Transcript 5-11-10 1-D Kinematics
1-D Kinematics
• • Kinematics – Kinematics is the branch of mechanics that describes the motion of objects without necessarily discussing what causes the motion.
– 1-Dimensional Kinematics (or 1 - Dimensional motion) refers to motion in a straight line.
Distance – The total length of the path traveled by an object is called distance .
– “How far have you walked?” is a typical distance question.
– The SI unit of distance is the meter (m).
• Displacement (Dx) – The change in the position of a particle is called displacement .
– ∆ is a Greek letter used to represent the words “change in”. ∆x therefore means “change in x”. It is always calculated by final value minus initial value.
– “How far are you from home?” is a typical displacement question.
– The SI unit for displacement is the meter. Calculation of displacement: ∆x = x f - x i
• Distance vs Displacement displacement 50m distance 100m • A picture can help you distinguish between distance and displacement.
• • • Another example distance displacement Walk east 100m Distance = 100m Turn and walk west 100m Displacement = 100m Distance = 200m Displacement = 0m
• • Questions Does the odometer in your car measure distance or displacement?
–
Distance
• Can you think of a circumstance in which it measures both distance and displacement?
–
If you go one direction and stop it will give both distance and displacement
• If Dx is the displacement of a particle, and d is the distance the particle traveled during that displacement, which of the following is always a true statement?
a) d = |∆x| b) d < |∆x| c) d > |∆x| d) d > |∆x| e) d < |∆x|
• • A particle moves from x = 1.0 meter to x = -1.0 meter.
What is the distance d traveled by the particle?
2m
-3 m -2 m -1 m 0 m 1 m 2 m 3 m • What is the displacement of the particle?
-2m, must show direction
• You are driving a car on a circular track of diameter 40 meters. After you have driven around 2 ½ times, how far have you driven, and what is your displacement?
C = 2πr C = 2π∙20m C = 125.66m
40m Distance = 2.5∙125.66m
= 314.16m
Displacement = 40m
• Average Speed – Average speed describes how fast a particle is moving. The equation is: s ave = d/t – where:
s ave = average speed d = distance
∆t = elapsed time The SI unit of speed is the m/s
Average speed is always a positive number.
• • Average Speed Average velocity describes how fast the displacement is changing. The equation is: v ave = ∆x/ ∆t – where:
v ave = average velocity
∆x = displacement ∆t = elapsed time The SI unit of speed is the m/s
Average velocity is + or – depending on direction.
• How long will it take the sound of the starting gun to reach the ears of the sprinters if the starter is stationed at the finish line for a 100 m race? Assume that sound has a speed of about 340 m/s.
∆t = ? ∆x = 100m v avg = 340 m/s v ave = ∆x/ ∆t ∆t= ∆x/ v ave ∆t= 100m / 340 m/s
= 0.294 s
• You drive in a straight line at 10 m/s for 1.0 km, and then you drive in a straight line at 20 m/s for another 1.0 km. What is your average velocity?
1.
∆t = ? v ave ∆x = 1km = 1000m = ∆x/ ∆t ∆t= ∆x/ v ave v avg ∆t= 1000m / 10 m/s
= 100 s
= 10 m/s
2.
∆t = ? ∆x = 1km = 1000m v avg v ave = ∆x/ ∆t ∆t= ∆x/ v ave = 20 m/s
3.
∆t= 1000m / 20 m/s
= 50 s
v ∆x = 2000m ave ∆t total = 150s v avg = ? m/s = ∆x/ ∆t = 2000 m /150s
= 13.3 m / s
• Qualitative Demonstrations 1) Demonstrate the motion of a particle that has an average speed and an average velocity that are both zero.
2) Demonstrate the motion of a particle that has an average speed and an average velocity that are both nonzero.
3) Demonstrate the motion of a particle that has an average speed that is nonzero and an average velocity that is zero.
4) Demonstrate the motion of a particle that has an average velocity that is nonzero and an average speed that is zero.
