Kinematics by Mathalino

Kinematics by Mathalino.com

Particle is a term used to denote an object of point size. A system of particles which formed into appreciable size is termed as body. These terms may apply equally to the same object. The earth for example may be assumed as a particle in comparison with its orbit, whereas to an observer on the earth, it is a body with appreciable size. In general, a particle is an object whose size is so small in comparison to the size of its path.

Rectilinear Translation (Motion Along a Straight Line) Motion with constant velocity (uniform motion)

s=vt

Motion with constant acceleration

vf=vi+at s=vit+12at2 vf2=vi2+2as

Free-falling body

v=gt h=12gt2 v2=2gh Note: From motion with constant acceleration, set vi = 0, vf = v, s = h, and a = g to get the free-fall formulas.

Motion with variable acceleration

a=dvdt v=dsdt

vdv=ads

Where s = distance

h = height v = velocity vi = initial velocity vf = final velocity a = acceleration g = acceleration due to gravity (g = 9.81 m/s2 in SI = 32.2 ft/s2 in English) t = time Note: • a is positive (+) if

v is increasing (accelerate). • a is negative (-) if v is decreasing (decelerate). • g is positive (+) if the particle is moving downward. • g is negative (-)if the particle is moving upward. Useful conversion factors: From

To

Multiply by

Kilometers per hour (kph)

Meter per second (m/sec)

1 / 3.6

Meter per second (m/sec)

Kilometers per hour (kph or km/hr)

Miles per hour (mph)

Feet per second (fps or ft/sec)

22 / 15

Feet per second (ft/sec)

Miles per hour (mph or mi/hr)

15 / 22

3.6

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Problem 1002 On a certain stretch of track, trains run at 60 mph (96.56 kph). How far back of a stopped train should be a warning torpedo be placed to signal an oncoming train? Assume that the brakes are applied at once and retard the train at the uniform rate of 2 ft/sec2 (0.61 m/s2). Solution: English System Initial velocity

vi=60 mph=60(2215) vi=88 ft/sec vf2=vi2+2as 02=882−2(2)s s=1936 ft

answer

SI units Initial velocity

vi=96.56 kph=96.56(13.6) vi=26.82 m/sec vf2=vi2+2as 02=26.822−2(0.61)s s=589.7 m

answer

Problem 1003 A stone is thrown vertically upward and return to earth in 10 sec. What was its initial velocity and how high did it go?

Solution:

Return in 10 seconds = 5 seconds upward + 5 seconds downward SI Units Going upward (velocity at the highest point is zero):

vf=vi−gt 0=vi−9.81(5) vi=49.05 m/sec

answer

Going downward (initial velocity is zero; free-fall):

h=12gt2 h=12(9.81)(52) h=122.625 m

answer

English System Going upward (velocity at the highest point is zero):

vf=vi−gt 0=vi−32.2(5) vi=161 ft/sec

answer

Going downward (initial velocity is zero; free-fall):

h=12gt2 h=12(32.2)(52) h=402.5 ft

answer

Problem 1004 A ball is dropped from the top of a tower 80 ft (24.38 m) high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft/sec (12.19 m/s). When and where do they pass, and with what relative velocity?

Solution: English System: h = 80 ft vA = 0 vB = 40 ft/s g = 32.2 ft/s2 From A to C (free-fall)

h1=12gt2 h1=12(32.2)t2 h1=16.1t2 From B to C (upward motion) From the formula s = vit + ½ at2

h2=vBt−12gt2 h2=40t−12(32.2)t2 h2=40t−16.1t2 A to C plus B to C is equal to height of the tower

h1+h2=h 16.1t2+(40t−16.1t2)=80 40t=80 t=2 sec h1=16.1(22) h1=64.4 ft They pass each other after 2 seconds at 64.4 ft from the top of the tower.

answer

Velocity at C of stone from A (after 2 seconds)

vC1=gt=32.2(2) vC1=64.4 ft/s Velocity at C of stone from B (after 2 seconds)

vC2=vB−gt==40−32.2(2) vC2=−24.4 ft/s → the negative sign indicates that the stone is moving downward Relative velocity:

vr=vC1+vC2=64.4−24.4 vr=40 ft/sec

answer

SI Units: h = 24.38 m vA = 0 vB = 12.19 m/s g = 9.81 m/s2 From A to C (free-fall)

h1=12gt2 h1=12(9.81)t2 h1=4.905t2 From B to C (upward motion) From the formula s = vit + ½ at2

h2=vBt−12gt2 h2=12.19t−12(9.81)t2 h2=12.19t−4.905t2 A to C plus B to C is equal to height of the tower

h1+h2=h 4.905t2+(12.19t−4.905t2)=24.38 12.19t=24.38 t=2 sec h1=4.905(22) h1=19.62 m They pass each other after 2 seconds at 19.62 m from the top of the tower.

answer

Velocity at C of stone from A (after 2 seconds)

vC1=gt=9.81(2) vC1=19.62 m/s Velocity at C of stone from B (after 2 seconds)

vC2=vB−gt==12.19−9.81(2) vC2=−7.43 m/s → the negative sign indicates that the stone is now moving downward Relative velocity:

vr=vC1+vC2=19.62−7.43 vr=12.19 m/sec

answer

Problem 1005 A stone is dropped down a well and 5 sec later, the sounds of the splash is heard. If the velocity of sound is 1120 ft/sec (341.376 m/s), what is the depth of the well?

