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Published byAmia Holden Modified over 4 years ago

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**I recommend writing down items in these yellow boxes**

Chapter 2: Motion in One Direction Motion Is Relative Displacement vs. Distance Velocity vs. Speed Velocity: Average and Instantaneous Acceleration Free Fall I recommend writing down items in these yellow boxes

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**“Motion Is Relative” – What does this mean?**

Motion of objects is always described as relative to something else. For example: Motion takes place over time and depends upon the frame of reference (precise location of objects in space).

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**Displacement – (your thoughts?)**

The length of the straight line drawn from its initial position to the object’s final position. Displacement: the change in position of an object Δx = xf – xi Δx = Change in position on the x-axis Δ = Change xf = final position (on x-axis) xi (xo) = initial position (on x-axis)

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**Displacement xf xi Δx = xf – xi**

Δx = Change in position on the x-axis xi xf Δx = xf – xi = 7.0 cm – 2.0 cm = 5.0 cm

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**Displacement xi xf Δx = xf – xi**

Δx = Change in position on the x-axis xi xf Δx = xf – xi = 1.2 cm – 6.5 cm = -5.3cm Positive or negative depends on the direction

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**Positive or negative depends on the direction**

Displacement Δy = yf – yi yf Δy = Change in position on the y-axis Δy = yf – yi = 10.0cm – 6.1 cm = 3.9cm yi Positive or negative depends on the direction

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**Displacement = Distance ???**

xi xf xf xi

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**Displacement = Distance ???**

xi xf xi xf xf Distance does not have positive or negative, it is the total distance traveled, NOT just the final outcome (xf – xi).

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**Velocity – (your thoughts?)**

Δx change in distance Δt change in time vavg = = = m/s Δx xf – xi Δt tf – ti vavg = = d t v = = m/s NOTE: For most problems, the initial values are going to be zero. We typically start at time zero, then begin. Our displacement also typically starts at a position of zero and moves a certain “distance”.

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**Velocity = Speed ??? (your thoughts?)**

Average Velocity Constant Velocity Instantaneous Velocity

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**Velocity (v) is a description of**

Δx change in distance Δt change in time v = = = m/s d t v = = m/s Velocity (v) is a description of the instantaneous speed of the object what direction the object is moving (north, left, right, up, down, pos x-axis, neg x-axis) Vectors have magnitude and direction Velocity is a vector quantity. It has magnitude: instantaneous speed direction: direction of object’s motion (typically positive and negative x-axis)

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**Speed (s) and Velocity (v)**

Constant speed is steady speed, neither speeding up nor slowing down. Constant velocity is constant speed and constant direction (straight-line path with no acceleration). Motion is relative to Earth, unless otherwise stated.

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**P.I.E.S.S ??? Δd change in distance Δt change in time v = = = m/s d t**

Example: A flying disc leave’s Mr. Pearson’s hand with a velocity of 15 m/s. Ignoring wind resistance, the disc travels for 10 seconds before hitting the ground. How far did the disc travel? Ignoring wind resistance = no friction = constant velocity v = 15 m/s d t v = d = t * v t = 10 s = 10 s * 15m/s d = ____ m v = 15 m/s d = 150 m P.I.E.S.S ???

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**Velocity t0 t1 t2 t3 Δx xf – xi Δt tf – ti vavg = = x0 = 0.0 cm**

1) What is the displacement between x1 and x2? 2) What is the average velocity over the entire movement from x0 to x3? 3) What is the average velocity between t0 and t1? 4) What is the average velocity between x2 and x3?

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Δx xf – xi Δt tf – ti vavg = = Δx Δt distance (m) Δx Δt time (s)

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**Δx xf – xi vavg = = Δt tf – ti Instantaneous distance (m) Velocity**

(speedometer) distance (m) Δx xf – xi Δt tf – ti vavg = = time (s)

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**Instantaneous Velocity**

Instantaneous Velocity - is the velocity at any specific instant. Example: When you ride in your car, you may speed up and slow down - (this will affect your average velocity over a given distance and time) Your instantaneous speed is given by your speedometer (your speed at that moment).

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Average Speed The entire distance covered divided by the total travel time Doesn’t indicate various instantaneous speeds along the way. 200 km 2 h = 100 km/h Example: What is your average speed if you drive a total distance of 200 km in 2 hours?

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**Average Speed CHECK YOUR NEIGHBOR**

The average speed of driving 30 km in 1 hour is the same as the average speed of driving 30 km in 1/2 hour. 30 km in 2 hours. 60 km in 1/2 hour. 60 km in 2 hours. 20

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**Average Speed CHECK YOUR ANSWER**

The average speed of driving 30 km in 1 hour is the same as the average speed of driving 30 km in 1/2 hour. 30 km in 2 hours. 60 km in 1/2 hour. 60 km in 2 hours. 60 km 2 hour = 30 km 1 hour Explanation: Average speed = total distance / time So, average speed = 30 km / 1 h = 30 km/h. Same Now, if we drive 60 km in 2 hours: Average speed = 60 km / 2 h = 30 km/h 21

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Acceleration Formulated by Galileo based on his experiments with inclined planes.

