represented by a straight line drawn perpendicular to AB from that point in AB which corresponds to the time in n A K L M n B question, the line containing as many units of length as the point possesses units of velocity. Let the time t be divided. into n equal intervals, and let the straight line AB be divided into corresponding portions at the points K, L, &c. Through A, K, L,..... B, let lines AC, KN, LP,... BD, be drawn perpendicular to AB, and representing the velocity of the moving point at the corresponding times. Draw the straight line Ckl... E parallel to AB. Then, if u be the initial velocity of the point, and fits constant acceleration, AC =u, KN=u+ft 2ft =u+f. AK, LP=u+ =u+f. AL, and so on. Therefore kN = f. Ck, IP = f. Cl, and so for the other points. Hence C, N, P,...D all lie in a straight line. Draw the straight line CD passing through each of the points N, P, &c. Complete the inner and outer series of parallelograms as in the figure. Then, if the moving point were to move during each interval with the velocity which it has at the beginning of the interval, the space passed over during any interval represented by LM will, since LP represents the velocity during that interval, contain as many units of length as the parallelogram PM contains units of area; and, this being true for each of the other intervals, it follows that the number of units of length which would be passed over by the point in the time represented by AB, that is, in t units of time, if it moved during each interval with the velocity which it actually has at the beginning of the interval, would be equal to the number of units of area contained in the sum of the inscribed parallelograms. Similarly the space passed over by the point in the same time, if it moved during each interval with the velocity which it actually has at the end of the interval, would be represented by the sum of the areas of the outer series of parallelograms. And this is true, however great may be the number of intervals into which the time t is divided. But the actual space passed over by the moving point must be intermediate between these two; and when the number of intervals into which the time is divided is in-. definitely increased, the sums of the areas of the inner and outer series of parallelograms ultimately coincide with the area of the figure CABD. Hence the number of units of length passed over by the point in the time represented by AB is equal to the number of units of area in CABD. But the figure CABD is made up of the rectangle CABE and the triangle CED. Therefore its area is equal to Now CA represents the initial velocity and therefore contains u units of length, DB represents the velocity at the end of time t and therefore contains u+ft units of length; hence DE contains ft units of length, and AB or CE contains t units of length. Hence the area CABD contains ut +5.ft.t, 1 2 1 2 or utft units of area. Therefore the space passed over in t units of time by a point starting with initial velocity u, and moving with a constant acceleration ƒ in the direction of motion, contains ut+ft units of length. 1 2 If the point start from rest u=0, and the figure will contain no rectangle corresponding to CABE. The space passed over in the time represented by AB will then be represented by the area of the triangle CED, and is there Since DE contains ft units of length, and CE contains t DE units, the ratio is numerically equal to f, and therefore EC the acceleration is represented geometrically in the figure by the tangent of the angle DCE. (Compare Newton, Lemma x.) 74. If a material particle be under the action of a constant force in the direction of its motion, its acceleration will be constant, and the above investigation determines the space passed over by the particle in any given time, substituting P m for ƒ the expression when m is the mass of the particle and P the force acting upon it. As an example we may take the following. A particle is allowed to fall from rest under the action of gravity only, find the space moved through by the particle in 4 seconds, supposing g=32 when a foot and a second are units. Here the only force acting on the particle is its weight, which is constant and equal to mg. Hence it will move with uniform acceleration g. The space passed over in 4 seconds will therefore beg. 42 ft. = 8g ft. = 256 ft. 1 75. If we wish to find the space passed over in any particular second, the nth for example, by a point moving with uniform acceleration, we may find the space passed over in n seconds and subtract from it that passed over in n - 1 seconds, or we may find the velocity at the beginning of the nth second, and then find the space passed over in 1 second by a point starting with this initial velocity and moving under the given acceleration. For example, let it be required to find the space passed over by a particle falling freely, during the 7th second of its fall, g being supposed equal to 32. The space passed over in 7 seconds by a particle falling from rest is 1 29. 72ft. 16 x 7 ft. That passed over in 6 seconds is 16 x 62 ft. The difference, or 16 x 13 ft., is the space passed over during the 7th second. Or we may proceed thus. The velocity at the end of the 6th second is 32 × 6 ft. Hence the particle at the beginning of the 7th second has an initial velocity of 32 x 6 ft., and will therefore during that second pass over a distance equal to 1 (32 × 6) +59 ft., or 32 × 63 ft., putting t=1 in the formula 1 ut += ft2. 2 76. If a point move from rest with a constant acceleration f in the direction of motion, and if v represent its velocity at the end of t units of time, and s the space passed over by the point during that time, then we have These three equations are very important, and should be remembered. If it be a material particle of mass m that is in motion, we obtain by multiplying each side of equation (3) by m and dividing by 2, Now mf is the force which must act on the particle of mass m to produce the acceleration f; and since s is the space moved through by the particle in the direction of the force, mf. s is the work done upon the particle by the force, and we thus see that the kinetic energy of the particle is equivalent to the whole amount of work that has been done upon it since the commencement of the motion. 77. If a point start with velocity u, and move with a constant acceleration f in the direction of motion, and if v v = u + ft... represent its velocity at the end of t units of time, and s the space passed over by it during that time, we have (1), (3). s = ut + 5 ft2 Squaring (1) and multiplying (2) by 2f, we see that v2 = u2 + 2fs. that is If in this case it be a material particle of mass m in motion, then, multiplying each side of equation (3) by m, dividing by 2 and transposing, we get Here as before mf. s represents the work done upon the particle by the force producing acceleration, and the expression on the left-hand side of the equation represents the increase of the kinetic energy of the particle. Hence the increase of the kinetic energy of the particle during any time is equivalent to the amount of work done upon it during that time by the force producing acceleration. If the direction of the acceleration of the moving point be opposite to that of its initial velocity, we have only to write -f for f in the above expressions, and the equations will still be true. 78. As illustrations of this portion of the subject we may take the following Examples. Ex. 1. A heavy particle is projected vertically upwards with velocity u: supposing its weight to be the only force acting upon it, it is required to completely determine the motion. The particle will in this case move with a constant acceleration g downwards. We must therefore write -g for ƒ in f the equations of the preceding Article. Its velocity v at the end of the time t will therefore be given by |