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Unformatted text preview: l=t$t=tﬁrno
ll Ill.l nit 3] all .‘1ll El
11 ‘t (9) Tangent. at P1 whose
x (m) slope is 112: 21.0 nits FIGURE 1]! Example 23.
(a) Engine traveling on a straight track.
(b) Graph otxvst: J: = Alt2 + B. Given 1: as a function of t. A jet engine moves along an
experimental track (which we call the J: aids) as shown in Fig, 2—13a. We will treat
the engine as if it were a particle. Its position as a function of time is given by the
equation I = Al2 + B, where A = 2.1tlm1’s2 and H = 2.80m, and this equa
tion is plotted in Fig 2—13b. (1:) Determine the displacement of the engine during
the time interval from t1=3.005 to I»; = 5.00s. {b} Determine the average
velocity during this time interval. (17) Determine the magnitude of the instanta
neous velocityI at t = 5.00s. APPROACH We substitute values [or t1 and t2 in the given equation for x to obtain
x1 and 1:3. The average velocity can be found front Eq. 2—2. We take the deriva tive of the given 1: equation 1with respect to t to find the instantaneous velocity,
using the formulas just given. SOLUTION [a] At :1 = 3.00s, the position (point P1 in Fig.2—13b] is 11. = .41? + B = (atom/1mm? + 2.30111 = 21.7111.
At t; = 5.00 s, the position (P1 in Fig. 2—13h) is x1 = (2.11::111/11=}{s.tio1)l + 2.80m = 55.3111.
The displacement is thus I; — :1 = 55.3rn — 21.7m = 33.6m. {h} The magnitude of the average velocity can then be calculated as 11.11 x1  x1 33.6m ”=E=l2t1 Ems
This equals the slope of the straight line joining points P1 and P2 shown in
Fig. 2—13b.
(e) The instantaneous velocity at t = t1 = 5.003 equals the slope of the tangent
to the curve at point P: shown in Fig 2—13b.We could measure this slope off the graph to obtain 1);. But we can calculate an more precisely for any time i. using
the given formula x=A12+B, = 15.8 m/s. which is the engine’s position x as a function of time t. We take the derivative of
x with respect to time (see formulas at bottom of previous page): _£=£ 1 =
11—!” d!(ar+a) am. We are given A = 2.10nt/s3, so for .t = t2 = 5.00s.
113 = 2.41 = 2(2.10mfs2](5.00s.) = atom/1. Acceleration given I“). A particle is moving in a straight
line so that its position is given by the relation J: = (2.1K! mfszit2 + (1mm), as
in Example 2—3. Calculate (it) its average acceleration dun'ng the time interval from
t, = 3.0th to II = 5.00 s, and {it} its instantaneous acceleration as afunction of time. APPENCH To determine acceleration, we ﬁrst must ﬁnd the velocity at 1‘1 and 33;
by differentiating x: t: = drfdf. Then we use Eq. 2—5 to ﬁnd the average
acceleration, and Eq. E—IS tn find the instantaneous acceleration.
SOLUTION {a} The velocity at any time t is
if): d u = E = E [massive]:1 + ass :11] = (4.21:: Infszjt,
as we saw in Example 2—312. Therefore, at t1 = Sﬂﬂ s, 1:1 = {4.20 misijfiiﬂtl s} =
12.6 mfs and at I; = 5H} s, u; = 21.!) me. Therefore, 2 21.0 nu's — 12.6 refs — = = — = 2
a as sass — sins 4‘10"”5'
[b] With it = (42$! misﬁt, the instantaneous acceleration at any time is
_ ﬂ = i 2 = 2
a — m {it [(4.3] mf's )t] 4.2L! n1fs . The acceleration in this case is constant; it does not depend on time. ﬁgure 2—13
shows graphs of (a) .I vs. I [the same as Fig. 2—13h}, {In} 11 vs. II, which is linearly
increasing as calculated above, and (c) e vs. I, which is a horizontal straight line
because a = constant. Rurlwav design. You are designing an airport for small
planes. Cine kind of airplane that might use this airfield must reach a speed before takeoff of at least 218 mg’s {IUD Irmfh}, and can accelerate at 2.02] mfsz.
(a) If the runway is 150 m long, can this airplane reach the required speed [or
takeoﬁ‘? {b} If not, what minimum length must the runway have?I ﬁPPHﬂACH The plane’s acceleration is constant, so we can use the kinematic
equations for constant acceleration. In [a], we want to find a, and we are given: Km Walled
an = t] u
a. = n
x =15tlm
e = liltingts1 SDLUTIDH {a} Of the above four equations, Eq. 2—12c will give us 1: when we
know 1.1.3., o, x, and x0:
1:: = 1:3 + Eafix — xﬁ}
= u + statiomissﬂisomj = enemies in = "tittitlv‘lniglfs2 = 24.5my’s. This runway,r length is not sufficient.
