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98. When a point is moving with given variable velocities, we can find its actual velocity at any instant by finding the resultant of the velocities which it has at that instant.

For, if we consider another point Q moving with uniform velocities such that at that instant it has the same simultaneous velocities as the point P, then Q will, at that instant, be moving in the same direction with the same speed as the first point P.

99. The Parallelogram of Velocities is sometimes stated in the following form, which is called

The Triangle of Velocities. When a point has two simultaneous velocities, the resultant velocity may be found as follows: Draw a line OA to represent one of the velocities; from A draw a line AR to represent the other velocity; the third side OR of the triangle OAR represents the resultant velocity of the two simultaneous ve locities OA, AC.

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I00.

This proposition is only another way of stating the parallelogram of velocities.

For in the parallelogram OARB'it is clear that if OA, OB represent two simultaneous velocities, then AR represents a velocity equal and parallel to that represented by OB.

To find the resultant of any number of velocities represented by straight lines OA, OB, OC..., we first find OR' the resultant of OA and OB; then we find OR" the resultant of OR and OC; and so on.

Or we may proceed thus. Draw from A a line AR' parallel and equal to OB; from R a line R'R" parallel and equal to'OC; and so on: the line joining O to the last point R represents the resultant of all the velocities. L. D.

6

IOI.

When a point at one instant has a certain velocity and at another instant has a different velocity, then some velocity must have been added to its original velocity in the interval.

The velocity which has been added may be found by the parallelogram of velocities; for the new velocity is the resultant of the original velocity and of the velocity added.

Example i. A point in a certain interval has its velocity changed from 20 velos in a given direction to 20 velos in a direction making an angle 60° with the first direction; find the additional velocity.

B

Let OA represent the original velocity; let OR represent the new velocity; join RA; draw OB parallel and equal to AR.

Then OB represents the additional velocity required.

The angle AOR=60° and OA=OR, therefore AOR is an equilateral triangle; hence the angle ROB=ORA=60°.

Therefore OB=0A.

Thus, the additional velocity is a velocity of 20 velos in a direction making an angle 120° with the original velocity.

Example ii. A stone is thrown with 33 velos from a train, at right angles to its direction of motion, while the train is going 30 miles an hour; find the initial velocity of the stone relatively to the ground.

Relatively to the ground the stone has two simultaneous velocities of 33 velos and 44 velos respectively, at right angles to each other.

Therefore, by the parallelogram of velocities, their resultant is a velocity of 55 velos making with the direction of the train an angle whose tangent is $.

Example iii. A carriage is going at the rate of 15 miles an hour; a man jumps off, giving himself an additional horizontal velocity of 10 miles an hour; find his velocity relative to the ground when he jumps in a direction making (i) an angle of 60', (ii) an angle of 150°, with the direction in which the carriage is going.

Draw the parallelogram of velocities AOBR; draw RN perpendicular to OA; then OR is the required velocity relative to the ground. Then

OA=15, OB= AR=10.

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In case (i) NAR=AOB= 60°;

AN=AR cos 60°= 4x10;

NR=AR sin 60° =İN 3 10; therefore,

OR-=(0A+AN)2 + NR?

=(15+5) +(V3 x 5)

= 475=(218...). Therefore the velocity required is 21.8... miles per hour.

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In case (ii) NAR=30°; NA=AR cos 30o = 113 x 10;

NR=AR sin 30o=10; therefore,

OR=(0A - NA)2 + NR?

=(15-543)* + 5
=(15-5 1°7320...)2 + 25

= (8°18...)2
Therefore the velocity required is 8:18...miles per

hour. NOTE. When the resultant velocity of a point is zero, the point is at rest.

For the distance due to all its velocities after any interval is also EXAMPLES. XXVII. Find the resultants of the following 12 pairs of velocities: 1, 12 velos and 16 velos mutually at right angles, 2. 20 velos and 15 velos mutually at right angles,

zero.

3. 30 miles an hour and 40 miles an hour at right angles to each other.

4. 2 velos and 8 velos mutually at right angles. 5. a velos and b velos at right angles to each other. 6. a ft. per second and m miles per hour mutually at right angles.

3 velos and 3 velos inclined at an angle of 120°. 8. a velos and a velos inclined at an angle of 60°. 9. 3 velos and 31/2 velos inclined at an angle of 45°. 10. 10 velos and 20 velos inclined at an angle of 60°. 11. 40 velos and 2013 velos inclined at an angle of 150". 12. 20 velos and 25 velos inclined at an angle whose sine is s.

13. Find the distance in 10 seconds due to 3 simultaneous velocities of 2 velos, 3 velos and i velo making the two angles of 30° and 15° with each other,

14. A particle is given a velocity of 50 velos in a direction making an angle 45° with the horizon; resolve this velocity into two; one horizontal and the other vertical..

15. A particle has velocity 100 velos making an angle whose sine is is with the horizon; what is the resolved part of the velocity in a horizontal direction?

16. From a ship sailing at the rate of 25 miles an hour a bullet is fired with a muzzle velocity of 1100 feet per second in a direction at an angle whose sine is 1; with the direction of the ship's motion; find the initial velocity of the shot relative to the water.

17. A football moving with 20 velos receives a kick, after which its direction is changed through a right angle but its speed is unaltered; what velocity was given it by the kick?

18. A particle travelling northwards with 20 velos, receives an additional velocity of 20 velos so that the resultant velocity is 20 velos ; in what direction was the additional velocity?

19. The direction of a point's motion is altered by 45° while its speed is unchanged; what is the alteration in its velocity?

20. A cricket ball having 30 velos is struck by a bat so that its velocity is changed to 40 velos in a direction making an angle 150° with its former direction; what velocity was given to the ball by the bat?

21. A stone thrown from a train has a certain velocity given it at right angles to the path of the train, so that relatively to the ground it has 60 velos in a direction making 6oo with the path of the train; find the velocity of the train. 22.

A particle is projected with a velocity whose vertical and horizontal components are 50 velos and 10 velos respectively; the vertical velocity is diminished at the rate of 32 velos per second, the horizontal velocity is unaltered; find (the magnitude and direction of) the velocity of the particle after i second. What was the direction in which it was projected?

23. A spherical shot is rolling directly across the smooth horizontal deck of a ship with a velocity 19 velos; find where it would strike the side of the ship supposing the ship, which is going to miles an hour, to be suddenly stopped when the shot is 20 ft. from the side.

24. A balloon rising vertically with a velocity of 10 miles an hour is carried by the wind over a horizontal distance of 100 yds. in 20 seconds; find the velocity of the balloon.

25. A man tricycling westwards at the rate of 8 miles an hour, feels the wind to be due south of him; on increasing his speed to 16 miles an hour the wind appears to be south-west; find the velocity of the wind.

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