-PHY10T2KINEMATICSCALULUS

KINEMATICS

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Kinematics is the study of how things move.

Kinematics (from Greek , kinein, to move) is the branch of classical mechanics that describes the motion of objects without consideration of the causes leading to the motion

Linear Motion : Linear or translational kinematics is the description of the motion in space of a point along a line, also known as trajectory or path. This path can be either straight (rectilinear) or curved (curvilinear).

KINEMATICS

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http://en.wikipedia.org/wiki/Ancient_Greek_language

SUBSCRIPTS

Terms with the subscript “O” – denotes initial or starting or original value or your initial reference value

Terms with the subscript “F” – denotes final or end value at a certain condition (time or displacement)

XO : means initial position in x-axis

VF : means end reference velocity

KINEMATICS

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Displacement (Distance) : The change of position. Units : m, cm, km, & ft

Horizontal Displacement

𝒔 = 𝑿𝑭 − 𝑿𝑶

Vertical Displacement

𝒉 = 𝒀𝑭 − 𝒀𝑶

(𝑿𝑶 , 𝒀𝑶)

(𝑿𝑭 , 𝒀𝑭)

Actual Displacement

|𝐝| = |𝒔|𝟐 + |𝒉|𝟐

θ = tan−1 |𝒉|

|𝒔|

𝒔

𝒉 𝒅

θ

@𝒕𝑶 @𝒕𝑭

KINEMATIC QUANTITIES

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Velocity (Speed) – The rate of change of position with respect to time (interval) . Units : km/hr, m/s, ft/s Time Interval – The time difference between two events. Units : sec, min, hrs

Average Horizontal Velocity

𝑽𝑿 = 𝒔

𝒕=

𝑿𝑭 − 𝑿𝑶

𝒕𝑭 − 𝒕𝑶

Average Vertical Velocity

(𝑿𝑶 , 𝒀𝑶)

(𝑿𝑭 , 𝒀𝑭)

Actual Velocity

|𝑽| = |𝑽𝑿|𝟐 + |𝑽𝒀|𝟐

θ = tan−1 |𝑽𝒀|

|𝑽𝑿

|

𝑽𝑿

𝑽

θ

@𝒕𝑶 @𝒕𝑭

𝑽𝒀 = 𝒉

𝒕=

𝒀𝑭 − 𝒀𝑶


𝑽𝒀


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MOTION HAVING CONSTANT VELOCITY EXAMPLES

1. A hiker travels a flat road in a straight line for 40 mins with an average velocity of magnitude 1.25 m/s. What distance does he cover during this time ?

Given Find

t= 40 mins, = 2400 sec

v = VX = 1.25 m/s s – distance covered

The scalar form of the equation is just :

VX = s/t hence s = VXt

s = VXt = ( 1.25 m/s)(2,400 s) = 3000 m = 3 km


We will consider : Average Horizontal Velocity because of the flat surface where motion occurred.

𝑽𝑿 = 𝒔

𝒕=



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MOTION HAVING CONSTANT VELOCITY EXAMPLES

2. A car odometer reads 22,487 km at the start of a trip and 22,891 km at the end. The trip took 4 hours. What was the car’s average speed?

a. in km/hr? b. in m/s?

The scalar and practical form of the equation for this problem is :

VX = (XF – XO) /t

a. VX = (XF – XO) /t = (22891km – 22487km) / (4 hr) = 101 km/hr

b. VX = (101 km/hr) 𝟏 𝒎/𝒔

𝟑.𝟔 𝒌𝒎/𝒉𝒓 = 28.056 m/s


We will consider again : Average Horizontal Velocity in the assumption that the overall motion of the car is horizontal.