[b] How we want to find the minimum length of runway, .14: — xﬂ, given
a: = 213 mills and a = 2.01} mfsz. So we again use Eq. 2—12c, hut rewritten as
{ )_ nz—trﬁ _ (2?.ﬂmfsf—[i
" 1” 2a static naval]
A EDDm runway is more appropriate for this plane. MUTE We did this Example as if the plane were a particle, so we round off our
answer to 20¢] n]. = 193 to. Falling from I tuner. Suppose that a hall is dropped [to = ll) from a tower Tiltlm high. How far will it have fallen after a time
t1 = Littler; = hilt] si and r, = 3.00 s? Ignore air resistance. AFFIDACH Let us take it as positive downward, so the acceleration is
a = g = +9.8llmfsz. We set on = ii and ya = (I. We want to ﬁnd the position v of the hall after three different time intervals Equation 2—12h, with .r replaced
by y, relates the given quantiﬁes (t, a, and an) to the unknown y. SDLUHUN We set t = t1 = 11le in Eq.2—12h: 321 = anti + set? = [l + set? = %{9.8ﬂmj51][l.ﬂﬂs)2 = 4.90111.
rIhe hall has fallen a distance of 4.90m during the time interval t=ll to
II = 1.00 s. Similarly. alter 2.00s (= t1}. the halls position is gs = as = %[9.3[lmiszl(2.ﬂﬂs}2 = 19am.
li‘inallv1 after 3.00s [= t3). the hall’s position is (see Fig. 2—29) h = so; = emancipation? = 44.1111. Thrown down from a trim Suppose the hall in Example 214
is thrown downward with an initial velocit].r of 3.01} mfs, instead of being dropped.
(a) What then would be its position after Mills and son s? (it) What would its
speed he alter tiltls and 2.011s? Compare with the speeds of a dropped hall. APPROACH Again we use Eq. 2—12hI but now on is not zeroI it is on = Sill] mfs.
SOUND“ (a) At t = 1.005, the position of the ball as given by Eq. 2—le is
y = out + to? = (3.Dﬂm/s}[1.ﬂils} +%[9.8[Im,’s1}{1.ﬂlls}3 = 1.90m.
At t = son a, {time interval t = I] to t = 2.00 s],the position is
y = not +%ot1 = (soc monsoon +§[s.som{51}(2nos}= = 25.6w.
As expected, the hall falls farther each second than if it were dropped with on = ll.
(it) The velocity is obtained from Eq. 2—12a7
o = on + at
= Iiilllnifs + (Eddmfsglﬂmsj = 12.3mfs [at a = lilils]
= son this + (9.8ﬂmfszl{2.tlﬂs) = memo. [sirI = sons]
in Example 2—14, when the hall was dropped (on = D], the first term (on) in
these equations was acre, so
a = [l + at
= [9.80 misllllﬂlls) = 9.30 mfs [at t1 = 1.005]
= [v.somrstjaoos) = 19.5mfs. [at a = one s]
HDTE For both Examples 2—14 and 2—15, the speed increases linearly in time by 9.3!} mfs during each second. But the speed of the downwardlyr thrown hall at any
instant is always 3.00 nifa (its initial speed} higher than that of a dropped ball. FIGURE 219 Example 214. {a} An object dropped from a tower
falls with progressively greater
speed and covers greater distance
with each successive second. [See also Fig. 2—25.] (h) Graph of 3; vs r. Acceleration
dueto
gravity .
__ ____y=ﬂ
yl=4£lﬂm
(storms
i " ﬂ. v2=l9om
H [II [After 2.005]
y3=44.l m
(Afteriiﬂilsj m Ball tltl'own ward. I. A person throws a hall upward into the
air with an initial velocity of 15.0 mfs. Calculate {a} how high it goes. and {it} how
long the hall is in the air before it comes back to the hand. Ignore air resistance. APPROACH We are not concerned here with the throwing action, but only,r with the motion of the hall eﬂer it leaves the thrower’s hand (Fig. 2—31)} and until it
comes back to the hand again. Let us choose 1' to be positive in the upward direc—
tion and negative in the downward direction. [This is a diﬁerent convention from
that used in Example; 2—14 and 2—15, and so illustrates our options.) The acceleration
due to gravity is doomward and so will have a negative sign, a = —g = —9.80 ms“.