• You are a particle located at the origin. Demonstrate how you can move from x = 0 to x = 5.0 and back with an average speed of 0.5 m/s.
d = 10m t = ? s s= 0.5 m / s s ave = d/t t= 10 m / 0.5 m/s t= d/s ave
= 20 s
Cover the total distance in 20s • What the particle’s average velocity for the above demonstration?
– 0 m/s because velocity takes into account the is a sign change average velocity – in the positive x is +0.5 m/s – in the negative x is – 0.5 m/s
• Graphical Problem x (m) 0 m t (s) Demonstrate the motion of this particle
• Graphical Problem x(m) 0 m t (s) Demonstrate the motion of this particle
• Graphical Problem x (m) 0 m A ∆t B ∆x t (s) v ave = ∆x/ ∆t What physical feature of the graph gives the constant velocity from A to B?
• Graphical Problem: Determine the average velocity from the graph.
x (m) B 2.0
A ∆x 1.0
∆t 0 -1.0
2.0
4.0
6.0
t (s) -2.0
v ave = 1 m/ 3 s
= 1 / 3 m/ s
• Graphical Problem x (m) 0 m t (s) Demonstrate the motion of these two particle
• Graphical Problem X (m) 0 m t (s) Demonstrate the motion of these two particle
• Graphical Problem X (m) 0 m t (s) What kind of motion does this graph represent?
• Graphical Problem x (m) A
∆x ∆t
0 m B v ave = ∆x/ ∆t T (s) Can you determine average velocity from the time at point A to the time at point B from the raph?
•
Graphical Problem: Determine the average velocity
between 1 and 4 seconds.
x (m)
∆x
2.0
∆t
1.0
0 -1.0
-2.0
v ave = ∆x/ ∆t 2.0
4.0
6.0
t (s) v ave = 1 m / 4 s
= 0.25 m/s
• Instantaneous Velocity – The velocity at a single instant in time.
– If the velocity is uniform, or constant, the instantaneous velocity is the same as the average velocity.
– If the velocity is not constant, than the instantaneous velocity is not the same as the average velocity, and we must carefully distinguish between the two.
• Instantaneous Velocity X 0 m v ins = ∆x/ ∆t
∆t
t
∆x
Draw a tangent line to the curve at B. The slope of this line gives the instantaneous velocity at that specific time.
• Determine the instantaneous velocity at 1 second.
X 2.0
∆x ∆t
1.0
0 2.0
4.0
6.0
t -1.0
-2.0
v ave = ∆x/ ∆t v ave = 1 m/ 1 s
= 1 m/s
• Acceleration (a) – Any change in velocity over a period of time is called acceleration.
– The sign (+ or -) of acceleration indicates its direction.
– Acceleration can be… • • • speeding up slowing down turning
• • • Questions If acceleration is zero, what does this mean about the motion of an object?
it means the velocity is constant at 0 m/s or some velocity greater than 0 m/s The object has a constant speed, and it is moving in a straight line Is it possible for a racecar circling a track to have zero acceleration?
No because the race car is changing direction. When an object changes direction it changes velocity, so it has an acceleration present.
• Uniform (Constant) Acceleration – In Physics B, we will generally assume that acceleration is constant.
– With this assumption we are free to use this equation: a = ∆v / ∆t • – The SI unit of acceleration is the m/s 2 .
Acceleration in 1-D Motion has a sign!
– If the sign of the velocity and the sign of the acceleration is the same, the object speeds up.
– If the sign of the velocity and the sign of the acceleration are different, the object slows down.
Qualitative Demonstrations 1) Demonstrate the motion of a particle that has zero initial velocity and positive acceleration.
2) Demonstrate the motion of a particle that has zero initial velocity and negative acceleration.
3) Demonstrate the motion of a particle that has positive initial velocity and negative acceleration.
4) Demonstrate the motion of a particle that has negative initial velocity and positive acceleration.
