DIMENSIONS OF THE MEASURE OF VELOCITY, ETC. 19 Art. 32, these sets of equations are equivalent to one another, and the only advantage of one particular set over another is, that the solution is effected with greater ease. In a few cases, the circumstances of the motion can be determined without the aid of the differential calculus; and we shall now proceed to determine certain particular cases of motion, when possible, without such assistance. 37. Referring to Arts. 4, 14, in which we have v= limit 3, or we conclude that the measure of a velocity is of 1 dimension in (linear) space, and of -1 in time. Also by Arts. 19, 23, in which a= or limit v t v the measure of an acceleration is t' of 1 dimension in velocity, and of -1 in time; and therefore is, on the whole, of 1 dimension in space, and of 2 in time. The measures of space and time of course are of no dimensions in time and space respectively. This evidently accounts for the fact that b2 appears in the transformation of Art. 27, while b only appears in that of Art. 6, and for a appearing equally in both. The reader would do well to apply these considerations to the analytical expressions in the subsequent Articles, by which many results, that at first sight might appear surprising, will be found to be perfectly consistent: and if they are carefully borne in mind, they will be very useful in preventing errour in any investigation. CHAPTER III. OF THE MOTION OF A POINT AFFECTED BY A CONSTANT ACCELERATION, THE DIRECTION OF WHICH IS ALWAYS THE SAME. 38. THIS is evidently the simplest kind of motion affected by acceleration conceivable, and it will subdivide itself into two heads, according as the direction of the acceleration is or is not coincident with the initial direction of motion. I. First, then, let the direction of the acceleration be coincident with the direction of motion: here it is plain that the path will be a straight line. 39. If the point move from rest and pass over a space s in time t, the measure of the acceleration being a, then will For if we divide the time t into n equal intervals 7, the velocities at the end of these intervals will be α. τη a. 27, α.3τ, a. NT. (Art. 21.) Suppose now the point to move for each interval 7 with the velocity it has at the end of that interval: then the whole space passed over would be aT.T+aα.2T. T+ α.3T . T + ... +a.NT.T +n) (1+ Again, suppose the point to move for each interval 7 with the velocity it has at the beginning of that interval; then the whole space passed over would be 0.T+αT.T+α.2T.T+ .... + a . (n − 1) t . t ...... But it is manifest that the space really passed over must lie between these two magnitudes; and since when n is indefinitely increased, they both become we must have s = at2 2 40. The velocity at the end of this time t will be at (see Art. 21). Also we shall have by eliminating t, v2=2as. 41. Next, suppose the point t be initially moving with a velocity v. Then as the mode of reasoning used in the "parallelogram of velocities" will apply here, we shall have final velocity = initial velocity + that due to the acceleration, i. e. v=v'+at. Also the space passed over space due to the initial velocity +that due to the acceleration, If the motion be retarded, the sign of a is changed. By means of these equations all the circumstances of the motion are determined. 42. II. The second case will be when the direction of the constant acceleration is not coincident with the initial direction of motion. Here it is evident that the motion will take place altogether in one plane, viz. that which passes through the initial direction of motion, and that in which the acceleration takes place: let this be the plane of the paper, and let A be the initial position of the moving point, AT the direction of the motion at A, yA the direction of the constant acceleration a, v' the initial velocity which is in direction of AT. Also let TA make with Ax at right angles to Ay the angle TAx=. Let P be the position of the moving point at any time t reckoned from the beginning of the motion. Draw PM parallel to TA meeting yA produced in M; wherefore TAMP is a parallelogram, TA is evidently a tangent to the path at A, therefore the above equation shews that the path is a parabola whose axis is parallel to yA the direction of the acceleration, the concavity being turned in that direction, and the distance of A from the 43. The position of the focus S is determined as follows: SAB=90°- SAY = 90°-2. TAY As we have drawn the figure cos 24 is negative, . 2 is > 90°. 44. If V be the vertex, the direction of motion there is parallel to Ax, and therefore the velocity in direction of Ay=0. Therefore if t be the time which has elapsed when the moving point reaches V, we must have |