As the hall rises, its speed decreases until it reaches the highest point (B in Fig. 2—30].
where its speed is zero for an instant; then it descends, with increasing speed. SULUTIUN [a] We consider the time interval from when the ball leaves the
thrower’s hand until the hall reaches the highest point. To determine the
maximum height, we calculate the position of the ball when its velocity.r equals
zero to = (It at the highest point}. At r = I] {point A in Fig. 2—3:!) we have
Jig. = I]. on = 15.1] mfs, and a = 5.Eiirn{'sz. At time t {maximum height],
1.! = D, a = 9.ﬂﬂ mfsf, and we 1wish to ﬁnd 3;. We use Eq. 2—12c, replacing it
with y: a2 = a3 + Zay. We solve this equation for y: i — at = a — {15o mtg)?
2a 2[9.B[mj's1} The hall reaches a height of 11.5 In above the hand. {it} Now we need to choose a different time interval to calculate how long the
hall is in the air before it returns to the hand. We could do this calculation in two
parts by ﬁrst determining the time reguired for the hall to reach its highest point,
and then determining the time it takes to fall back down. However. it is simpler
to consider the time interval for the entire motion from A to B to C (Fig. 2—3D}
in one step and use Eq. 2—12h. We can do this because y represents position or
displacement, and not the total distance traveled. Thus. at both points A and CL
y = ft. We use Eq. 2—12h with a = —9.Bll nifs2 and find 11 = 11.5111. y = r = at+wt+iatz ll = ll + (15.0 mfalt + ﬂ—Elﬁilmfszltz.
This equation is readily factored (we factor out one I}: {1511me — memoir): = c. There are two solutions: 15.1] Infs
t=D dt=—=3.tlE.
an 4.9[1 this2 5
The ﬁrst solution {I = [t]: corresponds to the initial point (A) in Fig. 230, when
the hall was ﬁrst thrown tom 3; = [1. The second solution, t = ans 5. corresponds
to point C, when the ball has returned to y = [1. Thus the hall is hi the air 3.065. NUT E We have ignored air resistance, which could he signiﬁcant, so our result is
only.r an approximation to a real, practical situation. Ball thrown upward at edge of cliff. Suppose that the person of Examples 216, 218, and 219 is standing on the edge of a eliﬂ, so
that the hall can fall In the base of the cliff Sﬂﬂm below as in Fig. 2—32. [st] How long does it take the hall to reach the base of the cliﬁ‘? [b] What is the total
distance traveled by the ball? Ignore air resistance (likely to he signiﬁcant, so
our result is an approximation). APPROACH We again use Eq. 2—12h, but this time we set 3;: = —5ﬂ.t] in, the
bottom or the curt, which is. sooth below the initial position (ya = o]. SOLUTION (ti) We use Eq. 2—12h with it = 9.Bﬂnt,lsz, tin = liﬂmfs, p0 = i},
and y = 5Et.lim: y = )h + tint + that:
5t].[itn = t] + (15.0mm)!  H9130 mfszitl.
Rewriting in the standard form we have
(esomxsihi — {lSﬂmith  {seem = it. Using the quadratic formula. we find as solutions t = 5.0“! s and t = 2.01 s.
The ﬁrst solution, t =51)”, is the answer we are seeking: the time it takes
the hall to rise to its highest point and then fall to the base of the cliff.
To rise and fall back to the top of the cliff took 3.66s (Example 2—115]; so it
took an additional 2.01 s to tail to the base. But what is the meaning of the other
solution, t= —2.IZ11 s? This is a time before the throw. when our calculation
begins, so it isn't relevant herefr (it) From Example 246, the hall moves up 11.5 to, falls 11.5 in back down to the
top of the cliff, and then down another Sillirn to the base of the cliff, for a total
distance traveled oi T31] rn. Note that the displacement, however. was 40.0111. Figure 2—33 shows the 1! its. 1‘ graph for this situation. Integrating a timevarying acceleration. An experimental
vehicle starts from rest [on = [2] at t = ﬂ and accelerates at a rate given by
rt = (1m mfssh. 1What is {it} its velocity and [it] its displacement 2.00s later? APPROACH We cannot use Eqs. 2—] 2 because a is net constant. We integrate the
acceleration a = [111de over time to ﬁnd 1: as a function of time; and then
integrate t! = :11de to get the displaeement. SOLUTIUIN From the definition of acceleration, a = deﬁcit, we have
tit: = a tit. We take the integral of both sides from t: = [l at t = D to ‘treloeitgtr it at an arbi—
tram»r time t: til .I'
Jae Jen‘t
El El 1: = thi't'llﬂ mfﬂt cit
n [Tilt] mf§}(%2) I t} 2 {am whim; — o) = (soomﬁhz. At t = 2.00 s. t: = [3.50 mfs3}(2.t}t] s? = 14.0 nuts. {in} To get the displacement, we assume x“ = t] and start with a = deﬁcit which
we rewrite as air = it sit. Then we integrate from a: = It] at t = t] to position xattimet:
1! l
de = Jodi
e e
1'3 Ellis 1m:
x = Jﬁﬁﬂmfsﬂtzdt = [3.Sﬂmfs3JE = 9.33m.