• Practice Problem: A 747 airliner reaches its takeoff speed of 180 mph (80.5 m/s) in 30 seconds. What is its average acceleration?
a = ? mph/s ∆v = v f – v i = 180 mph – 0 mph = 180 mph ∆t = t f – t i = 30 s – 0 s = 30 s a = ∆v / ∆t a = 180mph / 30 s = 6mph / s a = ? m/s 2 ∆v = v f – v i = 80.5 m / s– 0 m / s = 80.5 m / s ∆t = t f – t i = 30 s – 0 s = 30 s a = ∆v / ∆t a = 80.5 m / s / 30 s = 2.683 m / s 2
• A horse is running with an initial velocity of 11 m/s, and begins to accelerate at –1.81 m/s 2 . How long does it take the horse to stop?
a = -1.81 m/s 2 ∆v = v f – v i = 0 m /s – 11 m /s = -11 m /s ∆t = ? s ∆t = ∆v /a ∆t = -11 m /s / -1.81 m / s 2
= 6.1 s
• Graphical Problem v (m/s) 0 m T (s) Demonstrate the motion of this particle. Is it accelerating?
• Graphical Problem v (m/s) 0 m T (s) Demonstrate the motion of this particle. Is it accelerating?
• Graphical Problem v (m /s) 0 m A a = ∆v/ ∆t ∆t B ∆v t (s) What physical feature of the graph gives the acceleration from A to B?
• Graphical Problem: Determine the acceleration from the graph.
v (m/s) 2.0
1.0
0 -1.0
-2.0
a= ∆v/ ∆t 2.0
∆t
4.0
∆v
6.0
t (s) a = 1 m /s / 1 s = 1 m/s 2
• Graphical Problem: Determine the displacement of the object from 0 to 4 seconds..
v (m/s)
Area of a triangle = ½ b ∙h
2.0
1.0
h
0 -1.0
2.0
h b
4.0
6.0
t (s) -2.0
b A triangle = ½ b ∙h = ½ ∙ 2 ∙2 = 2 m A triangle = ½ b ∙h = ½ ∙ 2 ∙2 = - 2 m disp = 2 m – 2m = 0 m
• • Need to do this example below 5-14-10 1-D kinematics slide 41.xlsx
x (m) 0 t (s) 0 2.3
Pinewood Derby 9.2
20.7 36.8 57.5
1.0
2.0
3.0
4.0
5.0
•
On your graph paper, do the following.
a) Draw a position vs time graph for the car.
b) Draw tangent lines at three different points on the curve to determine the instantaneous velocity at all three points.
c) On a separate graph, draw a velocity vs time graph using the instantaneous velocities you obtained in the step above.
d)From your velocity vs time graph, determine the acceleration of the car.
distance - time graph x m
40 30 20 10 0 0 70 60 50 1 2 3
t (s)
4 5 6 x (m)
• Uniformly Accelerating Objects • You see the car move faster and faster as • time elapses.
For each successive ∆t, • ∆x is bigger.
The velocity vs time graph reflects the increasing velocity
What are the signs of velocity and acceleration? How is speed changing?
This object is moving with positive velocity and acceleration. Speed is increasing.
This object is moving in the negative velocity and acceleration. Speed is increasing.
This object is moving with negative velocity and positive acceleration. Speed is decreasing.
• Draw Graphs for Stationary Particles x v a 0 m t 0 m / s t 0 m / s 2 position vs time position vs time position vs time t
• Draw Graphs for Constant Non-zero Velocity x v a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Draw Graphs for Constant Non-zero Acceleration increasing and positive velocity x v a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Draw Graphs for Constant Non-zero Acceleration decreasing and negative velocity x v a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Draw Graphs for Constant Non-zero Acceleration increasing and negative velocity x v a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Draw Graphs for Constant Non-zero Acceleration x a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Kinematic Equations for uniformly accelerating objects v = v 0 x = x 0 + at + 1/2at 2 x = v 0 t + 1/2at 2 x = x 0 + v 0 t + 1/2at 2 On the AP equation sheet On the AP equation sheet v 2 = v 0 2 + 2a(∆x) v 2 = v 0 2 + 2a(x – x 0 ) On the AP equation sheet
• What must a particular Olympic sprinter’s acceleration be if he is able to attain his maximum speed in ½ of a second?
