Position location of the object on the x-axis or y-axis with respect to the originunit: m, km, cm, mm, ft, mi

Displacement change in position

Velocity rate of change of positionunit: m/s, cm/s, kph, ft/s, mph

Acceleration rate of change of velocityunit: m/s2, cm/s2, ft/s2

v=d xdt

x= x f x i

a=d vdt

v2v1i a1

For motion in a straight line, there are only 2 directions to consider: towards +x and towards x (towards +y and towards y). For simplicity, towards +x direction is considered positive while towards x direction is considered negative.

It must be noted that negative acceleration does not necessarily mean deceleration. It simply means that the acceleration is heading in the negative direction. An object will slow down only when the acceleration's direction is opposite that of the motion.

The kind of motion experienced by an object depends upon the acceleration.

An object with variable acceleration is said to be undergoing motion with non-uniform acceleration.

An object with constant zero acceleration is said to be undergoing uniform motion.

Whereas an object with constant non-zero acceleration is said to be undergoing uniformly accelerated motion.

Motion with non-uniform acceleration occurs when the acceleration is not constant and varies with respect to another quantity. Typically, acceleration is a function of time.

Depending upon the equation and values given by the problem or situation, your analysis and solution will start eq. 1, eq. 2, or eq. 3.

a=a(t )=d vdtconstant (1)

v=v (t)=d xdt

(2)

x= x (t ) (3)

EXAMPLE 1An object starting at the origin (x = 0) undergoes motion described by the following velocity function:

where A = 3.0 m/s3B = 5.0 m/sC = 0.50 rad/s

a) What is the object's position at t = 5.0s? b) At what times will it have minimum & maximum acceleration?

v (t )=At2B sin Ct

v (t )=At2B sin Ct=d xdt

d x=At2 dtB sin Ct dtd x=A t2 dtB sin Ct dtx=A (1

3t3)B (cosCt )(1

C)

=A3

t3+BC

cosCt

x(t=5)=33(5)3+5

0.5cos(0.5(5))

=125+10 cos(2.5)=116.98856384 mx(t=5)=116.989m

a=d vdt

=ddt(At2Bsin Ct )

=A ddt(t2)B d

dt(sin Ct )

=2AtB cos Ct (ddt(Ct ))

a=2AtBC cosCt

amaxd adt

=0=2ABC (sinCt )(C )

2A+BC2 sinCt=0

sin Ct=2ABC2

=2(3)5(0.5)2

sin Ct=4.8NOT ALLOWED !There is no time value at w /cacceleration is maximum.

amincos=max=1=Ct=4 ,2 , 0,2 ,4 ,Ct=0 t=0

t a=max=NO VALUE

EXAMPLE 2A car travels in a straight line with the following position function:

where b = 2.4 m/s2 and c = 0.12 m/s3a) Calculate the averate velocity for the time interval ti = 5.0s to tf = 10.0s. b) Determine the rate at which acceleration changes with respect to position.

x(t)=bt 2ct3

x(t)=bt 2ct3

v=d xdt

=2bt3ct2

v i(t=t i)=2(2.4)(5)3(0.12)(5)2

=15m / sv f (t=t f )=2(2.4)(10)3(0.12)(10)

2

=14 m /s

vave=v i+v f

2=15+14

2=14.5m /s

vave=14.50m / s

v=d xdt

=2 bt3 ct2

a=d vdt

=2b6 ct

d adt

=06 c=6 c

t a=min=0.00

d adtd xdt

=d adtd x dt

=d ad x

= 6c2bt3ct2

d ad x

= 6c3ct22bt

An object with uniform motion has a constant acceleration of zero.

a=d vdt=constant=0

v=constant=d xdt=v i= v f= vave

v= x t

=x f x it ft i

Motion with uniform acceleration occurs when the acceleration is constant at a non-zero value.

The following strategies may be used when analyzing the motion of objects undergoing uniform acceleration:

Strategy 1Use Eq. (a) first. If not enough, use Eq. (b). This strategy is applicable for most problems in kinematics.

a=d vdt=constant0 (a)

v=d xdt

(b)

EXAMPLE 3A car travels in a straight line speeding up from 15.0 kph to 60.0 kph in 30 minutes. Determine the car's displacement.v0=15 kph=4.166666667m / sv=60 kph=16.66666667 m/ st0=0 t=30 min=1,800 s

a=d vdt

a dt=d v

a0t

dt=v0v

d v a(t0)= vv0

a=vv0

t=12.5 m / s

1,800 s=0.00694444 m/ s2

v=v0+a t=d xdt

d x=v0 dt+a t dt

x0x

d x=v00t

dt+at0t

t dt

x x0= x= v0(t0)+a(12)(t202)

x=(4.166666667)(1,800)+(0.006944444)(0.5)(1,800)2

=18,750 m x=18,750.00 m=18.75km

Strategy 2When the initial time t0 = 0 and the final time t = not given, Eq (a) and Eq. (b) may be combined to get rid of t.