U El In sum, at r = 211th, ti = 14.0m/s and x = 9.33 in. three short trips. An airplane trip involves three legs, with
two stopovers, as shown in Fig. 3—14a. The first leg is due east for till} km; the second leg is southeast (45“) for 440m; and the third leg is at 53" south of west,
for 550 ion, as shower. What is the plane’s total displacement? APPROACH We follow the steps in the Problem Solving Strategy above SOLUTION 1. Draw a diagram such as Fig. 3—14a. where 131 , ﬂ), and ﬁg represent the three
legs of the trip, and ﬁg is the planet total displacement. 2. Choose ates: Ares are also shorten in Fig. 3—14a: n is east, y north. 3. Resolve oomponents: It is imperative to draw a good diagram. The components
are drawn in Fig. SMb. Instead of drawing all the vectors starting from a
common origin, as we did in Fig. 3—13h, here we draw them “tailtotip” style,
which is just as valid and may make it easier to see. 4. Calculate the components; ﬁ1:D1x — +D1cosil" = D1 = $ka
D1}. = +D1sintl° = llltui nan” = +D2cos45° = +[44tlltm](tl'ltt't]= +311ltrn
ox! = oases: = —[ttatm){o.)=rar —311m1
li D351)” = Dgeos53” = [55[tlrm 0.602]: 331lnn
D3]. =  D3 sin 53“ =  (550 km)(ll.l'99]= 439 Iron.
We have given a minus sign to each component that in Fig. 3—14]: points in the  x or  y direction. The components are shown in the Table in the margin.
5. Add the components: We add the x eomponents together, and we add the FIGURE 3" Example 3—3. m y components together to obtain the x and y components of the resultant:
Iilﬂll film} Dr: Du+Dn+D3g = ﬁﬁﬂkm+311km—331ltm = IStltlltm
rt1 are a = ”or +02? +03? = dim —311km—439km = —tsotm.
91 3‘11 '3“ The x and y components are hill] km and 450 km, and point respectively to
W the east and soutthis is one way to give the answer.
W 6. Magnitude and direction: WeI can also give the answer as
=\/ D§+ [$632+ {—Tjtljzkm = 95ka
land = g: = Salim = 1. 25 see = 51°.
Thus, the total displacement has magnitude 960 lrrn and points 51“I below the
53 Cl'lAPTER 5 .1: axis (south of east], as was shown] in our original sketch, Fig. 3—l4a. Poslﬂon given as a function of time. The position of a
particle as a function of time is given by i = [(5.0 Ito’s]! + {ﬁﬂmfszh‘zli + [[10 cu) — [lilmfsajtali where r is in meters and t is in seconds. [o] What is the particle's displaoenient
between :1 = lils and I: = 3.0 s? (b) Determine the particle‘s instantaneous
veintil)" and acceleration as a function of time. [a] Evaluate ir’ and i at t = 3.0 s. APPRDACH For {a}, we ﬁnd iii" = f;  f1, inserting £1 = ills for ﬁnding i=1 i
and E1 = 3.133 for ii. For [it], we take derivatiVEs {Eqs 3—9 and 3—11], and for
{e} we substitute t = 311s into our results in {b}. sotunou {a} a: :1 = 2.0a,
a] = [(5.0 mfsﬂlﬂs} + {asmxstttassﬂt + [{Tﬂm} — {asmxﬁgznnﬂj
= [34m]i — {1am}.
Similarly, at t; = 311s, i1 = {15m + 54m}i + ("film — s1 mji = {69m}i — (Tami.
Thus as = szs1 = {ﬁQmMmji+[T4rn+l?ru]j = [35m}i(5Trujj. That is, ex = 35 In, and n} = 5".r' m.
(b) To ﬁnd 1velocity+ we take the derivative of the given it with respect to time+
noting [Appendix 3—2) that d[t1]fdt = 2:, and atrijfat = 3:1: it = g = [5.u mfs + (12 111(51):]? + [a — [9.umfs3}t=]i.