v 0 = 0 m/ s t = ½ s v = v 0 + at a = v / t a = v / (1/2)
a = 2v
v = at
• A plane is flying in a northwest direction when it lands, touching the end of the runway with a speed of 130 m/s. If the runway is 1.0 km long, what must the acceleration of the plane be if it is to stop while leaving ¼ of the runway remaining as a safety margin?
t unk v 0 m/ s V 0 130 m/ s ∆x 750 m v 2 = v 0 2 + 2a(∆x) 0 = v 0 2 + 2a(∆x) a = v 0 2 / (2(∆x)) a = (130 m/s) 2 / (2(750m))
= - 11.23 m /s 2
• On a ride called the Detonator at Worlds of Fun in Kansas City, passengers accelerate straight downward from 0 to 20 m/s in 1.0 second.
a) What is the average acceleration of the passengers on this ride?
t v v 0 ∆x a 1.0s
20 m/ s 0 m/ s ?
? m/s 2 v = v 0 + at v = 0 + at a = 20 m/s / 1s
= 20 m/s 2
v = at a = v / t b) How fast would they be going if they accelerated for an additional second at this rate?
t = 2 s a = 20 m/s 2 v = 0 + at v = at v = 20 m/s 2 ∙ 2 s
= 40 m /s
c) Sketch approximate x-t, v-t and a-t graphs for this ride.
v x a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Air bags are designed to deploy in 10 ms. Estimate the acceleration of the front surface of the bag as it expands. t = 2 ms = 0.002 s a = ?
v = v 0 + at v 0 = 0 m / s v = at a = v / 0.002s
= 500 v
• You are driving through town at 12.0 m/s when suddenly a ball rolls out in front of you. You apply the brakes and decelerate at 3.5 m/s 2 .
a) How far do you travel before stopping?
t unk v 0 m/ s v 0 12 m/ s ∆x unk a -3.5 m/s 2 v 2 = v 0 2 (∆x) = - v 0 2 + 2a(∆x) / (2a) 0 = v 0 2 = - (12 m / s) 2 + 2a(∆x) / (2∙ - 3.5 m/s 2 )
= 20.57 m
b) When you have traveled only half the stopping distance, what is your speed?
∆x = 20.57 / 2 = 10.29 m v 2 = v 0 2 + 2a(∆x) v = √((12 m/s) 2 + 2 ∙ -3.5 m/s 2 ∙ (10.29 m))
= 8.49 m / s
•
Practice Problem -- continued
c) How long does it take you to stop?
t v v 0 ∆x a ? s 0 m/ s 12 m/ s 20.57 m -3.5 m/s 2 v = v 0 + at 0 = v 0 + at at = v 0 t = -v 0 / a = -(12 m/s) /- 3.5 m/s 2
= 3.43 s
•
Practice Problem -- continued
d) Draw x vs t, v vs t, and a vs t graphs for this. x v a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• • • • • Free Fall Free fall is a term we use to indicate that an object is falling under the influence of gravity, with gravity being the only force on the object. Air resistance is considered negligible.
Gravity accelerates the object toward the earth the entire time it rises, and the entire time it falls.
Freefall is like any other type of accelerated motion; it’s just that the acceleration is pretty well defined. The acceleration due to gravity near the surface of the earth has a magnitude of
9.8 m/s 2
.
The direction of this acceleration is
DOWN
.
• You drop a ball from rest off a 120 m high cliff. Assuming air resistance is negligible, a) how long is the ball in the air?
t v v 0 ∆y ? s ? m/ s 0 m/ s -120 m a -9.8 m/s 2 y = y 0 + 1/2at 2 y 0 = 0m t = √ (2y / a) y = 0 + 1/2at t = √ (2 ∙ -120m/ -9.8 m/s 2 )
Exact same equation Or = 4.94 s
2 y = v 0 t + 1/2at 2 y = 0 t + 1/2at 2 t = √ (2y / a)
• You drop a ball from rest off a 120 m high cliff. Assuming air resistance is negligible, b) what is the ball’s speed and velocity when it strikes the ground at the base of the cliff?
t 4.94s
v ? m/ s v 0 0 m/ s ∆y 120 m a -9.8 m/s 2 v = v 0 + at v = 0 + at v = at = - 9.8 m/s 2 ∙ 4.94s
= - 48.12 m / s Velocity = - 48.12 m / s speed = 48.12 m / s
• You drop a ball from rest off a 120 m high cliff. Assuming air resistance is negligible, c) sketch approximate x-t, v-t, a-t graphs for this situation.
x v a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Symmetry in Free Fall – When something is thrown straight upward under the influence of gravity, and then returns to the thrower, this is very symmetric.