Solution

tstone+tsound=5 For tstone (free-falling body):

h=12gt2 t=2hg−−−√

For tsound (uniform motion):

h=vt t=hv Thus,

2hg−−−√+hv=5

English System

2h32.2−−−−√+h1120=5 2h32.2−−−−√=5−h1120 2h32.2=(5−h1120)2 10h161=25−h112+h21254400 11254400h2−1832576h+25=0 h=88759.73 and 353.31 For h = 88 759.73 ft

t=2(88759.73)32.2−−−−−−−−−−√ t=74.2 sec >5 sec

(not okay!)

For h = 353.31 ft

t=2(353.31)32.2−−−−−−−−√ $t = 4.68 \, \text{ sec } okay!)

Thus, h = 353.31 ft

→

answer

SI Units

2h9.81−−−−√+h341.376=5 2h9.81−−−−√=5−h341.376 200h981−−−−−√=5−125h42672

200h981=(5−125h42672)2 200h981=25−625h21336+15625h21820899584 156251820899584h2−16267756976872h+25=0 h=27065.05 and 107.64 For h = 27 065.05 m

t=2(27065.05)9.81−−−−−−−−−−√ t=74.2 sec >5 sec

(not okay!)

For h = 107.64 m

t=2(107.64)9.81−−−−−−−−√ $t = 4.68 \, \text{ sec } okay!)

Thus, h = 107.64 m

→

answer

Problem 1007 A stone is dropped from a captive balloon at an elevation of 1000 ft (304.8 m). Two seconds later another stone is thrown vertically upward from the ground with a velocity of 248 ft/s (75.6 m/s). If g = 32 ft/s 2 (9.75 m/s2), when and where the stones pass each other? Solution: English:

h=1000 ft vi2=248 ft/s g=32 ft/s2 Stone dropped from captive balloon (free-falling body):

h1=12gt2=12(32)t12 h1=16t12 Stone thrown vertically from the ground 2 seconds later

s=vi2t2−12gt22 h2=248t2−12(32)t22 h2=248(t1−2)−16(t1−2)2 h1+h2=h 16t12+[248(t1−2)−16(t1−2)2]=1000 16t12+248(t1−2)−16(t12−4t1+4)=1000 16t12+248t1−496−16t12+64t1−64=1000 16t12+248t1−496−16t12+64t1−64=1000 312t1=1560 t1=5 sec The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon. answer

h2=248(5−2)−16(5−2)2 h2=600 ft The stones will meet at a point 600 ft above the ground.

answer

SI Units: h=304.8 m vi2=75.6 m/s g=9.75 ft/s2 Stone dropped from captive balloon (free-falling body):

h1=12gt2=12(9.75)t12 h1=4.875t12 Stone thrown vertically from the ground 2 seconds later

s=vi2t2−12gt22 h2=75.6t2−12(9.75)t22 h2=75.6(t1−2)−4.875(t1−2)2 h1+h2=h 4.875t12+[75.6(t1−2)−4.875(t1−2)2]=304.8 4.875t12+75.6(t1−2)−4.875(t12−4t1+4)=304.8 4.875t12+75.6t1−151.2−4.875t12+19.5t1−19.5=304.8 4.875t12+75.6t1−151.2−4.875t12+19.5t1−19.5=304.8 95.1t1=475.5 t1=5 sec The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon.

answer

h2=75.6(5−2)−4.875(5−2)2 h2=182.925 m The stones will meet at a point 182.925 m above the ground.

Answer

Problem 1008 A stone is thrown vertically upward from the ground with a velocity of 48.3 ft per sec (14.72 m per sec). One second later another stone is thrown vertically upward with a velocity of 96.6 ft per sec (29.44 m per sec). How far above the ground will the stones be at the same level? Solution: English:

s=vit+12at2 h=vit−12gt2 h=vit−12(32.2)t2 h=vit−16.1t2 For the first stone:

h1=48.3t−16.1t2 For the second stone

h2=96.6(t−1)−16.1(t−1)2 h2=96.6(t−1)−16.1(t2−2t+1) h2=96.6t−96.6−16.1t2+32.2t−16.1 h2=−16.1t2+128.8t−112.7 h1=h2 48.3t−16.1t2=−16.1t2+128.8t−112.7 80.5t=112.7 t=1.4 s h=48.3(1.4)−16.1(1.42) h=36.064 ft

answer

SI Units

s=vit+12at2 h=vit−12gt2 h=vit−12(9.81)t2 h=vit−4.905t2 For the first stone:

h1=14.72t−4.905t2 For the second stone

h2=29.44(t−1)−4.905(t−1)2 h2=29.44(t−1)−4.905(t2−2t+1) h2=29.44t−29.44−4.905t2+9.81t−4.905 h2=−4.905t2+39.25t−34.345 h1=h2 14.72t−4.905t2=−4.905t2+39.25t−34.345 24.53t=34.345 t=1.4 s h1=14.72(1.4)−4.905(1.42) h=10.994 m

answer

Problem 1009 A ball is shot vertically into the air at a velocity of 193.2 ft per sec (58.9 m per sec). After 4 sec, another ball is shot vertically into the air. What initial velocity must the second ball have in order to meet the first ball 386.4 ft (117.8 m) from the ground?

Solution: English:

s=vit+12at2 a=−g=−32.2 ft/s2 s=386.4 ft Thus,

386.4=vit−16.1t2 First ball:

386.4=193.2t–16.1t2 t2−12t+24=0 t=9.46 and 2.5 Use

t=9.46 s

Second ball:

386.4=vi(t−4)−16.1(t−4)2 386.4=vi(9.46−4)−16.1(9.46−4)2 vi=158.67 ft/s

answer

SI Units

s=vit+12at2 a=−g=−9.81 m/s2 s=117.8 ft Thus,

117.8=vit−4.905t2 First ball:

117.8=58.9t–4.905t2 4.905t2−58.9t+117.8=0 t=9.47 and 2.54 Use

t=9.47 s

Second ball:

117.8=vi(t−4)−4.905(t−4)2 117.8=vi(9.47−4)−4.905(9.47−4)2 vi=48.36 m/s

answer

Problem 1010 A stone is thrown vertically up from the ground with a velocity of 300 ft per sec (91.44 m/s). How long must one wait before dropping a second stone from the top of a 600-ft (182.88-m) tower if the two stones are to pass each other 200 ft (60.96 m) from the top of the tower?

Solution: SI Units

Stone from the ground:

s=vit+12at2 h1=vi1t−12gt2 182.88−60.96=91.44t−129.81t2 4.905t2−91.44t+121.92=0 t=17.19 sec and 1.44 sec

Stone from the top of the tower: Let t2 = time to wait before dropping the second stone

h=12g(t−t2)2 With t = 17.19 sec

60.96=12(9.81)(17.19−t2)2 t2=13.67 sec With t = 1.44 sec

60.96=12(32.2)(1.44−t2)2 t2=−2.08 sec (meaningless) Use t2=13.67

sec

answer

English Stone from the ground:

s=vit+12at2 h1=vi1t−12gt2 600−200=300t−1232.2t2 16.1t2−300t+400=0 t=17.19 sec and 1.44 sec

Stone from the top of the tower: Let t2 = time to wait before dropping the second stone

h=12g(t−t2)2 With t = 17.19 sec

200=12(32.2)(17.19−t2)2 t2=13.67 sec With t = 1.44 sec

200=12(32.2)(1.44−t2)2 t2=−2.08 sec (meaningless) Use t2=13.67

sec

answer

Problem 1011 A ship being launched slides down the ways with constant acceleration. She takes 8 sec to slide (the first foot | 0.3048 meter). How long will she take to slide down the ways if their length is (625 ft | 190.5 m)?

Solution:

s=vit+12at2 where vi=0 Thus

s=12at2 English Units

SI Units

1=12a(82)

0.3048=12a(82)

a=0.03125 ft/sec2

a=0.009525 m/sec2

625=12(0.03125)t2

190.5=12(0.009525)t2

t=200 sec

t=200 sec

t=3 min 20 sec

answer

t=3 min 20 sec

answer

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Problem 1012 A train moving with constant acceleration travels 24 ft (7.32 m) during the 10th sec of its motion and 18 ft (5.49 m) during the 12th sec of its motion. Find its initial velocity and its constant acceleration

Solution: English:

vf=vi+at

24=vo+10a → equation (1) 18=vo+12a → equation (2) Equation (1) minus equation (2)

6=−2a a=−3 ft/sec2

answer

From equation (1)

24=vo+10(−3) vo=54 ft/sec

answer

SI Units

vf=vi+at

7.32=vo+10a → equation (1) 5.49=vo+12a → equation (2) Equation (1) minus equation (2)

1.83=−2a a=−0.915 m/sec2

answer

From equation (1)

7.32=vo+10(−0.915) vo=16.47 m/sec

answer

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Problem 1019 The motion of a particle is given by the equation and

s=2t4−16t3+2t2 where s is in meter

t in seconds. Compute the values of v and a when t=2 sec.

Solution:

s=2t4−16t3+2t2 v=dsdt v=8t3−12t2+4t a=dvdt a=24t2−t+4 When t = 2 sec

v=8(23)−12(22)+4(2) v=70 m/s

answer

a=24(22)−2+4 a=98 m/s2

answer

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