𝑽𝑿 = 𝒔

𝒕=



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Given : xF = 22,891 km xO = 22,487 km t = 4 hrs

Acceleration – The rate of change of velocity with respect to time (interval). Units : m/s2 , km/hr2 , ft/s2

Average Horizontal Acceleration

𝒂𝑿 = ∆𝑽𝑿

𝒕=

𝑽𝑭𝑿 − 𝑽𝑶𝑿


Average Vertical Acceleration

𝑽𝑶𝑿

𝑽𝑭𝒀

Actual Acceleration

|𝒂| = |𝒂𝑿|𝟐 + |𝒂𝒀|𝟐

θ = tan−1 |𝒂𝒀|

|𝒂𝑿|

𝒂𝑿

𝒂

θ

@𝒕𝑶 @𝒕𝑭

𝒂𝒀

𝒂𝒀 = ∆𝑽𝒀

𝒕=

𝑽𝑭𝒀 − 𝑽𝑶𝒀


𝑽𝑶𝒀

𝑽𝑭𝑿


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CALCULUS BASED SOLUTIONS

Let : Position (x) be a time dependent quantity : thus x(t) Velocity (v) is the first derivative : v(t) = dx/dt Acceleration (a) is the next derivative : a(t) = dv/dt = dx/d2t It is therefore possible to reverse this order : You have to at least start with a(t) Acceleration (a) be a time dependent quantity : thus a(t) Velocity (v) is the first integral : v(t) = VO + ʃ a(t)dt Position (x) is the next integral : x(t) = XO + ʃ v(t)dt = XO + ʃʃa(t)d2t Where VO and XO are the initial position and velocity values at t = 0 Instantaneous Values are obtained by substituting the exact value of time (t) in the equations above

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3. A hobbyist is testing a new model rocket engine by using it to propel a cart along a model railroad track. He determines that its motion along the x-axis is described by the equation : x(t) = (0.160 m/s2) t2 . Compute for the magnitude of the instantaneous velocity of the cart at time t = 3 seconds.

Given Find

x(t) = (0.160 m/s2)t2 v(t) @ t = 3sec

Solution

x(t) = (0.160)t2

dx/dt = v(t) = 2(0.16)t

v(t) = (0.32 m/s2)t

v(3) = (0.32 m/s2)(3 s)

v(3) = 0.96 m/s

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CALCULUS BASED SOLUTIONS

4. A car has velocity v given as a function of time

v(t) = A + Bt2 , where A = 4 m/s & B = 0.2 m/s3 Calculate :

a. The average acceleration for the time interval tO = 0 to tF = 5 sec.

b. The magnitude of the instantaneous acceleration for t = 0 and t = 5 sec.

Solution

a. a – average from tO= 0, tF = 5 sec

v(t) = 4 + 0.2 t2

aX = (vFX – vOX)/(tF – tO)

vOX = v(0) = 4 + 0.2(0)2 = 4 m/s

vFX = v(5) = 4 + 0.2(5)2 = 9 m/s

aX =(9 m/s – 4 m/s) / (5 s – 0 s)

aX= + 1 m/s2

b. a(t) – instantaneous for t = 0 & t = 5sec

v(t) = 4 + 0.2 t2

a(t) =dv/dt = 0 + 2(0.2)t

a(t)= 0.4t

b.1 a @ t = 0 , a(0)= 0.4(0) = 0

b.2 a @ t = 5 , a(5)= 0.4m/s3(5s) = 2 m/s2

COMPARING AVERAGE AND

INSTANTANEOUS VALUES

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VELOCITIES & POSITION BY INTEGRATION

Determining Local Maximum (and Minimum) Values :

For Velocity : Set a(t) = 0, solve for value of t, then substitute it to v(t)

For Position : Set v(t) = 0, solve for value of t, then substitute it to x(t) 8/3

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Solution :

(a) Via integration : Velocity is :

v(t) = vo + 0∫t [At – Bt2] dt = vo + ½ At2 – (1/3)Bt3

v(t) = 0+ ½ (1.5m/s3)t2 – (1/3)(0.12m/s4)t3 , at t = 0 vo = 0 (at rest)

v(t) = (0.75m/s3)t2 – (0.04m/s4)t3

5. Example :

The acceleration of a motorcycle is given by : a(t) = At – Bt2, where A = 1.5 m/s3 and B = 0.12 m/s4 .