– The object spends half its time traveling up; half traveling down.
– Velocity when it returns to the ground is the opposite of the velocity it was thrown upward with.
– Acceleration is 9.8 m/s 2 and directed DOWN time the object is in the air!
the entire
Objects Launched from the Ground Ex: water rocket **Force of water accelerates rocket off the ground. Once the water is gone, gravity takes over:
3s; v=0m/s, a=9.8m/s 2 2s; v=9.8m/s a=-9.8m/s 2 4s; v=-9.8m/s a=-9.8m/s 2 1s; v=19.6m/s a= a=-9.8m/s 2 0s; v i =29.4m/s a=-9.8m/s 2 5s; v=-19.6m/s a=-9.8m/s 2 6s; v f =-29.4m/s a=-9.8m/s 2
• If object is thrown up, the upward downward velocities and displacements are equal and opposite.
v top = 0 m/s, a top = -9.8 m/s 2
v 1 > v top , a = -9.8 m/s 2
v 2 > v 1 , a = -9.8 m/s 2 v 3 > v 2 , a = -9.8 m/s 2
• You throw a ball straight upward into the air with a velocity of 20.0 m/s, and you catch the ball some time later.
a) How long is the ball in the air?
?s
t v 0 m/ s v 0 20 m/ s ∆y ? m a -9.8 m/s 2 v = v 0 + at 0 = v 0 + at at = -v 0 t = -v 0 / a = - (20 m/s) /- 9.8 m/s 2
= 2.04 s
b) How high does the ball go?
v 2 = v 0 2 (∆y) = -v 0 2 + 2a(∆y) / (2a) 0 = v 0 2 = - (20 m / s) 2 + 2a(∆y) / (2∙ - 9.8 m/s 2 )
= 7.35 m
•
Practice Problem -- continued
• c) What is the ball’s velocity when you catch it?
- 20 m/s, due to the symmetry of the ball motion.
d) Sketch approximate x-t, v-t, a-t graphs for this situation.
x v a 0 m t 0 m / s t 0 m / s 2 t position vs time position vs time position vs time
• Reflex Testing – Who’s has the quickest reflexes in Physics B? We’re dying to know.
– Using only a meter stick, test the reflexes of each person in your group by dropping the meter stick between his or her thumb and forefinger. One trial only per person, please!
– We will then identify the top 3 quickest students, and discuss the results.
• Ball A is dropped from rest at the top of a cliff of height h as shown. Using g as the acceleration due to gravity, derive an expression for the time it will take for the ball to hit the ground.
A v = v 0 + at h v = 0 + gt v = gt t = v / g
y = v 0 t + 1/2at t = √ (2h / g) 2 Derive an equation using g, h h = 0 t + 1/2gt 2
• Ball B is projected vertically upward from the foot of the cliff with an initial speed of v o . Derive an expression for the maximum height y max reached by the ball.
Derive an equation using g, y max
h
v 0 v 2 = v 0 2 + 2a(∆y) 0 = v 0 2 + 2g(y max ) y max = -v 0 2 / (2g)
B
• Ball A is dropped from rest at the top of the cliff at exactly the same time Ball B is thrown vertically upward with speed v 0 from the foot of the cliff such that Ball B will collide with Ball A. Derive an expression for the amount of time that will elapse before they collide.
y A = v 0 t + 1/2at 2 v B = v 0B + at y A = 0 + 1/2at 2 0 = v 0B + at a = 2y A / t 2 a= - v - v 0B / t = 2y A / t t 2 / t = 2ya /- v 0B 2 0B / t t = - 2y A / v 0B