The aeceleration is [keeping only two signiﬁcant ﬁgures):
it = g = [limiﬂi —[1Smfs3]lti. Thus a; = unit's2 is eonstant; but a}, = —[lﬂmfs3jt depends linearly on
time, increasing in magnitude with time in the negative 3! direction.
{c} We substitute t = lﬂs into the equations we just derived for i and i:
a = [Sﬂmfs + Eﬁmfsﬁ — (s1 mfsji = {41 mmi — [s1 mm
a = (12 maul — [s4 mat)3“.
Their magnitudes at t = 3.[s are at = “V l[411n,u’s]u2 + {31 nt,.:"s]l2 = 91 refs, and
a = "v [12 ruff): + {54 Intel]1 = 55 111,332. Driving off I cliff. A movie stunt driver on a motorcycle
speeds horizontally off a Stillmhigh cliff. How fast must the motorcycle leave
the cliff top to land on level ground below, S‘llﬂm from the base of the cliff where
the cameras are? Ignore air resistance APPROACH We explicitly follow the steps of the Problem Solving Strategy above.
SOLUTION 1. and 2. Read, choose the object, and draw a diagram. Our object is the
motorcycle and driver, taken as a single unit.The diagram is shown in Fig. 3—23. 3. Choose a coordinate system.We choose the y direction to be positive upward,
with the top of the clilf as Jii = t}. The x direction is horizontal with x0 = U
at the point where the motorcycle leaves the elilt 4. Choose a time Interval. We choose our time interval to begin [t = 0) just as
the motorcycle leaves the cliff top at position In, = 0, )3 = 0; our time
interval ends just betore the motorcycle hits the ground below. 5. Examine .r and y motions. to the horizontal (x) direction, the acceleration
ex = i], so the velocity is constant. The value of in when the motorcycle
reaches the ground is x = +901) In. In the vertical direction, the accelera—
tion is the acceleration due to gravity, a}, = g = 9.30 mfsz. The value of
y when the motorcycle reaches the ground is y = —5{].Um. The initial
velocity is horizontal and is our unknown, on; the initial vertical velocity is
zero, ow = 0. 6. List knowna and unknowna. See the Table in the margin. Note that in addition
to not knowing the initial horizontal velocity a”, [which stays constant until
landing}, we also do not know the time t when the motorcycle reaches the
ground. 7. Apply relevant equations. The motorcycle maintains constant oJr as long as it is in the air. The time it stays in the air is determined by the y motion—when it
hits the ground. So we ﬁrst find the time using the y motion, and then use this
time value in the 1' equations Tb find out how long it takes the motorcycle to reach the ground below, we use Eq. 2—12b (Table 3—2) for the vertical (y)
direction with ya = 0 and ”I“ = 0:
y = g] + 1.5.0! + §riyt1
= n + e + ﬂ—gjtz
or
y = eat".
We solve fort and set y = 5CI.D tn: 2 2 50.0m
r= .i—y = ,il—g = 3.19s.
'3' 9.Sﬂmy’s
To calculate the initial velocity1 the: we again use Eq. 2—12b, but this time for the horizontal (x) direction, with a; = D and I.) = I]: a: = x0+trmt+ilaxt2 = t] + amt + t]
or
.t: = vied.
Then
x 90.0m
= — = = 28.2 ,
v” r 3.195 "1"5 which is about 100 kmfh (roughly 60 mifh). NOTE In the time interval of the projectile motion, the only acceleration is g in
the negative y direction. The acceleration in the x direction is zero. FIGURE 313 Example 3—6. a=n=0
x = Sillﬂm
y = —5{i.tint
ox =
a}. = g = §Lﬂlllmtis2
3}”! = 0 Unknown 1's!) FIGURE 3!!! Example 3—7. @ansms APPLIED Sp arts J" vy=0atthispoint _________ ﬁLer‘: vyn _!
r‘ 7*“ ti=E=s.l A kicked football. A football is kicked at an angle it.) = 3?.0"
with a velocity of 20.0 mls, as shewn in Fig, 3—24. Calculate (a) the maximum
height, (b) the time of travel before the football hits the ground, (e) hov.r far away
it hits the ground, ((1) the velocity vector at the maximum height, and (e) the
acceleration vector at maximum height. Assume the ball leaves the foot at
ground level, and ignore air resistance and rotation of the ball. APPROACH This may seem difﬁcult at ﬁrst because there are so many questions.
But we can deal with them one at a time. We take the y direction as positive
upward, and treat the x and y motions separately. The total time in the air is again
determined by the y motion. The x motion occurs at constant velocity. The y
component of velocity varies, being positive (upward) initially. decreasing to zero
at the highest point, and then becoming negative as the football falls. SOLUTION W...
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 Fall '15
 hossu
 Physics

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