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**vo = velocity initial (if starting from rest vo = ZERO) **

Acceleration (a) - rate at which velocity changes over time Δv vf - vo change in velocity m/s Δt tf - to time required for change s a = = = = = m/s2 a = = = m/s2 v m/s t s vf = velocity final vo = velocity initial (if starting from rest vo = ZERO) tf = time final to = time initial NOTE: Usually we don’t use tf - to just t = required time vo = velocity at the starting time At time = 0 sec

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**Your car’s acceleration is 2.5 km/h/s.**

Δv vf - vo change in velocity m/s Δt t time required for change s a = = = = = m/s2 Example: You car’s speed right now is 40 km/h. (vo = 40 km/h) Your car’s speed 2 s later is 45 km/h. (vf = 45 km/h t = 2s ) Your car’s change in speed is 45 – 40 = 5 km/h. (change = Δv) Δv vf - vo km/h – 40 km/h 5 km/h Δt t s s a = = = = = 2.5 km/h/s Your car’s acceleration is 2.5 km/h/s.

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**Δv vf - vo change in velocity m/s Δt t change in time s **

a = = = = = m/s2 Example: A flying disc leave’s Mr. Pearson’s hand with a velocity of 15 m/s. After 3 seconds, wind resistance slows down the disc to 11 m/s. What is the acceleration of the disc? , Change in velocity results in Acceleration vo = 15 m/s vf - vo t a = 11 m/s – 15 m/s 3 s = vf = 11 m/s -4 m/s 3 s = t = 3 s a = 1.3 m/s2 The negative means that the acceleration is in the direction opposite to the motion of the object = resulting in slowing a = = = m/s2 v m/s t s

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**x vf = 8 m/s Δv vf - vo change in velocity m/s a = = = = = m/s2**

Δt t change in time s a = = = = = m/s2 Example: If a different disc has an acceleration of m/s2. What will the velocity be after 2 seconds of flight if Mr. Pearson releases the disc at 13 m/s. Don’t wait for the teacher, get to work vf = 8 m/s Is this answer REASONABLE? Explain x a = = = m/s2 v m/s t s

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Acceleration When you are driving a car, what are three things you can do to cause acceleration?. Acceleration involves a change in velocity: A change in speed (faster or slower), or A change in direction, or Both. NOTE: You do not feel velocity You can feel a CHANGE in velocity You DO feel ACCELERATION

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**Acceleration CHECK YOUR NEIGHBOR**

An automobile is accelerating when it is slowing down to a stop. rounding a curve at a steady speed. Both of the above. Neither of the above. 28

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**Acceleration CHECK YOUR ANSWER**

An automobile is accelerating when it is slowing down to a stop. rounding a curve at a steady speed. Both of the above. Neither of the above. Explanation: Change in speed (increase or decrease) is acceleration, so slowing is acceleration. Change in direction is acceleration (even if speed stays the same), so rounding a curve is acceleration. Acceleration occurs due to a change in either speed or direction (or both). When a car slows down it changes its speed, so it is accelerating. When a car rounds a curve, although its speed is steady it is accelerating because it is changing direction. 29

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**Acceleration CHECK YOUR NEIGHBOR**

Acceleration and velocity are actually the same. rates but for different quantities. the same when direction is not a factor. the same when an object is freely falling. 30

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**Acceleration CHECK YOUR ANSWER**

Acceleration and velocity are actually the same. rates but for different quantities. the same when direction is not a factor. the same when an object is freely falling. Explanation: Velocity is the rate at which distance changes over time, Acceleration is the rate at which velocity changes over time. Δv vf - vo change in velocity m/s Δt t change in time s a = = = = = m/s2 Δd change in distance Δt change in time v = = = m/s 31

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**Acceleration Galileo increased the inclination of inclined planes.**

Steeper inclines gave greater accelerations. When the incline was vertical, acceleration was max, same as that of the falling object. When air resistance was negligible, all objects fell with the same unchanging acceleration. Unchanging = CONSTANT Acceleration Which ball hits the ground first? Which ball has the highest vf ? Which ball has the highest vo ?

