Example iv. A point having 32 celos starts from rest; how far will it go in the 4th second of its motion? and how far in the 12th second? At the middle of the 4th second, that is, after 3 secs., it has 3×32 velos, or 112 velos; hence, during the 4th second its average velocity is 112 velos; therefore, in that second it passes over 112 × 1 ft., or 112 ft. Again, at the middle of the 12th second; that is, at the end of II× 32 velos, or 378 velos; therefore, II secs. it has in the 12th second its average velocity is 378 velos; so that in that second it passes over 378 ft. Example v. A point which has u velos, has u' velos added on in t seconds; what is its acceleration? When I velo is added in 1 second the point has i celo; Ι I therefore, when I velo is added in t seconds the point has celos, and when u velos are added in t seconds the point has u' t Therefore the required acceleration is celos. Hence 20. The measure of the acceleration of a point is the ratio of the number of velos added in an interval to the number of seconds in that interval. In other words, the acceleration of a point is measured by the rate at which its velocity increases per unit-interval of time or, by the number of units of velocity added on per unit-interval. 21. When the velocity of a point is decreasing uniformly it is said to have negative acceleration. When the acceleration of a point is opposite in direction to its velocity it is often called retardation. Suppose a point to have 32 velos and 1 second after to have 30 velos, we say it has - 2 celos. This is consistent with the actual result and is a very convenient way of speaking. This point 5 seconds later will have (32 - 2 × 5) velos. EXAMPLES. VII. In these examples the motion in each case is uniformly accelerated in one straight line. 1. A point has 16 velos and 32 celos, find (i) its velocity after 2 seconds, (ii) its velocity after 5 seconds, (iii) how far it goes in the first 4 seconds, (iv) how far it goes in the first 20 seconds, (vi) when it was at rest. 2. A point has 200 velos; 5 seconds later it has 300 velos; find its acceleration. 3. A point passes over 144 ft. in 3 seconds and has 32 celos; shew that it started from rest. 4. A point passes over 144 ft. in 3 seconds and has 32 celos; how far will it go in the next 4 seconds? 5. In a certain interval of 5 seconds a point passes over 250 ft. and in the next 8 seconds it passes over 608 ft.; find its acceleration. 6. A point starts from rest, and in the roth second of its motion it passes over 304 ft.; find its acceleration and the distance it passes over in the 5th second of its motion. 7. A point has 30 velos and 5 celos; find (i) its velocity after 3 seconds, (ii) when and where it comes to rest, (iii) the distance passed over in 5 seconds, (iv) the ratio of the distance passed over in the first 3 seconds to that passed over in the next 3 seconds, (v) when it has gone half of the distance which it goes in the first 6 seconds. 8. In a certain interval of 3 secs. a point passes over 288 ft.; in next 2 secs. it passes over 32 ft.; find 9. A point passes over h feet in t seconds; in the next 27 seconds it passes over k feet; find its acceleration. 10. A point passes over h feet in 2 seconds; after an interval of t seconds it is observed to pass over k feet in 2 seconds; find its accele ration. L. D. 2 22. PROP. Let a point have u velos at the beginning of a certain interval of t seconds; let it have a celos; and let s feet be the distance it passes over in the t seconds; then (i). s=ut+at2 For, the average velocity of the point for the t seconds is that which it has at the middle of the interval; that is, the velocity which it has after seconds. But a velos are added to its velocity per second ; therefore, ( times a) velos are added on in 1⁄2 seconds; so that its average velocity for the seconds is (u + at) velos. Therefore, the distance passed over in the t seconds is hence, [(u + at) xt] feet; s = ut + at2. Q. E. D. 23. In future, when using the letters s, u, a or t we shall always suppose them to have the meaning given them in Art. 22. 24. The results of Arts. 17 and 22 may also be proved by the method of Examples V. 6 as follows: To prove, with the usual notation, that s = ut +}at3. Let the interval t seconds be divided into n equal intervals each of 7 seconds; so that, nr=t. The velocities at the be The velocities at the end ginning of these n intervals of these n intervals are re Therefore the sum of the distances passed over by the moving point in these n intervals is [Art. 14] greater than [UT + (U + AT) T + (U + 2αT) T + + (u + n · - I аT) T] feet, ... +(u+nar) and is less than [(u + аT) T + (u + 2аT) T + ... + (u + naт) T] feet. and less than an n and s is less than {unr + an3r® + } an°r"} but nr is t, so that that is, Let n be increased without limit; then s lies between two numbers each of which differs from ut + at by a number ate n which is diminished without limit. Therefore s cannot be different from ut + at. Q. E. D. 25. Let v velos be the velocity of the point at the end of the interval of t seconds, then 26. These three results are of great importance; viz. it being understood, that, in an interval of t seconds, a point which has a celos has u velos at the beginning, and v velos at the end of the t seconds; and during the t seconds passes over s feet. Example i. How long will a point which has initially 400 velos and 32 celos, take to go a distance of 1600 feet from its initial position? Let the required interval be x seconds; then since So that, either x= 20, or x=5. Thus the required interval is either 5 seconds, or 20 seconds. The point has - 32 celos; that is, its velocity is decreasing. In 5 seconds it passes over 1600 ft.; then it goes on, comes to rest, and returns to the same point, after 15 seconds more. Example ii. A point has initially 20 velos and it has 5 celos; how far will it go in attaining a velocity of 100 velos? Let the required distance be x feet; then since therefore, or, that is, as = {v2 – }u2, 52= 5000 – 200. The required distance is 960 feet. Example iii. A point is moving with a velocity of 15 miles per hour, and has an acceleration of 1 mile per minute per minute; how far does it go in the next minute? 15 miles per hour=22 velos. I [Ex. ii, p. 3.] By an acceleration of 1 mile per minute per minute, is meant that an additional velocity of one mile per minute is produced in each minute. Now, a mile per minute=178° x 3 ft. per second, that is, 88 velos. |