Unit 3 Kinematics Equations

Unit 3 Kinematics Equations

Objectives:Learn the 4 motion equations for 1 dimensional motion when acceleration is constant

a = ΔvΔt

1. Kinematics Equation 1 Motion at Constant Acceleration

but since "Δ" means change

Δv = v - vo and

Δt = t - to if we always let to = 0, Δt = t

Solving for "v"

This equation tells us how an object's velocity changes as a function of time.

a = v - vo

t

at = v - vo

V - vo = at

t

Problem-Solving Strategy

The strategy involves the following steps:1.Construct an informative diagram of the physical situation.2.Identify and list the given information in variable form.3.Identify and list the unknown information in variable form.4.Identify and list the equation that will be used to determine unknown information from known information.5.Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information.6.Check your answer to insure that it is reasonable and mathematically correct.

Eg1. Starting from rest, you accelerate at 4.0

m/s2 for 6.0s. What is your final velocity?

Step1: =0,

step2: this is motion problem, use equation V=

Step3: substitute the related variables into equationV=0+ (4.0m/s2)(6.0s)=24m/s

eg2 Starting from rest, you accelerate at 8.0 m/s2 for

9.0s. What is your final velocity?

Step1: =0,


Step3: substitute the related variables into equationV=0+ (8.0m/s2)(9.0s)=72m/s

Eg3. You have an initial velocity of 5.0 m/s. You then

experience an acceleration of -1.5 m/s2 for 4.0s;

what is your final velocity?

Step2: =5.0m/s,


Step4: substitute the related variables into equationV=5.0m/s+ (-1.5m/s2)(4.0s)=-1.0m/s

Step1: interpret the motionV

a

Eg4. You have an initial velocity of -3.0 m/s. You then

experience an acceleration of 2.5 m/s2 for 9.0s;

what is your final velocity? Step1: interpret the motion

V

a

Step2: =-3.0m/s,


Step4: substitute the related variables into equationV=3.0m/s+ (2.5m/s2)(9.0s)=+19.5m/s

Eg5. How much time does it take to accelerate from an initial velocity of 20m/s to a final velocity of

100m/s if your acceleration is 1.5 m/s2?

Eg6. How much time does it take to come to rest if your initial velocity is 5.0 m/s and your acceleration is

-2.0 m/s2?

Eg7. An object accelerates at a rate of 3 m/s2

for 6 s until it reaches a velocity of 20 m/s. What was its initial velocity?

Eg8. An object accelerates at a rate of 1.5 m/s2 for 4 s

until it reaches a velocity of 10 m/s. What was its initial velocity?

2. Kinematics Equation 2

Motion at Constant Acceleration

If velocity is changing at a constant rate, the average velocity is just the average of the initial and final velocities.

And we learned earlier that

Some problems can be solved most easily by using these two equations together.It shows displacement as function of time.

𝑥−𝑥0=(𝑉+𝑉 0)

2∆ 𝑡

∆ 𝑥=(𝑉+𝑉 0)

2∆ 𝑡

𝑉=(𝑉+𝑉 0)

2

𝑉=(∆ 𝑥)∆𝑡

(𝑉 +𝑉 0)2

=(∆𝑥 )∆ 𝑡

∆ 𝑥=(𝑉+𝑉 0)∆ 𝑡

2

52 Starting from rest you accelerate to 20 m/s in 4.0s. What is your average velocity?

53 Starting with a velocity of 12 m/s you accelerate to 48 m/s in 6.0s. What is your average velocity?

54 Starting with a velocity of 12 m/s you accelerate to 48 m/s in 6.0s. Using your previous answer, how far did you travel in that 6.0s?

Step2: =12.0m/s, V

step3: this is motion problem, use equation

Step4: substitute the related variables into equation

V

a

∆ 𝑥=(𝑉+𝑉 0)

2∆ 𝑡

Step1: interpret the motion


We can combine these three equations to derive an equation which will directly tell us the

position of an object as a function of time.

v = v + vo

2

Δx tv =

x - xo = ½ (v + vo)t x - xo = ½vt + ½vo

x = xo + ½vo + ½vt

x = xo + ½vo + ½(vo + at)t

x = xo + ½vo + ½vo + ½a( )2

x = xo + vo + ½a( )2

v = vo + at

∆ 𝑥=𝑉 0 ∙∆ 𝑡+12𝑎¿

𝑉=(∆ 𝑥)∆𝑡

eg9 An airplane starts from rest and

accelerates at a constant rate of 3.0 m/s2

for 30.0 s before leaving the ground. How far did it move along the runway?