The motorcycle is at rest at the origin at t=0. (a) Find the position and velocity as a function of time. (b) Calculate the maximum velocity and maximum displacement it attain


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Continuation :

and Position will be : x(t) = xo + 0∫t [(0.75m/s3)t2 – (0.04m/s4)t3] dt x(t) = 0 + (1/3)(0.75m/s3)t3 – (1/4)(0.04m/s4)t4 , at t = 0 xo = 0 (at origin) x(t) = (0.25m/s3)t3 – (0.01m/s4)t4

5. Example :

The acceleration of a motorcycle is given by : a(t) = At – Bt2, where A = 1.5 m/s3 and B = 0.12 m/s4 .

The motorcycle is at rest at the origin at t=0. (a) Find the position and velocity as a function of time. (b) Calculate the maximum velocity and maximum displacement it attain


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Continuation :

(b.1) Maximum Velocity :

Set a(t) = 0, then solve for t :

a(t) = 1.5t –0.12t2 = 0

t(1.5 – 0.12t) = 0

t = 0 and t = (1.5/0.12) = 12.5 s

Subs in v(t)

v(12.5) = (0.75m/s3)(12.5s)2 – (0.04m/s4)(12.5s)3 ‘

vmax = 39.0625 m/s

x(t) = (0.25m/s3)t3 – (0.01m/s4)t4

Working Equations :

a(t) = (1.5m/s3)t – (0.12m/s4)t2

v(t) = (0.75m/s3)t2 – (0.04m/s4)t3


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Continuation :

(b.2) Maximum Displacement:

Set v(t) = 0, then solve for t :

v(t) = 0.75t2 –0.04t3 = 0

t2(0.75 – 0.04t) = 0

t = 0 and t = (0.75/0.04) = 18.75 s

Subs in x(t)

x(18.75) = (0.25m/s3)(18.75s)3 – (0.01m/s4)(18.75s)4

xmax = 411.987 m

x(t) = (0.25m/s3)t3 – (0.01m/s4)t4

Working Equations :

a(t) = (1.5m/s3)t – (0.12m/s4)t2

v(t) = (0.75m/s3)t2 – (0.04m/s4)t3


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EQUATIONS for PURELY HORIZONTAL MOTION (x-axis) Where:

VFX = VOX + aXt VF? – final velocity

s = VOXt + ½ aXt2 VO? – initial velocity

VFX2 = VOX

2 + 2aXs s – horizontal

displacement/distance traveled

Note : if aX = 0 ; V = constant a – constant acceleration

EQUATIONS for PURELY VERTICAL MOTION (y-axis) h – vertical displacement/ height

VFY = VOY + gt g – gravitational acceleration =

9.8 m/s2

h = VOYt + ½ gt2 Sign Convention (from origin) :

s : (+) → & h : (+) ↑

VFY2 = VOY

2 + 2gh V : (+) → OR (+) ↑

Note : aY = g

If object is free-fall or dropped VO = 0

At highest point reached velocity is zero

a : (+) if speeding up OR (+) ↑,

hence g = – 9.8 m/s2

FORMULA TABLE FOR MOTION HAVING

CONSTANT ACCELERATION

6. A car is traveling along a straight road at 10 m/s. It accelerates uniformly for 25 seconds until it is moving at 35 m/s. What was the acceleration?

