to the boat; the former is usually spoken of as the absolute velocity, and the latter the relative velocity. Or, suppose the case of a man standing on the deck of a boat moving south with a velocity v while the wind blows from the east with a velocity v1. It is required to find the velocity of the wind with respect to the man, or, in other words, the apparent direction and velocity of the wind as ob

ย

FIG. 105

V1

served by the man. Referring to Fig. 105,

we represent the velocity (see Art. 85) of the boat with respect to the earth by v, and the velocity of the wind with respect to the earth by v1, then V represents the velocity of the wind with respect to the man; that is, the wind appears to the man to be coming from the southeast. The velocity V was obtained by reversing the arrow representing the velocity of the boat and finding the resultant of this reversed velocity and the velocity v, of the wind.

If v be considered as the velocity with respect to the earth of a man walking across the deck of a steamer moving with a velocity v, then represents the velocity of the man with respect to the boat.

Problem 118. An ice boat is moving due north at a speed of 60 mi. per hour, the wind blows from the southwest with a velocity of 20 mi. per hour. What is the apparent direction and velocity of the wind as observed by a man on the boat?

Problem 119. A man walks in the rain with a velocity of 4 mi. per hour. The rain drops have a velocity of 20 ft. per second in a direction making 60° with the horizontal. How much must the man incline his umbrella from the vertical in order to keep off the

rain: (a) when going against the rain, (b) when going away from the rain? If he doubles his speed, what change is necessary in the inclination of his umbrella in (a) and (b)?

Problem 120. The light from a star enters a telescope inclined at an angle of 45° with the surface of the earth. The velocity of light is 186,000 mi. per second and the earth (radius 4000 mi.) makes one revolution in 24 hr. What is the actual direction of the star with respect to the earth? This displacement of light due to the velocity of the earth and the velocity of light is known as aberration of light.

Problem 121. A man attempts to swim across a river, 1 mi. wide, which is flowing at the rate of 4 mi. per hour. If he can swim at the rate of 3 mi. per hour, what direction must he take in swimming in order to reach a point directly across on the opposite shore?

Problem 122. A train is moving with a speed of 60 mi. per hour, another train on a parallel track is going in the opposite direction with a speed of 40 mi. per hour. What is the velocity of the second train as observed by a passenger on the first?

mi.

Problem 123. A man in an automobile going at a speed of 40 per hour is struck by a stone thrown by a boy. The stone has a velocity of 30 ft. per second and moves in a direction perpendicular to the direction of motion of the automobile. With what velocity does the stone strike the man?

Problem 124. A locomotive is moving with a velocity of 40 mi. per hour. Its drive wheels are 80 in. in diameter. What is the tangential velocity of the upper point of the wheels with respect to the frame of the locomotive? What is the tangential velocity of the lowest point?

CHAPTER X

CURVILINEAR MOTION

85. Representation of Velocity and Acceleration. It has been shown (Art. 71) that velocity is measured in terms of feet per second, miles per hour, or in general in terms of the units of distance and units of time. The velocity is, moreover, in a given direction and may accordingly be represented by an arrow just as forces may be so represented. It follows then that velocity arrows may be

added algebraically if parallel and if such addition is not inconsistent with the problem. They may be resolved into

components or

combined to form

X resultants (see

Art. 11 and Art.

12). In case the body moves in a curve it is often desirable, instead of dealing with the resultant velocity along the tangent, to deal with the components of that velocity along the two coördinate axes. Thus if v is the velocity along the tangent (Fig. 106),

v,

v cos a and y = v sin a are

the component velocities along the axes x and y respectively. In a similar way if we know the velocity of a body along the x-axis, v, and the velocity along the y-axis, vy, we find the resultant velocity to be v=√ v2+v and its direction with the x-axis such that

Accelerations have been seen to be measured in terms of units of distance and units of time, in particular in terms of feet per (second)2 (see Art. 72). An acceleration may be represented by an arrow, the length of the arrow representing the number of feet per (second)2 and the direction of the arrow giving the direction of the acceleration. Since arrows represent accelerations, the acceleration arrows may be treated just as velocity arrows. That is, they may be added algebraically if parallel, or added and subtracted geometrically if intersecting. Or we may say that the parallelogram law holds for accelerations. Referring to Fig. 106, it is seen that the resultant acceleration, a, of the body moving in the curve y=f(x) is directed toward the concave side of the curve in the direction of the resultant force. This is evident from Newton's Law, which states that the acceleration is proportional to the resultant force and in the same direction. Let a be the resultant acceleration, then the accelerations along the two axes are ax=a cos 0 and ay = a sin 0, respectively. In a similar way if we know the accelerations along the two axes, a, and ay, the resultant acceleration a = √ až + az, and its direction is given by the equation

α

86. Tangential and Normal Accelerations. Suppose a body (Fig. 106) moves in any curve y=f(x) and that at a certain point P it has a resultant velocity v and a resultant acceleration a, v acts along the tangent, which at this point makes an angle a with the x-axis and a acts along the line of action of the resultant force on the concave side of the curve. It is seen from the figure that vx = v cos α, vy = v sin α, ax = a cos 0, a1 = a sin 0, so that v = v2 √v2+ v2 and a = √ a2 + a2.

It is usually convenient in curvilinear motion to consider the acceleration along the tangent and normal; that is, a, and an Since the tangent and normal are at right an angles, it is evident that a = √a+a2. When it is remembered that v acts along the tangent, it is evident

at

that the tangential acceleration a1 =

dv

dt

or,

since

= cos a and y =

It now remains to find the normal acceleration. It has

been shown that a = √a2+a2; therefore, an

Va2+ a2; therefore, a„ = √a2 — a2. Substituting the value of a2 = (a2+ a2), and the value of

a, just found, we have as the value of the normal acceleration,