Step2: =0m/s, a


Step4: substitute the related variables into equation

V

a

∆ 𝑥=𝑉 0 𝑡+12𝑎¿

eg10 A Volkswagen Beetle moves at an initial velocity of 12 m/s. It coasts up a hill with

a constant acceleration of –1.6 m/s2. How

far has it traveled after 6.0 seconds?

eg11A motorcycle starts out from a stop sign and

accelerates at a constant rate of 20 m/s2. How

long will it take the motorcycle to go 300 meters?

Step2: =0m/s, a


Step4: substitute the related variables into equation solve t from above equation:

Step5: apply to real life physics, we can only take t=5.5sec.

V

a

∆ 𝑥=𝑉 0 𝑡+12𝑎𝑡2


eg12A train pulling out of Grand Central Station accelerates from rest at a constant rate. It covers 800 meters in 20 seconds. What is its rate of acceleration?

eg13A car has a initial velocity of 45 m/s. It accelerates for 4.8 seconds. In this time, the car covers 264 meters. What is its rate of acceleration?

eg14 A Greyhound bus traveling at a constant velocity starts to accelerate at a constant

2.0 m/s2. If the bus travels 500 meters in

20 seconds, what was its initial velocity?


We can also combine these equations so as to eliminate t:

Use the relation, when a=constantWe can prove:

𝑥−𝑥0=𝑉 +𝑉 0

2∆ 𝒕=(𝑉+𝑉 0

2 )(𝑉 −𝑉 0

𝑎 )

𝑉 −𝑉 0=𝑎∆ 𝒕 ∆ 𝒕=𝑉 −𝑉 0

𝑎

eg15A car accelerates from rest to 30m/s while traveling a distance of 20m; what was its acceleration?

V

a

𝑉 2=𝑉 02+2𝑎∆ 𝑥


Step2: =0m/s, V


Step4: substitute the related variables into equation solve t from above equation: a

eg16

You accelerate, from rest, at 10m/s2 for a

distance of 100m; what is your final velocity?

eg17Beginning with a velocity of 25m/s, you accelerate at a rate of 2.0m/s2. During that acceleration you travel 200m; what is your final velocity?

V

a


Step2: =25m/s, a


Step4: substitute the related variables into equation solve = We know the velocity should be positive to the right,

v=+38m/s

𝑉 2=𝑉 02+2𝑎∆ 𝑥

eg18

You accelerate from 20m/s to 60m/s while traveling a distance of 200m; what was your acceleration?

eg19

A dropped ball falls 8.0m; what is its final velocity?

eg20

A ball with an initial velocity of 25m/s is

subject to an acceleration of -9.8 m/s2;

how high does it go before coming to a momentary stop?

5. Mixed Kinematics Problems


We now have all the equations we need to solve constant-acceleration problems.

Sample ProblemFinal Velocity After Any DisplacementA person pushing a stroller starts from rest, uniformlyaccelerating at a rate of 0.500 m/s2. What is thevelocity of the stroller after it has traveled 4.75 m?

Chapter 2

Sample Problem, continued

1. DefineGiven:

v0 = 0 m/s = ?a = 0.500 m/s2

x = 4.75 m= ?

Diagram: Choose a coordinate system. The most convenient one has an origin at the initial location of the stroller, as shown above. The positive direction is to the right.

Chapter 2Sample Problem, continued

2. PlanChoose an equation or situation: Because the

initial velocity, acceleration, and displacement are known, the final velocity can be found using the following equation:

2 2 2f iv v a x

2 2f iv v a x

Rearrange the equation to isolate the unknown: Take the square root of both sides to isolate vf .

Chapter 2Sample Problem, continued

Tip: Think about the physical situation to determine whether to keep the positive or negative answer from the square root. In this case, the stroller starts from rest and ends with a speed of 2.18 m/s. An object that is speeding up and has a positive acceleration must have a positive velocity. So, the final velocity must be positive.

3. CalculateSubstitute the values into the equation and solve:

4. EvaluateThe stroller’s velocity after accelerating for 4.75 m is 2.18 m/s to the right.

2 2(0 m/s) 2(0.500 m/s )(4.75 m)fv

2.18 m/sfv

Free Fall: Acceleration Due to Gravity

Free Fall

All unsupported objects fall towards Earth with the same acceleration. We call this acceleration the "acceleration due to gravity" and it is denoted by g.

g = 9.8 m/s2

Keep in mind, ALL objects accelerate towards the earth at the same rate.

g is a constant!

