(4) A person travelling eastward, at the rate of 4 miles an hour, observes that the wind seems to blow directly from the north, and that, on doubling his speed, it appears to come from the north-east: determine the direction of the wind, and its velocity.

The wind blows from the north-west with a velocity of 4√2 miles an hour.

(5) Two bodies move uniformly along two straight lines from their point of intersection, their velocities being inversely proportional to their masses: shew that their centre of gravity describes the line bisecting the angle between them, and determine its velocity.

If u, v, denote the velocities of the two bodies, and w the angle between their paths, the required velocity is equal to

(6) ABC is a triangle: two spheres start together from A, B, their centres moving along AC, BC, with velocities which would carry them separately to C in the same time; find the distance each has gone through when they meet.

If a, b, c, be the sides of the triangle, and r, s, the respective radii of the spheres, the required distances are equal to

(1) Through how many inches does a heavy particle, let fall from rest, descend in the first half-second of its motion? If it were to move uniformly during the next half-second with the velocity thus acquired, through what space would it move during that interval?

see that the number of inches in the descent is equal to

The space which would be described by the particle in one second, with the velocity acquired in one second, is equal to 32.2 feet: hence the space which would be described in one second, with the velocity acquired in half a second, is equal to 16.1 feet, and therefore the space which would be described in half a second, with the velocity acquired in half a second, is equal to 8.05 feet.

(2) If a body be projected vertically upwards, with a velocity 8g, to find the time in which it will rise through the height 14g.

Let t denote the required time: then

Hence the body will arrive at the height 14g after 2 seconds, and, after reaching its greatest altitude, will descend to the same point, the whole interval between its projection and second arrival at this point being 14 seconds.

8

(3) A body, falling to the ground, is observed to pass throughths of its original height in the last second: find the height.

9

Let h the original height, and t = the whole time of the descent: then

(4) A body, falling in vacuum under the action of gravity, is observed to describe 144.9 feet and 177.1 feet in two successive seconds; to determine the accelerating force of gravity, and the time from the beginning of the motion.

Let u represent the velocity of the body at the beginning of the two seconds during which its motion is observed: then, by the formula for falling bodies,

s = Vt + ge2,

putting t = 1, Vu, and s = 144.9, we have

Again, at the beginning of the second of the two seconds, the velocity of the body is u+g: hence, putting, in the standard formula, t = 1, s = 177·1, and Vu+g, we obtain

Let t denote the number of seconds from the beginning of the motion to the beginning of the first of the two seconds: then

(5) A constant force acts upon a body from rest during 3 seconds and then ceases: in the next 3 seconds it is found that the body describes 180 feet: to find both the velocity of 17

W. S.

the body at the end of the 2nd second of its motion and the numerical values of the accelerating force, (1) when a second, (2) when a minute is taken as the unit of time.

Let V denote the velocity acquired at the end of 3 seconds, and f the accelerating force, a second being taken as the unit of time.

Then

and therefore

3V=180,

V = 3f,

ƒ= 20 feet.

Also the velocity, at the end of two seconds, is equal to 2ƒ= 40 feet.

The velocity which would be acquired from rest in one minute, under the action of the accelerating force, is equal to 60f1200 feet: hence the body would describe in one minute, with the velocity acquired in one minute, a space equal to 72000 feet. Therefore, if a minute instead of a second be taken as the unit of time, the accelerating force will be equal to 72000 feet.

(6) A particle descends by gravity and describes, in the nth second of its fall, a space equal to p times the space described in the last but n: to find the whole space fallen through from

rest.

=

=

Let s the whole space fallen through and the whole number of seconds.

The velocity acquired at the beginning of the nth second is (n-1)g: hence the space described in the nth second is equal to

Again, the velocity acquired at the beginning of the last second but n, that is, at the end of the (t-n − 1)th second, is equal to (t-n-1)g, and, accordingly, the space described, in the last second but n, is equal to

=

(t − n − 1) g + 1 g (2t2n-1) g.

Hence, by the hypothesis,

(2n-1) g = p (2t2n-1) g,

2pt2n-1+ (2n + 1) p;

and therefore

s =

=

9 8p2

\gt2 = 3/23 {2n − 1 + p (2n +1)}2.

(7) A body is projected vertically upwards with a velocity 4g: after two seconds suppose gravity to cease to act for one second, and then to be doubled: to find the greatest height to which the body ascends, and the velocity when it returns to the point of projection.

At the end of two seconds the body has an upward velocity equal to 4g - 2g = 2g, its height being equal to

4g x 2-gx4 = 6g.

At the end of three seconds, it still has an upward velocity 2g, its height being equal to

6g+ 2g = 8g.

If h denote the additional height acquired, before the motion

Thus the greatest altitude at which the body arrives, above the point of projection, is equal to

6g+2g+g=9g.

Let V denote its velocity, when it again reaches the point of projection: then

and therefore

V2 = 2 × 2g x 9g,

V = 6g.

(8) A particle, projected in the direction of a uniform force with a velocity u, after describing a space s acquires an additional velocity v: it acquires a second additional velocity v, after describing an additional space 28: to find the ratio between u and v.