means acceleration.

Let the car move with an acceleration of 30 m/s² for one second. Now here is a simple problem for you. What is the distance traveled by the car in that one second during which it underwent that acceleration?
Nope, it's not 30 meters. The car achieved a velocity of 30 m/s only at the end of that second. So the distance covered by the car during that one should be less than 30 meters. But exactly how much?

Let us find out the answer through logical reasoning. At the beginning of that second the car was at rest and hence its velocity (which we call its initial velocity) is zero. Then the car was accelerated with an acceleration of 30 m/s² for a period of one second, which means the velocity was increased by 30 m/s. Since the initial velocity of the car was zero, the velocity at the end of that once second (which we call the final velocity) was 30 m/s.

Now to find the distance traveled by the car during that one second could be calculated if we know the velocity with which the car traveled during that one second. But since the car was accelerating during that one second, it did not have a constant velocity. Nevertheless, assuming that the acceleration was uniform (i.e, 30 m/s² throughout that one second), we can calculate the average velocity of the car during that one second which is the average of the initial velocity and final velocity i.e (0+30)/2 m/s, which is 15 m/s.

Hence, the distance traveled by the car in that one second of acceleration is 15 meters:-)

of course, this entire problem could have been solved in a split second by using the Newton's equations of motions. But the more we rely on mathematics to solve problems in physics, the less are the chances that we really understand the physics behind it. But yes, as the problem becomes complicated mathematics is the only way out to solve it.

There is one more pitfall into which the beginners in classical mechanics usually fall into. In the above problem we concluded that the distance traveled by the car during that second in which it accelerated was 15 meters. Now what do you think is the distance traveled by the car during the first half of that second and the second half of that second.  

Did you say, its 7.5 meters in both cases? Nope, it's not. Remember that the car was accelerating during that entire second and its velocity was constantly increasing from 0 in the beginning to 15 m/s at the end. Hence the distance traveled during the second half of the second should be naturally greater than that during the first half of the second.

It is left for you as an exercise to you to calculate the actual distance traveled during the first and second half of the second. You may well use the Newton's equations of motions to solve this now :-)

-by Gurudev
MADE IN INDIA
gurudevp@vsnl.net
On 1 September 2002

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