We know that if s represents distance from a certain point at the time t, then Dts v, and hence, in this example,

=

Dts = 4t.

The required function, s, is therefore one whose derivative with respect to t is 4t. One such function is 2ť2, and adding the constant of integration (page 302), we have

S 2t2 + C.

(1)

The value of the constant of integration may be determined if we know a pair of values of t and s, and these depend on when and where the ball starts to roll. We shall assume that t = 0 when the ball begins to roll. The distance s is measured from some chosen station O.

(a) The distance the ball will roll in 2 seconds is the same as its distance from the starting point after 2 seconds. Hence for this part of the problem we choose the station O at the starting point, so that s = O when t 0. Substituting these values in (1) we find that C = 0. Hence, substituting this value of C in (1), the distance from the starting point after t seconds is S 2t2.

=

Then at the end of 2 seconds, s = 2 × 22 = 8 feet.

(2)

(b) Take O at the upper end of the plane. Since the ball starts 4 feet from O, we have s = 4 when t 0. Substituting these values in (1), we find that C 4. Setting this value of C in (1), the distance from the upper end of the plane to the ball after t seconds is

After 1.5 seconds,

S= 2t2 + 4.

s =

2(1.5)2 + 4 = 8.5 feet.

(3)

That is, the ball is 8.5 feet from the upper end of the plane after 1.5 seconds.

(c) If O is taken at the lower end of the plane and the ball placed 12 feet above it, then s - 12 when t C (s being negative since the positive direction for s is the same as that of v, namely, down the plane). Substituting in (1) we find C= 12, so that the distance from the lower end of the plane at any time is

=

C-4

-18:

6.

9.

If t = 1, s - 10, that is, after one second the ball is 10 feet above the lower end of the plane. Graphically, the given relation v = 4t is represented by the straight line through the origin whose slope is 4. The graph of (1) is given for C = 0, 4, – 12,

FIG. 177.

2

3 t

which are the values of C in (2), (3), (4), respectively. The intercepts of these curves on the s-axis are the respective values of C, which give the distances from the station O to the ball when t = 0.

If the acceleration is given as a function of the time, t, by an equation of the form a = f(t), the velocity is obtained by integrating this equation and the distance by integrating the result. Two constants of integration are introduced in the process which are determined by given conditions as in

EXAMPLE 2. A balloon is ascending with a uniform velocity of 28 feet per second and at the height of 720 feet a ball is dropped. When will the ball strike the ground and with what velocity?

Let s be the height of the ball above the ground, and let t ball is dropped.

=

O when the

The direction of the acceleration of gravity, being toward the center of the earth, is opposite to the positive direction of s, and hence the acceleration is negative.

The initial conditions and required data are collected in the table. As soon as the ball leaves the balloon it is subject to the law

When the ball strikes the ground s = 0. Substituting this value of s in (5)

the negative value of t having no meaning in this problem. Substituting this value of t in (3)

1. A ball is rolled up a plane inclined at an angle of 10° with an initial velocity of 15 feet per second. Find the distance it rolls up the plane and the time that elapses before it returns to the starting point. (Suggestion: Find the component of the acceleration of gravity acting along the plane.)

2. Solve Example 2 of the preceding section if the balloon is descending. 3. A high jumper raises his center of gravity 3 feet. How long is he off the ground, and with what velocity does he light?

4. A baseball dropped from one of the windows of the Washington monument, 500 feet from the ground, has been caught. Compare the velocity with which it struck the catcher's hand with the velocity of 120 feet per second which is said to be the maximum velocity that a pitcher has imparted to a ball.

5. A man descending in an elevator whose velocity is 10 feet per second drops a ball from a height of 6 feet above the floor. How far will the elevator descend before the ball strikes the floor of the elevator?

6. In the preceding problem, suppose the elevator is ascending instead of descending.

7. A balloon is ascending with a velocity of 24 feet per second when a ball is dropped from it. The ball reaches the ground in 5 seconds. Find the height of the balloon when the ball was dropped. Determine the highest point that the ball reached.

8. An automobile reduces its speed from 35 miles an hour to 20 miles an hour in 8 seconds. If the retardation is uniform how much longer will it be before it will come to rest, and how far will it travel in this length of time?

9. How high will a ball rise if thrown vertically upward with an initial velocity of 60 feet per second?

10. A street car in going from one stop to another 400 feet distant is uniformly accelerated at the rate of 2 feet per second per second for a distance of 320 feet and then brought to rest with a uniform retardation. Find the time it took to go the 400 feet.

108. Motion in a Plane. If a particle moves along a curve in a plane, two rectangular axes are chosen and the position of the particle at any time determined by means of the coördinates of the particle.

If the coördinates of the particle on the axes are given by the equations

x = f(t) and y = F(t)

the position of the particle in the plane is determined.

(1)

The discussion of the motion of a particle in a plane is thus resolved into the discussion of two motions along straight lines. The components of the velocity, obtained by differentiating equations (1) are

Vx Dtx and v1y Dty.

(2)

They are represented in Fig. 178 by directed lines parallel to

the axes.

Since the components are perpendicular the magnitude of the resultant velocity, v, is given by the equation

and the direction of the velocity can be found from the equation

The components of the acceleration, obtained by differentiating equations (2) are

Ax

Divx

and ay Divy.

(5)

They are represented in Fig. 179 by directed lines parallel to the axes.

The magnitude of the resultant acceleration, a, is given by the equation

a

√ a2x + a2 y

and its direction can be found by means of the equation

(6)

(8)

EXAMPLE 1. If a particle moves in accordance with the law x t3,

y = ť2, find the equation of the path, the position of the particle when

2, and the magnitude and direction of v and of a at this point.

2

From the first equation t x and hence y = (x) = x, which is the equation of the path shown in Fig. 180. Differentiating the given equations with respect to t we obtain

Ꮎ

Ax= 6t, and ay

2

a = √36ť2 + 4,

tan = 2/6t = 1/3t.

(4)

(5)

(6)

12.64,

Substituting t = 2_in_equations (2), (3), (5), (6), we find v = 0 = 18°.43, a = 12.17, † = 9°.46, at the point where x = 8 and y 4. The inverse problem of determining the path of a particle, given the equations of the component accelerations, is illustrated in

EXAMPLE 2. Find the equation of the path of a projectile fired at an angle of 30° to the horizontal with a muzzle velocity of 1200 feet per second.

components of the initial velocity are 1200 cos 30°, and 1200 sin 30°, respectively.

If the resistance of the air be disregarded there will be no horizontal acceleration.

From the instant the projectile leaves the gun gravity is acting on it vertically in the direction opposite to the positive direction on the y-axis. Hence the acceleration equations of the motion are