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4. A point is at rest and velocity is given to it uniformly so that at the end of 1 second it has 32 velos; shew as in example 3, that in the 3rd second of its motion the distance which it passes over does not differ from 80 feet by more than 6 inches.

5. A point has u velos, and its velocity is increasing at the rate of 32 velos per second ; shew, by dividing the next 8 seconds into 8n equal parts, that the distance passed over in those 8 seconds cannot

128 differ from (8u+1024) feet by more than

feet.

n

6. A point has u velos, and its velocity is increasing at the rate of a velos per second ; shew, by dividing the next t seconds into n equal parts, and considering the velocity at the beginning and the end of these n intervals, that the distance passed over in 7 seconds cannot

at2 differ from (u +3 at) x t feet by more than feet.

2n

16. It is of the greatest importance that the student should mentally realize the motion of a point moving with uniformly increasing (or decreasing) velocity.

Consider again the case of two very long trains A and B moving on parallel lines in the same direction; suppose them to be passing a certain station H,—the train A with a constant velocity 20 velos,—the train B with a constantly increasing velocity which, when the engine is at H, is 4 velos.

At H suppose the engines are abreast; at first A will gain upon B; but it will gradually gain more and more slowly, until an instant arrives when the two trains are relatively at rest; after which time B will at once begin to gain upon A; and will gradually gain at a faster and faster rate.

Hence an instant arrives when the engines are again abreast; suppose this to occur at a place K; then each train has passed over the distance from H to K in the same interval; hence the average velocity of E, for the time occupied in going from H to K is the velocity of A, viz. 20 velos.

We proceed to prove that the instant at which the two trains are relatively at rest, is at the middle of the interval of time which they each take in going from H to K.

The proof given on the next page may be deferred if it is thought desirable. An alternative proof is given in

Art. 24.

17. PROP.

When a point is moving with uniformly increasing velocity, its average velocity for any given interval of time is equal to the velocity which the moving point has at the middle of that interval of time.

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Let the given interval of time t seconds be represented by the length of the finite straight line AB.

Divide the t seconds into any odd number n of smaller equal intervals, each containing 7 seconds.

Let DE represent the middle interval of 7 seconds; and let C be the middle instant of AB, and therefore of DE. Take a pair of the little intervals LM and NR, one on each side of C and equidistant from C.

Let the moving point at the instant D have v' velos; at the instant C, v velos; and at the instant E, 7" velos; let w be the number of velos added on in the interval LD.

At the instant L the moving point has (' w) velos; at the instant N it has (v + w) velos; at the instant M it has (0" —W) velos; at the instant R it has (2" + 2) velos. [Since the number of velos which it has increases uniformly with the time, and LD= ME=DN= ER.]

The distance passed over by the point in the intervals LM and NR together, is therefore greater than {(v' - w)

+(v'+w) T} feet, and less than {(1"-w)++ (v + w)} feet; that is, greater than 2TV' feet, and less than 2TV'' feet.

Also, in the middle interval DE, the distance passed over is greater than tv' ft. and less than tv" ft.

Hence, the whole distance passed over in all the n intervals is greater than nt x V' ft. and less than nt xv" ft. ; that is, greater than tv feet and less than tu" feet.

This is true however great the odd number n may be. Let n be increased without limit; then , is diminished without limit and ū and v" each approach v as their limit. Therefore the distance passed over in the whole interval t seconds (which distance is between tv feet and tu" feet) cannot be other than tv feet.

In other words, its average velocity for the time t seconds is v velos; and this is the velocity which the point has at C, the middle of that interval of t seconds.

Q. E. D. Example i. A point has 18 velos, and is moving with uniformly increasing velocity, so that i second later it has 20 velos; how far does it go in that second?

The average velocity for the 1 sec. is the same as the velocity which the point has at the middle of the interval, that is, after half a second.

Since in i second the velocity is increased by 2 velos,
therefore in 1 second the velocity is increased by i velo;
hence the average velocity for the 1 second is (18+I), or, 19 velos.

In i second it passes over 19 * 1 ft., or 19 feet.
Example ii. How far does the point in Example i. go in 8 seconds?

The middle of the interval of 8 seconds is 4 seconds after the time when it has 18 velos.

In i second the velocity is increased by 2 velos, therefore in 4 seconds the velocity is increased by 8 velos; hence, the average velocity for 8 seconds is (18+8) or 26 velos; therefore, the distance passed over is 26 x 8 ft.; that is, 208 feet.