a=d vdt

v=d xdt

av=

d vdtd xdt

=d vdt

dtd x

av= d v

d x

a d x= v d v

EXAMPLE 4A car travels 75.0 km speeding up from 15.0 kph to 60.0 kph. Determine the car's acceleration.v0=15kph=4.166666667 m /sv=60kph=16.66666667 m /sx0=0 x=75 km=75,000mt0=0 t=not given

av=

d vdtd xdt

a d x=v d v

a0xd x=v0

vv d v

ax0=12

v212

v02

ax cos0=12

v212

v02

a=v2v0

2

2x=260.4166668m /s

2(75,000 m)=0.001736111m/ s2

a=1.736103 m /s2

Strategy 3When the initial velocity v0 = 0 and the final time v = not given, Eq (a) and Eq. (b) may be combined to get rid of v.

a=d vdt

v=d xdt

a= ddt( d x

dt)

EXAMPLE 5A moving car accelerates from rest for 5.0 minutes traveling a total distance of 600.0 m. Determine the acceleration of the car.

v0=0 v=not givenx0=0 x=600 mt0=0 t=5min=300 s

a=d vdt=d

dt(d x

dt) a dt=d (d x

dt)

a0t

dt=d (d xdt )

a(t0)=d xdt

a tdt=d x

a0ttdt=0

xd x

a(12)(t202)= x0 a=2 x

t 2

a=2(600m)(300 s)2

=0.013333333m / s2

a=1.333102 m/ s2

In some cases, the system involves 2 or more moving objects. In the case of 2 objects, the following situations are commonly encountered:a) 2 objects moving towards each otherb) 2 objects moving away from each

otherc) 2 objects racing against each other

Whatever the case, analyze the motion of each object separately, with either the displacement or the time (or both) linking the motion of the objects together.

EXAMPLE 6An overspeeding car moving at a constant speed of 90.0 kph passes a parked police car, which then begins to move after the car. What should be the police car's acceleration such that it catches up with the car in 5.0 minutes?

v10 v2=constanta1

x1= x2 t1= t 2

v10=0 v1=not givenv2=90kph=25m/ s=constantx10= x10=0 x1= x2= xt10=t20=0t1=t2=t=5min=300 s

v2=constant=d xdt

d x=v2 dt

0x

d x=v20t

dt x0=v2(t0)x=(25 m / s)(300s)=7,500 m

a1=d v1dt

=ddt

d xdt

a1 dt=d (d xdt

)

a10t

dt= d(d xdt )

a1(t0)=d xdt

a1 tdt=d x

a10t

tdt=0x

d x

a1(12)(t202)= x0 a1=

2 xt2

a1=2(7,500 m)(300 s)2

=0.166666666 m /s2

a1=0.167m /s2

Freefall motion is a special case of uniformly-accelerated motion wherein motion is along the vertical axis and the only source of acceleration is gravity.

For locations on or near the surface of the earth, the acceleration due to gravity is given by

Freefall motion may be divided into 3 cases:a) dropped objectb) object thrown vertically downwardc) object thrown vertically upward

a=g

a=g ,towards y a y=gg=9.8m /s2

Dropped Object CaseThe object is simply released, let go, dropped, or falling from a particular height.

a=gv0=0 v=y0=0 y=

EXAMPLE 7A flower pot falls from a window 40.0 m above the sidewalk. How many minutes will it take for the flower pot to strike the sidewalk?v0=0 v=not giveny0=0 y=40mt0=0 t=not givena=9.8m/ s2

a=d vdt

a dt=d v

a0t

dt=0vd v a(t0)=v0

v=a t=d ydt

a tdt=d y

a0ttdt=0

40d y

a(12)(t202)=400

t2=2(40)

a=809.8

t=2.857142857 s=0.047619047min

t=0.048min

Object Thrown Vertically Downward CaseWhile its motion appears similar to the dropped object case, an object thrown vertically downward has non-zero initial velocity. It is thrown or fired giving it a downward initial velocity.

a=gv0= v=y0=0 y=

EXAMPLE 8Two students conduct an experiment on the roof of a building. One drops an egg. At the same time, the other student throws a baseball vertically downward giving it an initial speed of 5.0 m/s. If the difference between the times at which the two objects strike the ground is 0.5 s, what is the height of the building?v10=0 v1=not givenv20=5m / s v2=not giveny10= y20=0 y1= y2=ht10=t20=0 t1=not givent2=not givena1=a2=9.8m /s

