ÉîÒ¹ÑÇÖÞ¸£Àû¾Ã¾Ã

A useful applet illustrating the integral as an area under a curve is available  from the DESMOS website.

Integration is used in this section to derive algebraic expressions describing motion in two important one-dimensional situations, namely

     (a) motion with constant velocity   and

     (b) motion with constant acceleration

The results of such derivations [equation (2.14) for situation (a) and equations (2.15) to (2.20) for situation (b)] are printed in boxes in Understanding Physics. These expressions are examples of relationships that are used so often in physics that it may be worthwhile to learn the equations by rote. These are very exceptional cases, however; in the case of most relationships encountered in studying physics it is important, indeed essential, to be able to derive the appropriate expressions for yourself from first principles.

Mathematical and Physical Solutions 

At the end of this section we note that the equations which are derived to describe linear motion follow directly from the definitions of velocity and acceleration. They are not based on experimental observations and in this sense their derivation and application are exercises in applied mathematics rather than physics. 

The mathematical solutions of these equations may describe hypothetical situations which do not correspond to observed physical reality. We encountered an example at the end of Section 2.3 when, by solving simultaneous equations, we established that it was possible mathematically for the two trains M and N to be at the same position at the same time. However for this to happen we needed to establish whether the motion of train M, as described in Table 2.1, can be extended to negative values of x, something that can only be established by further physical observations. 

Problem 2.5.1 gives a further example. A sandbag is released from a balloon while it is rising and we are asked to determine the velocity of the sandbag when it hits the ground. By solving equation (2.20) we obtain two valid mathematical solution, namely v = + 41 m s^−1 and v = âˆ’ 41 m s^−1. Because the sandbag must be travelling downwards when it hits the ground we adopt the − 41 m s^−1 solution and reject the + 41 m s^−1 solution as unphysical. 

Investigating the rejected solution further, however, (see footnote to the solution of this problem) we find that it describes a situation in which, 3.15 s before it is released from the balloon, the sandbag is at ground level and travelling upwards at − 41 m s^−1. When it reaches the balloon it follows the observed motion. This is a valid mathematical solution to the equations of linear motion but we know that this did not happen -- it does not correspond to the observed motion which does not start until the sandbag is released from the balloon. 

We will often obtain alternative mathematical solutions to problems and must decide, from the observed physical data, which solution corresponds to the situation we are investigating.