In cases where the integration is excessively awkward or difficult, integra- integra-tion by graphical, numerical, or computer methods can be utilized

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The position coordinate of a particle which is confined to move along a straight line is given by s 2t3 — 241 + 6, where s is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0, (6) the ac-celeration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the interval from t = 1 s to i = 4 s.

Solution. The velocity and acceleration are obtained by successive differentia-tion of s with respect to the time. Thus,

[[/ = s] v = 6i2 - 24 m/s [a = v] a = 121 m/s2

(b) Substituting v 30 m/s into the expression for v gives 30 St2 — 24, from which the positive root is t 3 s, and the corresponding acceleration is

a = 12(3) = 36 m/s2 Ans.

(a) Substituting v 72 m/s into the expression for v gives us 72 = dt2 — 24, from which t ±4 s. The negative root describes a mathematical solution for t before the initiation of motion, so this root is of no physical interest. Thus, the desired result is

Aras.

(c) The net displacement during the specified interval is

i s = s4 — s1 or

is = [2(43) - 24(4) + 6] - [2(13> - 24(1) + 6]

= 54 m Ans.

which represents the net advancement of the particle along the s-axia from the position it occupied at t 1 s to its position at t ~ 4 s.

To help visualize the motion, the values of s, v, and a are plotted against the time t as shown. Because the area under the v-t curve represents displacement, we see that the net displacement from t — 1 s to 1 4 s is the positive area Asa 4

less the negative area As1_2.

Helpful Hints

(T) Be alert to the proper choice of sign when taking a square root. When the situation calls for only one an-swer. the positive root is not always the one you may need.

© Note from the graphs that the val-ues for t? are the slopes (s ) of the s-t curve and that the values for a are the slopes '.v) of the v-t curve. Sug-gestion: Integrate v dt for each of the two intervals and check the answer for As. Show that the total distance traveled during the interval t 1 s to / 4 s is 74 m.

2 8 C h a p t e r 2 K i n e m a t i c s o f P a r t i c l e s

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Helpful Hints A particle moves along the x-axis with an initial velocity o, 50 ft/sec at the

origin when f 0. For the first 4 seconds it has no acceleration, and thereafter it is acted on by a retarding force wliich gives it a constant acceleration ax —10

ft/sec2. Calculate the velocity and the x-coordinate of the particle for the condi- (T) Learn to be flexible with s> mbol

© tions of t 8 sec and t — 12 sec and find the maximum positive £-coordinate The position coordinate a is just as reached by the particle. vaEd as s.

Solution. The velocity of the particle after t 4 sec is computed from

© J dv = J a dfj J dvx = - 1 0 j dt vx = 90 - lOf ft/sec

and is plotted as shown. At the specified times, the velocities are

t = 8 sec, vx = 90 - 10(8) = 10 ft/sec

t = 12 sec, vx = 90 — 10(12) = - 3 0 ft/sec A] is.

The x-coordinate of the particle at any time greater than 4 seconds is the tance traveled during the first 4 seconds plus the distance traveled after the dis-continuity in acceleration occurred. Thus,

/ ds = jvdt X = 50(4) + J (90 - 10i) dt -512 + 90i - 80 ft

For the two specified times,

t = 8 sec, .v = - 5 ( 82) + 90(8) - 80 = 320 ft

t = 12 sec, * = —5(122) + 90(12) - 80 = 280 ft A/is.

The .r-coordinate for t 12 sec is less than that for t = 8 sec since the motion is in the negatives-direction after t 9 sec. The maximum positivex-coordinate is, then, the value of x for t 9 sec which is

f, sec

. = -5(92) + 90(9) - 80 = 325 ft A J I S .

These displacements are seen to be the net positive areas under the v-t graph up © Show\ the tol tl distance traveled to the values of f in question. by the particle in the 12 sec is 370 ft.

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The spring-mounted slider moves in the horizontal guide with negligible friction and has a velocity tk in the s-direction as it crosses the mid-position where s 0 and f = 0. The two springs together exert a retarding force to the motion of the slider, which gives it an acceleration proportional to the displace-ment but oppositely directed and equal to a = —k2s, where k is constant. [The constant is arbitrarily squared for later convenience in the form of the expres-sions.) Determine the expressions for the displacement s and velocity v as func-tions of the time t.

vVVVVVVW

VWWWW

Solution I. Since the acceleration is specified in terms of the displacement, the differential relation v du a ds may be integrated. Thus,

a ri™

(T) We have used an indefinite integral here and evaluated the constant of integration. For practice, obtain the same results by using the definite integral with the appropriate limits.

The plus sign of the radical is taken when v is positive (in the plus ,s-direct! on).

This last expression may be integrated by substituting v dsidt. Thus,

| - = = I dt + Cn a constant, or - sin 1 — = t + C.,

jv<? - ¿V k v^u

With the requirement of t 0 when s 0, the constant of integration becomes C2 = 0, and we may solve the equation for s so that

@ Again tiy the definite integral here as above.

s = — sin kt h ^Ans.

The velocity is v s, which gives

Ans.

Solution II. Since a s. the given relation may be written at once as

s + k^s = 0

This is an ordinaiy linear differential equation of second ol der for which the so-lution is well known and is

s = A sin Kt + B cos Kt

where A, B. and K are constants. Substitution of this expression into the differ-ential equation shows that it satisfies the equation, provided that K k. The ve-locity is v = s, which becomes

© This motion is called simple har-monic inotion and is characteristic of all oscillations where the restoring force, and hence the acceleration, is proportional to the displacement but opposite in sign.

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A Freighter is moving at a speed of 8 knots when its engines are suddenly (T) stopped. If it takes 10 minutes for the freighter to reduce its speed to 4 knots,

de-termine and plot the distance s in nautical miles moved by the ship and its speed v in knots as functions of the time t during this interval. The deceleration of the ship is proportional to the square of its speed, so that a -kv2.

Helpful Hints

(T) Recall that one knot is the speed of one nautical mile (6076 ft) per hour.

Work directly in the units of nauti-cal miles and hours.

Solution. The speeds and the time are given, so we may substitute the expres-sion for acceleration directly into the basic definition a dvidt and integrate.

Thus,

©

dt v2

1 . 1 .

• = ~kt v 8

rs-w

JS V^J0

1 + 8kt

Now we substitute the end limits of v = 4 knots and t ¿p = g hour and get

8 ,, 3 •-] .. .. 8

4 ==

1 + &H1/6)

^{k = • ~ mi}4 ¹

1 + 6 (

^Ans.

The speed is plotted against the time as shown.

The distance is obtained by substituting the expression for v into the defi-nition ii ds/dt and integrating. Thus,

8 1 + 6f

<fe

dt f' 8 dt fs

J A 1 + 6F J

U ds s = | In (1 + 6f) Alis.

The distance s is also plotted against the time as shown, and we see that the ship has moved through a distance ,s ' In (1 + : | In 2 = 0.924 mi (nautical) dur-ing the 10 minutes.

J

a ⁴

4 6 t, min 10

t, mill

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P R O B L E M S

In document Engineering Mechanics Dynamics 6th Edition (Page 35-40)