Example iii. A point, moving with uniformly decreasing velocity, passes over 294 ft. in 7 secs.; and at the beginning of the 7 seconds has 49 velos; find its velocity after 3 seconds more.

Since it passes over 294 ft. in 7 seconds, its average velocity [Art. 8] is 42

velos; its actual velocity at the middle of the 7 secs. is equal to this average velocity; hence, since at the beginning of the 7 seconds it has 49 velos, and 31 seconds later it has 42 velos, therefore its velocity decreases by 7 velos in 31 secs.; or, by 2 velos per second ;

hence, after to seconds it will have (49 – 20) or 29 velos.

EXAMPLES. VI.

In the following examples the velocity is uniformly increasing (or decreasing).

1. A point has 100 velos at 12 o'clock, and its velocity increases to 104 in 1 second; find

(i) how far it goes in 1 second,
(ii) how far it goes in 2 seconds,
(iii) how far it goes in the 30 seconds after 12. I' o'clock,

(iv) how far it goes between 12. I' and 12.2' o'clock.

2. At a certain time a point has 16 velos, and its velocity increases at the rate of 32 velos per second; shew that in the next three seconds it passes over 192 ft. (See Ex. V. 3.)

3. A point has 120 velos, and its velocity is decreasing at the rate of 32 velos per second; shew that in the next second it passes over 104

ft. (See Ex. V. 2.]

4. A point in a certain interval of 3 seconds passes over 300 feet ; in the next 4 seconds it passes over 764 ft. ; shew that its velocity is increasing at the rate of 26 velos per second.

5. A point in a certain interval of 5 seconds passes over 125 feet; in the next 8 seconds it passes over 304 feet; when did it start from rest?

6. A point is observed to pass over 120 ft., 129 ft., 138 ft. respectively in 3 consecutive seconds; is this consistent with uniformly increasing velocity?

7. Starting from rest, in the 5th second of its motion, a point passes over 144 ft.; find its velocity after 10 seconds from rest.

8. A point passes over 144 ft. in a certain interval of 2 seconds and comes to rest after 3 seconds more; how much further did it move?

9. A point starts from rest, and after 10 seconds it has 25 velos; how far does it go in the 20th second of its motion ?

10. A point passes over 24 ft. in a certain second ; at the end of that second it has 28 velos; how far does it go in the next second ? and when is it at rest?

11. A point passes over 144 ft. in two seconds and over 96 ft. in the next two seconds; when will it come to rest ? and how far does it move before doing so ?

12. A train goes to miles in a quarter of an hour, having started from rest; shew that if its velocity has been uniformly increasing, it is moving at the end of that quarter of an hour with a speed of 80 miles

Hver

per hour.

ACCELERATION.

18. DEF. When a point is moving with uniformly increasing velocity, its acceleration is that which varies directly as the increase of velocity in a given interval, and inversely as the interval required for a given increase of velocity. [Compare Arts. 3 and 4.]

Thus, the acceleration of a point is doubled, or trebled etc. when the velocity added in a given interval is doubled, or trebled etc.; and it is halved, or divided by three, etc. when the interval (in which a given velocity is added) is doubled, or trebled etc.

19. We shall choose as our unit acceleration the acceleration of a point which, moving with uniformly increasing velocity, has its velocity increased by i velo in the course of each second.

We shall call this unit acceleration a celo.

Thus, a celo is (a foot per second) per second.

NOTE. It is most important to notice that a point which has acceleration requires an interval of time in which to increase its velocity by any definite amount.

Example i. A point which has 6 velos added to its velocity in the course of each second, has six times the acceleration of a point having the unit acceleration; hence it has 6 celos. The same point has i velo added in the course of the įth part of a second.

Example ii. A point has at a certain instant 12 velos and it has 6 celos; how many velos will it have 10 seconds later?

In the course of i second 6 velos are added, therefore in 10 seconds 60 velos are added.

So that after 10 secs. the point has (12+60) velos, or 72 velos. Example iii. How far does the point in Ex. ii go in those 10 secs.

The average velocity of the point in the 10 seconds is that which it has at the middle of the interval; that is, (12+5 x 6) velos, or 42 velos; therefore, in 10 seconds the point goes 42 X 10 ft., or 420 feet.

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