VFX2 = VOX

2 + 2aXs VFX = VOX + aXt s = VOXt + ½ aXt2

Given :

VFX = 35 m/s VOX = 10 m/s t = 25 s

35 = 10 + aX(25)

35 = 10 + 25aX

35 − 10 = 25aX

25 = 25aX

(25/25) = aX

aX = +1 m/s2 Since the car’s speed increases aX is positive

Required : aX

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MOTION HAVING CONSTANT ACCELERATION

7. A car covers a distance of 50 m in 10 seconds while smoothly slowing down to a final speed of 2.5 m/s. A) Find the car's original speed. B) Find the car's acceleration. Given :

s = 50 m t = 10 s VFX = 2.5 m/s

Required :

VOX & aX

2.5 = VOX + aX(10) 50 = VOX (10) + ½ aX(10)2

2.5 = VOX + 10aX

VOX = 2.5 − 10 aX < eq. 1

50 = 10(2.5−10aX) + 50aX

50 = 10VOX + 50aX < eq. 2

50 = 25 −100aX + 50aX 50 − 25 = −100aX + 50aX

25 = −50aX

aX = −0.5 m/s2

a = 0.5 m/s2 (slowing down)

VOX = 2.5 − 10(– 0.5)

VOX = 2.5 + 5

VOX = 7.5 m/s

eq. 1 in eq.2 value of a in eq.1

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MOTION HAVING CONSTANT ACCELERATION

VFX2 = VOX

2 + 2aXs VFX = VOX + aXt s = VOXt + ½ aXt2

STN A

8. A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s2 for 8 sec. It runs at constant speed for 70 sec, and decelerates at 2.5 m/s2 until it stops at the next station. Find the total distance (traveled) covered.

STN B

0 1 2 3

xo =

to =

Vo =

x1 =

t1 =

V1 =

0

0

0

Δt = t = 8 sec

8 sec

Δt = t = 70 sec

a =+1.6 m/s2 a = 0 <constant speed> a = – 2.5 m/s2

From 0 to 1

s = Vot + ½ at2

s1 = V0t + ½ at2

s1 =(0)(8s) + ½ (1.6 m/s2)(8s)2

s1 = 51.2 m

51.2 m

VF = VO + at

V1 = V0 + at

V1 = 0 + (1.6 m/s2)(8s) = 12.8 m/s

12.8 m/s

s1 = s2 = s3 =

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STN A


STN B

0 1 2 3

xo =

to =

Vo =

x1 =

t1 =

V1 =

x2 =

t2 =

V2 =

x3 =

t3 =

V3 =

0

0

0 0

Δt = t = 8 sec

8 sec

Δt = t = 70 sec

78 sec


51.2 m

12.8 m/s

From 1 to 2

s = Vot + ½ at2

s2 = V1t + ½ at2

s2 =(12.8 m/s)(70s) + ½ (0)(70s)2

s2 = 896 m

VF = VO + at

V2 = V1 + at

V2 = 12.8 m/s + (0)(70s) = 12.8 m/s

896 m s1 = s2 = s3 =

12.8 m/s

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STN A


STN B

0 1 2 3

xo =

to =

Vo =

x1 =

t1 =

V1 =

x2 =

t2 =

V2 =

x3 =

t3 =

V3 =

0

0

0 0

Δt = t = 8 sec

8 sec

Δt = t = 70 sec

78 sec


51.2 m

12.8 m/s

896 m

12.8 m/s

From 2 to 3

VF2 = VO

2 + 2as

V32 = V2

2 + 2as3

(0)2 = (12.8m/s)2 + 2(– 2.5 m/s2)(s3)

163.84m2/s2 = 5 m/s2(s3)

s3 = 32.768 m

sT = 51.2 m + 896m + 32.768 m

sT = 979.968 m ≈ 980 m

s1 = s2 = s3 = 32.768 m

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yO

yF

Δy or h

tO

tF

Δt

MOTION ALONG A STRAIGHT LINE (y-axis)

Case 1 : Object Going Downward

If object is released without initial velocity (Vo = 0) – it is considered a freely falling body

VF

VO

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Case 2 : Object Thrown Upward

y

yO

Δy or h

tH

tO

Δt1

y

yF

tH

tF

Δt2

At the highest level the velocity is zero

VO

VH = V = 0 VH = V = 0

VF

Going Up Going Down = Free Fall


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For Motion along Y-Axis

The displacement is now VERTICAL. We refer to it as HEIGHT.

