I I Ι 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 34 secs., it has 31 x 32 velos, or 112 velos; hence, during the 4th second its average velocity is 112 velos; therefore, in that second it passes over 112 x 1 ft., or 112 ft. Again, at the middle of the 12th second; that is, at the end of nsecs. it has II} x 32 velos, or 378 velos; therefore, 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 ; therefore, when i velo is added in t seconds the point has celos, t u and when u' velos are added in t seconds the point has celos. u Therefore the required acceleration is celos. Hence t 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. 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 x 5) velos. 20. 21. х EXAMPLES. VII. In these examples the motion in each case is uniformly accelerated in one straight line. (i) its velocity after 2 seconds, 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 3 (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 find (i) its acceleration, (v) when it has – 144 velos. 9. A point passes over h feet in t seconds; in the next at 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 acceleration. ft.; L. D. 2 22. S 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 s=ut+ at (i). 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 {t seconds. But a velos are added to its velocity per second ; therefore, (it times a) velos are added on in it seconds; so that its average velocity for the t seconds is (u + jat) velos. Therefore, the distance passed over in the t seconds is [(u + zat) t] feet; hence, s=ut + lat. 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. 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 + lat". Let the interval t seconds be divided into n equal intervals each of 7 seconds ; so that, nt = t. The velocities at the be- The velocities at the end ginning of these n intervals of these n intervals are reare respectively spectively u velos, (u + at) velos, (u + at) velos, (u + 2at) velos, (u + 2at) velos, (u + 3ar) velos, 24. : T (u + n - 1 ar) velos. (u + nat) velos. that is, {un n n n Therefore the sum of the distances passed over by the moving point in these n intervals is [Art. 14] greater than [ur + (u + at)++ (u + 2at)- + ... + (u + n - 1 at) 7) feet, and is less than [(u + ar) – +(u + 2at) + + ... + (u + nat) 7) feet. ( That is, the required number of feet s is greater than In {24T + (n − 1) a7"} and less than In {2 (u + at)- + (n − 1) ar”}; an°25 s is greater than unt + 1 an*r anor) and s is less than unt + 1 anore + 1 but nt is t, so that at s is greater than ut + Lat® – n at and s is less than ut + lat + 1 at? This is true however great n may be; and } diminishes n as n increases. Let n be increased without limit; then s lies between two numbers each of which differs from ut + jať by a number at which is diminished without limit. Therefore s cannot be different from ut + fať. Q. E. D. 25. Let v velos be the velocity of the point at the end of the interval of t seconds, then v velos =(u + at) velos or, v=U + at (ii), but, s=ut + jut", therefore, 2as = 2aut + aʼt, hence, 2as = d° + 224at + ato – tuo = (u + at) - u (iii). ( n = = 26. These three results are of great importance ; viz. V=U+at (i), s=ut+zat? (ii), as= }v - u (iv), 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 s=ut + fat, therefore, 1600=400x – 16x2, or, x2 – 25x + 100=0, or, (x – 20) (x – 5)=0. 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 5 far will it go in attaining a velocity of 100 velos? Let the required distance be x feet ; then since as=1v2 – ļu, therefore, 5x=1 (100– 20%), or, 5x = 5000 – 200. that is, x=960. 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 ? [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=176° ~ 3 ft. per second, = per that is, 88 velos. 15 miles |