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Free Fall Falling under the influence of gravity only - with no air resistance Freely falling objects on Earth accelerate at the rate of 10 m/s/s, i.e., 10 m/s2 a = 10 m/s2 = 9.8 m/s2 = m/s2 (sig figs) We say this is the acceleration due to gravity (g): a = g = 10 m/s = m/s2 Text book vs Real World (easier math)

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Free Fall—How Fast? a = Δv t The velocity acquired by an object starting from rest is So, under free fall, when acceleration is a = g = 10 m/s2, the speed is v0 = 0 m/s before release: t0 = 0 s v1 = 10 m/s after 1 s t1 = 1 s v2 = 20 m/s after 2 s. t2 = 2 s v3 = 30 m/s after 3 s. t3 = 3 s And so on. Acceleration = Change of 10 m/s per second

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**Gravity WITH Friction Δv a = t**

In this situation g still equals 10 m/s2, but the motion of the object is no longer strait down. Here, we must include friction and trigonometry. In this example: a = 4 m/s2 NOT 10 m/s2 v0 = 0 m/s before release: t0 = 0 s v1 = 4 m/s after 1 s t1 = 1 s v2 = 8 m/s after 2 s. t2 = 2 s v3 = 12 m/s after 3 s. t3 = 3 s And so on. Acceleration = Change of 4 m/s per second

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Free Fall—How Fast? A free-falling object has a speed of 30 m/s at one instant. Exactly 1 s later its speed will be the same. 35 m/s. more than 35 m/s 60 m/s. a = Δv t Δ v = a*t vf = vo + Δv vf = vo + a*t vf = 30m/s + (10m/s2 * 1s) vo = 30 m/s The object gains 10m/s after every second. So, at a given instance, if its speed is 30m/s, then a second later it will be 10m/s greater than that i.e. 30m/s + 10m/s = 40m/s So, Its speed a second later is 40m/s which is more than 35m/s. vf = 30m/s + 10m/s t = 1 s vf = 40m/s vf = ____ m/s a = g = 10 m/s2 Freely Falling 36

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Free Fall—How Far? The distance covered by an accelerating object -- starting from rest is: d = ½ *a*(t)2 m = ½ *(m/s2)*(s)2 Note that the seconds cancel So, under free fall, when acceleration is 10 m/s2, the distance is at t = 1 s; d = ½ *a*(t)2 = ½ (10 m/s2) (1 s)2 = 5 m at t = 2 s; d = ½ *a*(t)2 = ½ (10 m/s2) (2 s)2 = 20 m at t = 3 s; d = ½ *a*(t)2 = ½ (10 m/s2) (3 s)2 = 45 m And so on

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**Free Fall—How Far? CHECK YOUR NEIGHBOR**

What is the distance covered of a freely falling object starting from rest after 4 s? 4 m 16 m 40 m 80 m d = ½ *a*(t)2 d = ½ *(10 m/s2)*(4 s)2 d = (5 m/s2) * (16 s2) d = 80 m vo = 0 m/s (rest) t = 4 s d = ____ m a = g = 10 m/s2 Freely Falling 38

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**2*d = (t)2 d = ½ *a*(t)2 a t = (2*d/a)1/2 t = (2*34 m / 10m/s2)1/2**

EXAMPLE: A ball thrown straight up into the air begins to slow down (constant acceleration) until it stops in mid air at a height of 34 meters. How long does it take to reach this height? (NOTE: the same equation for free-fall is the same) d = 34 m 2*d = (t)2 a d = ½ *a*(t)2 t = _____ s a = g = 10 m/s2 t = (2*d/a)1/2 Freely Falling t = (2*34 m / 10m/s2)1/2 t = (68 m / 10m/s2)1/2 t = (6.8 s2)1/2 t = 2.6 s d = 34 m Is this answer REASONABLE?

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**Free Fall—How Far? d = ½ *a*(t)2 Δd change in distance**

at t = 1 s; d = ½ *a*(t)2 = ½ (10 m/s2) (1 s)2 = 5 m at t = 2 s; d = ½ *a*(t)2 = ½ (10 m/s2) (2 s)2 = 20 m at t = 2.6 s; d = ½ *a*(t)2 = ½ (10 m/s2) (2.6 s)2 = 34 m at t = 3 s; d = ½ *a*(t)2 = ½ (10 m/s2) (3 s)2 = 45 m Δd change in distance Δt change in time v = = = m/s d t v = = m/s Δv vf - vo change in velocity m/s Δt t change in time s a = = = = = m/s2 d = ½ *a*(t)2 = ½ *(m/s2)*(s)2 = m a = g = 10 m/s2 = 9.8 m/s2 (for freely falling objects)

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**d = ½ *a*(t)2 vf = vo + a*t a = g = 10 m/s2 Δv a = t**

Figure 3.8 t3 = 3 s v3 = 0 m/s t2 = 2 s v2 = 10 m/s t4 = 1 s t4 = 4 s v4 = -10 m/s (negative = direction) t1 = 1 s v1 = 20 m/s t5 = 2 s t5 = 5 s v5 = -20 m/s t0 = 0 s v0 = 30 m/s t6 = 3 s t6 = 6 s v6 = -30 m/s a = g = 10 m/s2 d = ½ *a*(t)2 a = Δv t vf = vo + a*t t7 = 7 s v7 = -40 m/s t7 = 4 s

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Ch. 2 Conceptual Practice

Ch. 2 Conceptual Practice

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