Δy = h

Acceleration here is a fixed value called the gravitational acceleration (g)

g= 9.8 m/s2 = 980 cm/s2 = 32 ft/s2


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General Equations for Linear Motion (Y-axis)

vF2 = vO

2 + 2gh

vF = vO + gt

h = vOt + ½ gt2

MOTION ALONG A STRAIGHT LINE (Y-AXIS)

Sign Convention for gravitational acceleration (g) – ALWAYS NEGATIVE

g = − 9.8 m/s2 = − 980 cm/s2

(metric system)

g = − 32 ft/s2

(english system)

For Height & Velocity – The sign convention would just tell if the object is going down (−) or up (+).

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Sample Problems :

1. If a flea can jump to a height of 0.75 m, what is its initial velocity as it leave the ground? For how much time is it in the air?

Given :

h = 0.75 m Required :

VO & time in air (T)

VF2 = VO

2 + 2gh VF = VO + gt h = VOt + ½ gt2

VO

h=0.75m

VH = 0

02 = VO2 + 2(-9.8)(0.75)

g = −9.8 m/s

0 = VO2 − 14.7

VO2 = 14.7

VO = 3.834 m/s

VF = VO + gt

0 = 3.834 +(-9.8)t

9.8t = 3.834

t = 3.834/9.8

t = 0.39 s

this is only going up, but this is also the same time going down, hence T = 2t = 0.78 s

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Sample Problems :

2. A brick is dropped from the roof of a building. The brick strikes the ground after 5 seconds.

a. How tall, in meters, is the building (magnitude only)?

b. What is the magnitude of the brick’s velocity just before it reaches the ground?

t = 5s

h = ?

VF = ?

VO = 0 h = vOt + ½ gt2

h = (0)(5) + ½ (-9.8)(5)2

h = −122.5 m

(−) sign here means below the reference or starting point.

h =122.5 m (down)

vF = vO + gt

VF = 0 +(-9.8)(5)

VF = −49 m/s

(−) sign here means going down.

VF = 49 m/s (downward)

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Sample Problems :

3. A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower’s hand with a speed of 15 m/s.

a. What is its speed in 5 seconds?

b. How far does it fall in 2 seconds?

c. What is the magnitude of the velocity after falling 10 m?

VF = ?

VO = -15 m/s a. VF in 5 sec

VF = 64 m/s (downward)

VF = -15 + (-9.8)(5)

h =(-15)(2) + ½ (-9.8)(2)2 = -49.6 m

b. h in 2 sec

vF = vO + gt

h = 49.6 m (down)

vF2 = vO

2 + 2gh

c. VF @ h = -10 m

vF2 = (-15)2 + 2(-9.8)(-10)

vF2 = 225 +196 = 421

vF = 20.51 m/s

VF = - 64 m/s

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PROJECTILE MOTION

- Motion along a curved path or trajectory

- Elements of BOTH straight line motion & freely-falling bodies apply here.

Consider a projectile (ball) thrown at an angle instead of horizontally

θ

“A Projectile is any body that is given initial velocity and then follows a path determined by the effects of gravitational acceleration & air resistance.”

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Projectile Motion – Curvillinear Translation

θ

Δymax or H

R

+y

−y

(0,0) Origin

+x

Vo

Vox

Voy

V1 V1y

Vy = 0

V

V2

V2y

V2x

θ

VF

VFx

VFy

* VF = Velocity @ impact ≠ 0

V1x

= Vx

VO = Initial/Project Velocity

V = Velocity @ Highest point

Vn = Velocity @ a certain point

Ex V1, V2

VFx = x-comp of VF

VOx = x-comp of VO

Vx = x-comp of V

Vnx = x-comp of V1x, V2x..etc

VFy = y-comp of VF

VOy = y-comp of VO

Vy = y-comp of V

Vny = y-comp of V1y, V2y..etc

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Projectile Motion – Consider X – components :

