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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.
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,
(v) when it will have 1100 velos,
(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,

3
(ii) when and where it comes to reșt,
(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 find

(i) its acceleration,
(ii) when it is at rest,
(iii) where it is after 3 more seconds,
(iv) when it has 144 velos,

(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
= 2,2 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
hour
= 22 velos.

that is, 88 velos.

15 miles

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