2

t1 t 2=t1t2=0.5 s

a1=d v1dt

a1 dt=d v1

a10t 1

dt1=0v1

d v1 a1(t 10)= v10

a1 t1=d y1dt1

a1 t1 dt 1=d y1

a10t 1

t1 dt1=0h

d y1

a1(12)(t1

20)=h0

t1=2 ha1 =2hg t 2=2hg 0.5a2=

d v2dt2

a20t 2

dt2=5v2

d v2

a2(t20)=v2+5 v2=a2 t25=d y2dt2

0h

d y2=a20t2

t 2 dt250t 2

dt2

h0=a2(12)(t2

20)5(t20)

2h=g t 22+10 t2

2hg=t2

2+10g

t 2

2hg=(2hg 0.5)

2

+10g(2hg 0.5)

t12=(t10.5)

2+10g

t15g

t12=t1

2t1+0.25+10g

t15g

10g

t1t1=5g0.25

t1=12.75 s=2hgh=

g t12

2h=796.5562496m

h=796.556 m

Object Thrown Vertically Upward CaseThe object is given an non-zero initial velocity directed vertically upwards or towards +y.

a=gv0=+ v=y0=0 y=

v1

v2

v top=0

tup t down

v2=v1 tup= t down

EXAMPLE 9A toy gun set on the floor fires a ping-pong ball vertically upwards with an initial speed of 7.5 m/s. At what times will the ball be 2.0 m above the floor?v0=7.5m/ s v=not giveny0=0 y=2mt0=0a=9.8 m / s2

a=d vdt

a0t

dt=7.5v

d v

a(t0)=v7.5 a t+7.5=v=d ydt

a0ttdt+7.50

tdt=0

2d y

9.8(12)(t 202)+7.5(t0)=20

4.9 t 27.5 t+2=0

t=(7.5)(7.5)24(4.9)(2)

2(4.9) t1=1.186649445 s t2=0.343962799 s

t1st time=0.344 st2nd time=1.187 s

Projectile motion occurs when freefall motion is given a horizontal component. Since gravity remains the sole source of acceleration, the object experiences uniformly-accelerated motion along the y-axis and uniform motion along the x-axis.

ax=0 a y=9.8m /s2

Analyze the motion along the x-axis separately from the motion along the y-axis, with time uniting the two motions.

An object undergoing projectile motion is referred to as a projectile.

The path taken by the projectile is referred to as the object's trajectory.

Lastly, the maximum horizontal distance traveled by the projectile is called its range.

EXAMPLE 10A marble rolls off the edge of a level table top with a speed of 1.25 m/s. If the height of the table is 1.00 m determine the velocity magnitude at which the marble strikes the floor.v x=1.25m / sv y0=0 v y=not givenx0=0 x=not giveny0=0 y=1mt0=0 t=not givenax=0 a y=9.8m /s

2

a y=d v ydt

a y dt=d v y

a y0t

dt=0v y

d v y a y(t0)=v y0

a y t=v y=d ydt

a y tdt=d y

a y0ttdt=0

yd y

a y (12)(t 202)= y0

t=2 ya y =2(1)9.8 =0.204081632 sv y=a y t=(9.8)(0.204081632)v y=1.999999994 m /sv=v x2+v y2=2.358495278m /s

v=2.358m /s

EXAMPLE 11A man on a balcony throws an apple to his friend on the balcony of an adjacent building. He throws it with an angle of 500 above the horizontal. If the buildings are 10.0 m apart and the man's balcony is 5.0 m above his friend's balcony, what should be the initial speed of the apple?v x=v0 cos50v y0=v0 sin50 v y=not givenx0=0 x=10 my0=0 y=5 mt0=0 t=not givenax=0 a y=9.8 m / s

2

ax=0 v x=d xdt

v x dt=d x

v x0t

dt=010

d x v x(t0)=100

t=10v x= 10

v0 cos 50

a y=d v ydt

a y dt=d v y

a y0t

dt=v y0v y

d v ya y (t0)=v yv y0

v y= v y0+a y t=d ydt

d y=v y0 dt+a y tdt

05

d y=v y00t

dt+a y0t

tdt

5=v y0(t0)+a y (12)(t202)

5=v0 sin50(10

v0 cos50)4.9( 100

v02 cos2 50

)

490v0

2 cos2 50=10 tan 50+5

v0= 490(cos2 50)(10 tan 50+5)v0=8.372631737m / s

v0=8.373m /s