Analyze using Motion along a Straight Line

θ

s

+y

−y

(0,0) Origin

+x

Vox

V

V2x

θ

VFx

V1x

= Vx

Vox = VO cosθ {x-component of VO}

Using Kinematics Equation (2) to determine s at any time (t)

s = VOXt + ½ aXt2

For a projectile , the HORIZONTAL component of the velocity is CONSTANT

= Vox

= Vox

= Vox

= Vox

Since VX ‘s are the same :

Hence aX = 0

s = VOXt

t

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θ

h

+y

−y

(0,0) Origin

+x

Voy

V1y

Vy = 0

V

V2y

θ

VFy

Voy = VO sinθ {y-component of VO}

For a projectile , the VERTICAL component of the velocity is NOT CONSTANT

Using Kinematics Equation (2) to determine h at any time (t)

h = VOYt + ½ aYt2

aY = g = − 9.8m/s2 = − 980cm/s2 = − 32 ft/s2

h = VOYt + ½ gt2

Projectile Motion – Consider Y – components :

Analyze using Freely Falling Bodies

t

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PROJECTILE MOTION EQUATIONS

VOX = VX = V1x = V2x = VnX

VnY = VOY + gt

X - Component Y - Component

VOX = VOcosθ VOY = VOsinθ

S = VOX t h = VOY t + ½ gt2

VnY2 = VOY

2 + 2gh

g= − 9.8 m/s2 = − 980 cm/s2 = − 32 ft/s2

VY = 0

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PROJECTILE MOTION EQUATIONS MAXIMUM HEIGHT [ H ]

h = VOYt + ½ gt2 h = H

H = VOYt + ½ gt2 VOY = VO sin θ

t = ? VY = VOY + gt

@ H , VY = 0

t = (VO sinθ) / g

0 = + VO sinθ + g t

0 = + VO sinθ − gt

(VOsin θ)(VOsin θ) g [(VO sinθ)]2

H = − g 2 g2

(VOsin θ)2 (VO sinθ)2

H = − g 2 g

H = VOYt + ½ gt2

H = VOYt − ½ gt2

RANGE [ R ]

s = VOXt s = R

R = VOXT VOX= VO cos θ

T = ? t = (VO sinθ) / g

R = {(VO cos θ)} {2(VO sinθ) / g}

R = VO2 [2(cos θ)(sinθ)] / g

R = [ VO 2 (sin 2θ) ] / g

but we are using g as – 9.8 m/s

but we are using g as – 9.8 m/s

(VOsin θ)2

H = 2g

T = 2t = (2VO sinθ) / g

sin (θ+β) = cos β sin θ + sin β cos θ

sin (2θ) = cos θ sin θ+ sin θ cos θ

Let θ = β

Recall : sine of sum of two angles

sin (2θ) = 2cos θ sin θ

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PROJECTILE MOTION EQUATIONS

MAXIMUM HEIGHT [ H ]

tH = VOY/g = (VO sinθ)/g

R = [VO2 (sin 2θ)] / g

Note :

g=+ 9.8 m/s2 = + 980 cm/s2 = + 32 ft/s2

Or

tR = 2tH = (2VO sinθ) / g

tR = R/VOX = R/(VO cosθ)

H = VOY2/(2g) = (VOsinθ)2/(2g)

TIME TO REACH THE MAXIMUM HEIGHT [ tH ]

RANGE [ R ]

TIME TO REACH THE END OF RANGE [ tR ]

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θ

+y

−y

(0,0) Origin +x

Vo

Vox

Voy

V1 V1y

Vy = 0

V

V2

V2y

V2x

θ

VF

VFx

VFy


V1x

= Vx

V3

V3y

V3x

V4 V4y

V4x

|V1|= |V2|

|V3| = |V4|

Due to its trajectory, the projectile passes again the same vertical level going down, Velocity at that level are equal in magnitude, with their vertical components, equal in magnitude but opposite in direction and their horizontal components perfectly equal.

|V1| = |V2| V1y = −V2y V1x= V2x

|V3| = |V4| V3y = −V4y V3x= V4x

COMMON LEVEL VELOCITY

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Projectile Motion – Projectile fired horizontally