We choose upward as positive, then the free fall object on earth surface acceleration is

a= 9.8 m/s2

Fact:The acceleration due to gravity on the surface of the Moon is 1.6249 m/s2, about 16.6% that on Earth's surface.

It speeds up(negative acceleration)

g = -9.8 m/s2

It stops momentarily.v = 0

g = -9.8 m/s2

An object is thrown upward with initial velocity, vo

It slows down.(negative acceleration)

g = -9.8 m/s2

What happens when it goes up?

What happens when it goes down?

What happens at the top?

It returns with itsoriginal velocity.What happens when it lands?

It speeds up.(negative acceleration)

a = -9.8 m/s2


a = -9.8 m/s2


It slows down.(negative acceleration)

a = -9.8 m/s2

It returns with itsoriginal velocity.

a

v0

On the way up:

a

v1

v1

a

v2

v2

a

a

v

a

av0

On the way down:

v1

v1v2

v2

v

vt = 0 s

t = 1 s

t = 2 s

t = 3 s t = 0 s

t = 1 s

t = 2 s

t = 3 s

v0 = 0

g = -9.8 m/s2

v(m/s)

For any object drop from certain height, this is what the velocity vs time graph looks like.

t (s)

v(m/s)

t (s)



g = -9.8 m/s2

It returns with itsoriginal velocity but in the opposite direction.

For any object thrown straight up into the air, this is what the velocity vs time graph looks like.

46 A ball is dropped from rest and falls (do not consider air resistance). Which is true about its motion?

A acceleration is constant

B velocity is constant

C velocity is decreasing

D acceleration is decreasing

c

c

c

c

Sample Problem

Falling ObjectJason hits a volleyball so that it moves with an initialvelocity of 6.0 m/s straight upward. If the volleyballstarts from 2.0 m above the floor, how long will it be in the air before it strikes the floor?

Choose upward as positive direction and 2.0m height as origin of y –axis :

0m

-2.0m

Sample Problem, continued1. Define

Given: Unknown:

v0 = +6.0 m/s t = ? a = –g = –9.81 m/s2

y = –2.0 m

Diagram: Place the origin at the Starting point of the ball

(y0 = 0 at t0 = 0).

Sample Problem, continued2. Plan Choose an equation or situation:

Both ∆t and v are unknown. Therefore, first solve for v using the equation that does not require time. Then, the equation for v that does involve time can be used to solve for ∆t.

Rearrange the equation to isolate the unknown: Take the square root of the first equation to isolate v. The second equation must be rearranged to solve for ∆t.

𝑉 2=𝑉 02+2𝑎∆ y 𝑉=𝑉 0+a∆ t

𝑉=±√𝑉 02+2𝑎∆ y ∆ 𝑡=

𝑉 −𝑉 0

𝑎


2 2 22 (6.0 m/s) 2(–9.81 m/s )(–2.0 m)f iv v a y

2 2 2 2 2 236 m /s 39 m /s 75 m /s –8.7 m/sfv

3. Calculate Substitute the values into the equation and

solve: First find the velocity of the ball at the moment that it

hits the floor.

Tip: When you take the square root to find vf , select the negative answer because the ball will be moving toward the floor, in the negative direction.


2 2

–8.7 m/s 6.0 m/s –14.7 m/s

–9.81 m/s –9.81 m/sf iv v

ta

1.50 st

Next, use this value of vf in the second equation to solve for ∆t.

4. EvaluateThe solution, 1.50 s, is a reasonable amount of time

for the ball to be in the air.

47 An acorn falls from an oak tree. You note that it takes 2.5 seconds to hit the ground. How fast was it going when it hit the ground?

48 A rock falls off a cliff and hits the ground 5 seconds later. What velocity did it hit the ground with? How high is the cliff?

49 A ball is thrown down off a bridge with a velocity of 5 m/s. What is its velocity 2 seconds later? How far did it fall?

50 An arrow is fired straight upward into the air and it reaches its highest point 3 seconds later. What was its velocity when it was fired? What is the maximum height it reached?

51 A rocket is fired straight up from the ground. It returns to the ground 10 seconds later. What was it's launch speed?

Check your understanding:http://www.physicsclassroom.com/class/1DKin/Lesson-6/Sample-Problems-and-Solutions

http://www.physicsclassroom.com/class/1DKin/Lesson-6/Sample-Problems-and-Solutions




Unit 3 Kinematics Equations

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