other: when and where will they meet, if they travel at the respective rates of 7 and 8 miles an hour? Since the rate or velocity of A is 7 miles an hour, and the rate or velocity of B is 8 miles an hour, therefore 7+ 8 = 15 miles. is the distance by which they approach each other in 1 hour, or their relative velocity: hence we have 15 mi. 300 mi. :: 1 hr. : 20 hrs.; or, 20 hrs. is the time in which they approach 300 miles towards each other and therefore meet: whence the required time is expressed by the whole distance divided by the sum of their velocities or rates per hour. Also, 7 × 20=140 miles is the distance travelled by A, and 8×20=160 miles is the distance travelled by B: and they meet at 140 and 160 miles from C and D respectively. Hence also, the distance between them after any assigned interval may be found. When two motions in a straight line are in opposite directions, the velocity of approach or the relative velocity is equal to the sum of the absolute velocities. (2) A, travelling at the rate of 12 miles an hour starts 15 miles behind B who travels only 10 miles an hour: find when A will overtake B and the distance then travelled by each. Here, the gain of A upon B is 12-10=2 miles in 1 hour which is their relative velocity: whence, 2 mi.: 15 mi. :: 1 hr. : 71⁄2 hrs., and 7 hrs. is the time in which A gains 15 miles upon B and therefore overtakes him: so that A has travelled 12 × 7 = 90 miles, B has travelled 10 × 7 = 75 miles: and the difference of these distances is 15 miles which is accordant with the enunciation of the problem. In cases like this, the velocity of approach or relative velocity is the difference of the absolute velocities, and the time is found by dividing the distance by it. The reasoning employed in these two instances is evidently applicable to any uniform motions whether in straight lines or curves, provided the distance be measured along the paths described. (3) Two couriers pass through a place at an interval of 4 hours travelling at the rates of 114 and 171⁄2 miles an hour: how long and how far must the first travel before he is overtaken by the second? The relative velocity = 17-11 = six miles : also, 11 × 4 = 46 miles = the distance between them, when the second passes through the given place: whence, as before, we have 46 = 6 73hrs., the time when the second after leaving the given place overtakes the first: and therefore the first has travelled in 113 hours, the distance 11 × 113-134 miles: and the second in 73 hours, the distance 174 × 73 134 miles. = (4) If 252 men in 5 days of 11 hours each, can dig a trench 210 yards long, 3 wide and 2 deep; in how many days 9 hours long, can 24 men dig a trench of 420 yards long, 5 wide and 3 deep? The solid content of the first trench is 210 × 3 × 2 1260 solid feet: and that of the second is 420 × 5 × 3 : 6300 solid feet. Now, 252 men in 55 hours dig 1260 solid feet: = = whence, 1 man in 55 hours digs 5 solid feet: and 24 men in 55 hours dig 120 solid feet: therefore, 24 men in 55 × 52 hours dig 6300 solid feet: 55 × 52/2 = 320 days of 9 hours each 9 and consequently, is the time required. (5) If 10 men in 3 days reap a field the length of which is 1200 feet and the breadth 800 feet; what is the length of a field whose breadth is 1000 feet which 12 men can reap in four days? Here, 10 men in 3 days reap 1200 × 800 square feet: and 1 man in 3 days reaps 120 × 800 square feet: 40 × 800 square feet: whence, 12 men in 1 day reap 480 × 800 square feet: and 12 men in 4 days reap 1920 × 800 square feet: but 1920 × 800 = 1536000 = 1536 × 1000 square feet: whence, it follows that the required length is 1536 feet. (6) If a pipe of 6 inches bore discharge a certain quantity of fluid in 4 hours: in what time will 4 pipes each of 3 inches bore, discharge twice that quantity? = If 1 denote the quantity of fluid discharged by the first pipe in 4 hours, we have quantity discharged by it in 1 hour: but the quantities discharged by the pipes are as the areas of their sections and therefore as the squares of their diameters: whence, quantity discharged by one of the second set in 1 hour :: 6 : 3 :: 4 : 1; and therefore the quantity discharged by one of these pipes in 1 hour =: hence, the quantity by 4 such pipes in 1 hour = 1; and therefore the quantity discharged by these 4 pipes in 8 hours = 2, or twice the quantity discharged by the first in 4 hours: that is, 8 hours is the time required. (7) If a beam which is 10in. wide, 8 in. deep and 5 ft. 6 in. long, weigh 8 cwt. 1qr.: find the length of another beam of the same kind, the end of which is a square foot, which shall weigh 1ton. The volume of the first beam = 10 × 8 × 66 solid inches, and that of the second beam = 12 × 12 × the required length = 144 solid inches, suppose: also, the weights are proportional to the volumes or masses, and therefore we have 8cwt. 20 cwt. :: 10 × 8 × 66: 144x: (8) If a ball, whose diameter is 2 inches weigh 5 lbs. ; what must be the diameter of another ball of the same substance which shall weigh 78.125 lbs. ? Since the weights are proportional to the volumes, and therefore to the cubes of the diameters, we have 5lbs. 78.125lbs. :: 23 (the required diameter); (9) A rectangular court the sides of which are 300 feet and 200 feet has a walk 20 feet wide cut off from it on every side: compute the area of the walk and compare it with that of the court. The area of the whole court = 300 × 200 60000 square feet: also, since the dimensions are diminished 20 feet on each side by the walk, the area of the remaining portion = (300-40) × (200-40)=260 × 160 = 41600 square feet: whence, the area of the walk 60000 - 41600 square feet and the walk therefore takes up 18400 184 46 = ths of the court. 18400 (10) Multiply 2 feet 1 inch by 1 foot 2 inches and explain the meaning of the terms of the result by a geometrical construction. Here by the rule of Article (212), we have To explain this result geometrically, if AB = 2 feet, Bb = 1 inch, AC=1 foot, Cc = 2 inches and the construction be completed as below: × = 2 square feet: Db + Dc = 1' x 1'+2′ x 2′ = 1′+4' 5 superficial primes : = 1' x 2' = 2 square seconds: that is, the entire product is 2. 5'. 2′′ superficial measure as before found: and the diagram shews clearly what is meant by each of the denominations of the result; namely, that a superficial prime is a foot in length and an inch in breadth; a superficial second is an inch in length and an inch in breadth, or a square inch, &c. Hence, it appears that the product of two numbers of feet retains their common denomination, whilst in the Duodecimal Subdivisions the denomination of the product of two quantities of different names is ascertained by the addition of the numbers expressing the denominations of the quantities themselves: and this conclusion is sometimes embodied in the form of a verbal rule. |