Δymax or H

+y

−y

(0,0) Origin

+x

VOy = Vy =0

VO

V1

V1y

V1x

θ

VF

VFx

VFy

Here VO = Vx

Δx


VO = Initial/Project Velocity

V = Velocity @ Highest point

Vn = Velocity @ a certain point

Ex V1, V2

VFx = x-comp of VF

VOx = x-comp of VO

Vx = x-comp of V

Vnx = x-comp of V1x, V2x..etc

VFy = y-comp of VF

VOy = y-comp of VO

Vy = y-comp of V

Vny = y-comp of V1y, V2y..etc

All previous equations are useful.

EXCEPT : Range & Max. Height

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Projectile Motion – Projectile beyond the range

H

+y

−y

+x

Vy =0

V

V2

V2y

V2x

θ

VF

VFx

VFy

R VFx

At point F : Set VF as your initial velocity. Then apply the principles of horizontally fired projectile, but note that VFy is NOT zero

F

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1. A bullet is fired at an initial velocity of 350 m/s and at an angle of 50° with the horizontal. Neglecting air resistance. Determine:

(a) The range, (b) The travel time of the bullet before hitting the ground

Given

Vo = 350 m/s, θ = 50° ,

Required

(a) Range, (b) travel time - T

Solution

(a) Range

R = Vo2 (sin 2θ) / g

R = (350 m/s)2 (sin 100°) / (9.8 m/s2)

R = 12,310.1 m

R = 12.3 km

(b) T

Using Highest Point

Where Vy = 0

Vy = Voy + gt

Voy = Vo sinθ = (+350 m/s) (sin 50°)

Voy = +268.116 m/s

Vy = Voy + gt

0 = +268.116 m/s + (-9.8 m/s2) t

t = (268.116 m/s) / (9.8 m/s2) = 27.359 s

T = 2t, T = 54.717 s

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2. A bullet is fired at an angle of 75° with the horizontal with an initial velocity of 420 m/s. How high can it travel after 2 seconds? How far horizontally did it travel after that same 2 seconds?

Given

Vo = 420 m/s, θ = 75° ,

Required

h & s after t = 2 sec

Solution

h = Voy t + ½ g t2

Voy = Vo sinθ = (+420 m/s) (sin 75°) Voy = +405.689 m/s h = (405.689 m/s)(2s) + ½ (-9.8m/s2)(2s)2

h = (405.689 m/s)(2s) + ½ (-9.8m/s2)(2s)2

h = + 791.778 m

s = Vox t

Vox = Vo cosθ = (+420 m/s) (cos 75°) Vox = +108.704 m/s s = (108.704 m/s)(2s) s = 217.407 m

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3. A plane is flying horizontally at 350 km/hr at an altitude of 420 m. At this instant, a bomb is released. How far horizontally from this point will the bomb hit the ground?

Given

Vo = 350 km/hr

H = 420 m

Required

S – distance traveled by the bomb horizontally from the point of release to the ground

S

H

420 m

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3. A plane is flying horizontally at 350 km/hr at an altitude of 420 m. At this instant, a bomb is released. How far horizontally from this point will the bomb hit the ground?

Given

Vo = 350 km/hr

H = 420 m

Solution

S = Voxt (for horzontally fired projectile) Vox = Vo = 350 km/hr Voy = Vy = 0 H = Voy t + ½ gt2

- 420 m = (0)t + ½ (-9.8m/s2)t2

- 420 = - 4.9 t2

t2 = (420/4.9) = 85.71 t = 9.26 sec t = 9.26 sec *(1 hr/3600 sec) t = 0.00257hr S = (350 km/hr)(0.00257 hr) S = 0.900 km = 900 m S

H

420 m

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4. A man in a hot-air baloon drops an apple at a height of 150 m. If the balloon is rising at 15 m/s, find the highest point reached by the apple?

