An elementary treatise on kinematics and dynamics / by James Gordon MacGregor.

  • MacGregor James Gordon, 1852-1913.
Date:
1902
Catalogue details

Licence: In copyright

Credit: An elementary treatise on kinematics and dynamics / by James Gordon MacGregor. Source: Wellcome Collection.

Provider: This material has been provided by the Royal College of Physicians of Edinburgh. The original may be consulted at the Royal College of Physicians of Edinburgh.

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    50. Examples. (1) A point moving in a circle of 40 ft. radius makes 45 revolu- tions in 20 seconds. Show that the mean speed is 565...ft. per sec. (2) A railway train runs from A to D, stopping at B and C. The distances are : A to B, 20 miles ; B to C, 5 miles ; C to B, 10 miles. It goes from A to B in 30 min., from B to C in 10 niin., and from C to D in 14 min. It remains 2 min. at B and 10 min. at C. Find the mean speed (a) during the whole time, (b) between the times of leaving A and C, and (c) between the time of leaving B and that of arriving at D. Ans. (a) 0-53..., (b) 0-48..., (c) 0'44... mile per min. (3) The distance (s feet, measured along the path) of a moving point from a given point in its path, at any time (t seconds after the instant chosen as zero) being given by the formula s = 4 + 5£, show that the mean speed for any interval and the instantaneous speed at any instant are both 5 ft. per sec. [To determine the instantaneous speed find the value of (s' - s)/(t' -1) where t and t' and therefore s and s' differ by indefinitely small quantities.] (4) The distance s of Example 3 being represented by the formula s=5t+6t2, show that the mean speed between the beginning of the 10th and the end of the 12th second is 131 ft. per sec, and that the instantaneous speed at the end of the 10th second is 125 ft. per sec. [To find the instantaneous speed at the end of t seconds, we have s'-s = 5(t'-t) + 6(t'2-t'i). Hence («' - *)/(<' 0 = 5 + 6 W + 0 = 5 +12<> since t and t' are indefinitely nearly equal.] (5) Compare the magnitudes of the foot-second and the mile-hour units of speed. The magnitudes of the units of length and time involved in these units of speed are : [L] = l ft., [T) = l sec, [Z']=l mile = 5280 ft., [T'] = l hour = 3600 sec Hence (47) the magnitude of the ft.-sec unit being [ F] and that of the mile-hour unit being [ V] we have [F]:[F'] = [Z]/[T] =