Given

Vo = 15 m/s,

θ = 90° ,

Required

Max height reached by the apple (H)

Solution

H = h1 + h

Vny2 = Voy

2 + 2g(h) Voy = Vo sinθ = (+15 m/s) (sin 90°) Voy = +15 m/s 0 = (15m/s)2 + 2(-9.8 m/s2)(h)

h = 11.468 m H = 11.468 m + 150 m H= 161.468 m

h1 = 150 m

h

H

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5. A ball is thrown from a tower 30 m high above the ground with a velocity of 300 m/s directed at 20° from the horizontal. How long will the ball hit the ground?

Given

Vo = 300 m/s, θ = 20°,

H = 30 m

Required

T – time for the ball to hit the ground

Solution (Short Version)

h = VOYt + ½ g t2 VOY = Vo sinθ = (+300 m/s) (sin 20°) VOY = +102.606 m/s -30 m = (102.606)(T) + ½ (-9.8m/s2)(T)2

4.9T2 – 102.606T – 30 = 0

Using QF T =21.228s

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Simultaneous Events

Two events (involving different objects) that happened at the same time and/or place

Kinematics analysis here would require to determine (find) common quantity (-ies) such as displacement and/or time

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Example 1

A ball was dropped from a 10 m tall building , at that same instant a stone was propelled vertically upward starting from the ground with an initial velocity of 25m/s.

(a) At what time (from the start) will both objects be at the same vertical height with respect to the ground?

(b) At that said time, how far does :

(b.1) the ball fall?

(b.2) the stone rise?

H = 10 m

VOB = 0

VOS = + 2.5 m/s

H = 10 m

hB = ?

hS = ?

tB = ?

tS = ?

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Example 1






H = 10 m

hB = ?

hS = ?

tB = ?

tS = ?

(a) When objects “meet” (even without collision) they will meet at that location, point or level at the same time

tB = tS = t

Ball (Freely-falling body)

hB = VOB tB + ½ gtB2

hB = (0.5)(– 9.8)t2

hB = – 4.9t2

Stone

hS = VOS tS + ½ gtS2

hS = (25)t + (0.5)(– 9.8)t2

hS = 25t – 4.9t2

In terms of height : we see that : (considering absolute values)

|H| = |hB| + hS

10 = +(4.9t2) + (25t – 4.9t2)

10 = 25t , t = 10/25

t = 0.4 sec

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Example 1






H = 10 m

hB = ?

hS = ?

tB = ?

tS = ?

Ball (Freely-falling body)

tB = t = 0.4 sec

hB = – 4.9(0.4)2

hB = – 0.784 m

Stone

tS = t = 0.4 sec

hS = 25(0.4) – 4.9(0.4)2

hS = + 9.216 m

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Example 2

You are on the roof of the Physics building 46 m above the ground. Your prof who is 1.8m tall, is walking along side the building at constant speed of 1.2 m/s. If you wish to drop an egg on your prof, where should the prof be (“S”) when you release the egg? Assume that the egg is in free-fall.

P6 BLDG

H = 46 m

VOE = 0

VP = 1.2 m/s

hP = 1.8 m

S = ?

The effective drop height is only

h = H – hP = 46 m – 1.8 m = 44.2 m

and again the time for the egg and the prof (head) to meet is the same :

tE = tP = t

h

Egg (Freely-falling body)

h = VOE tE + ½ gtE2

– 44.2 = (0)t + (0.5)(– 9.8)t2

– 44.2 = – 4.9t2

t = 3.0 seconds

Prof

S = VP tS

S = (1.2)t = (1.2)(3)

S = 3.6 m

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Example 3

A student is running at her top speed of 5m/s to catch a bus, which is stopped at the bus stop. When the student is still 40m from the bus, it starts to pull away, moving with a constant acceleration of 0.17 m/s2.

(a) For how much time and what distance does the student have to run at 5m/s before she overtakes the bus?

(b) When she reaches the bus, how fast is the bus moving?

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Documents

-PHY10T2KINEMATICSCALULUS