 # PHYS 201 Chapter 2: Kinematics in 1-D Distance Displacement Speed

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PHYS 201 Chapter 2: Kinematics in 1-D Distance Displacement Speed
Velocity Acceleration Equations of Kinematics Free Fall

Distance (scalar, units: mi, ft, m, km, …)
How far an object travels. Displacement (vector, units: mi, ft, m, km, …) The change in position. (Shortest distance between two points. The vector points from initial to final point).

Average Speed (scalar, unit: mi/h (mph), ft/s, m/s, km/h, …)
The rate of distance change. Velocity (vector, unit: mi/h (mph), ft/s, m/s, km/h, …) The rate of displacement change.

Acceleration (vector, unit: mi/h2, ft/s2, m/s2, km/h2, …)
The rate of velocity change.

CLICKER! A car is moving with a constant speed of 20 mi/h at the Richland avenue round about in circle. When the car reaches to its initial point, its displacement is (radius: r = 50 ft, circumference = 2pr) (1) 2pr (2) 0 (3) 2pr2 Initial and final points are the same. So displacement = 0.

CLICKER! (1) Velocity is constant. (2) Velocity is zero.
A car is moving with a constant speed of 20 mi/h at the Richland avenue round about in circle. Which of the following is true? (radius: r = 50 ft, circumference = 2pr) (1) Velocity is constant. (2) Velocity is zero. (3) Velocity is changing. Although the magnitude does not change, the direction changes. So the velocity is varying

CLICKER! (1) There is an acceleration. (2) Acceleration is zero.
A car is moving with a constant speed of 20 mi/h at the Richland avenue round about in circle. Which of the following is true? (radius: r = 50 ft, circumference = 2pr) (1) There is an acceleration. (2) Acceleration is zero. (3) Acceleration is constant. Velocity changes, so there is an acceleration.

Ex. 1 The distance between Athens and Columbus is 80 miles and it takes about 1 hr and 30 min to drive (of course, if you are not speeding!). What is the average speed you are driving?

Ex. 2 A car travels in a straight line with an average velocity of 60 mi/hr for 1 hr and 30 min, and then it travels an average of 45 mi/h for 30 min in a road construction area. (a) What is the total displacement? (b) What is the average speed for the total trip? (c) Draw a distance vs. time graph for (b).

Ex. 3 You bike towards East with 30 mi/h for 30 min, and then 15 mi/h towards North for 10 min. Find (1) the total distance covered (2) the average speed (3) the total displacement (4) the average velocity (5) draw a distance vs. time plot for this trip.

Equations of Kinematics for Constant Acceleration
Equations of Kinematics for Constant Acceleration V = V0 + at Dx = ½ (V0 + V) t Dx = V0t + ½ at2 V2 = V a Dx Conditions: must moves in a straight line. must be a constant acceleration.

Ex. 4 A car accelerates uniformly from rest to a speed of 25 m/s in 8
Ex. 4 A car accelerates uniformly from rest to a speed of 25 m/s in 8.0s. Find the distance the car travels in this time and the constant acceleration of the car. Define the +x direction to be in the direction of motion of the car.     (1) V = V0 + at (2) Dx = ½ (V0 + V) t (3) Dx = V0t + ½ at2 (4) V2 = V a Dx

Ex. 5 [Plane Landing] A jet plane lands with a velocity of +100m/s and with full brakes it accelerates at a rate of -5.0m/s2. From the instant it touches the runway, what is the minimum time needed before it can come to rest? Can this plane land on a small island airport where the runway is 0.80km long?     (1) V = V0 + at (2) Dx = ½ (V0 + V) t (3) Dx = V0t + ½ at2 (4) V2 = V a Dx

Free Fall Objects fall near the Earth surface moves with a constant acceleration due to gravity (if the air resistance is neglected). The acceleration due to gravity g = m/s2. All constant acceleration equations above can be applied. Following symbols are changed: ‘a’ = ‘g’, and ‘x’ = ‘y’.

Ex. 6 A ball is thrown up into the air at a speed of 25 m/s
Ex. 6 A ball is thrown up into the air at a speed of 25 m/s. How high does the ball travel?           (1) V = V0 + gt (2) Dy = ½ (V0 + V) t (3) Dy = V0t + ½ gt2 (4) V2 = V g Dy

CLICKER! E D C position B A time (1) A (2) D (3) C (4) A and D
This position versus time graph represents the position of a person as a function of time. At what point or points is the velocity constant (not changing)? (1) A (2) D (3) C (4) A and D (5) B and E (6) A, B, D and E E D C position B A time Slope represents the velocity. Anywhere slope is not changing, velocity is not changing! At what point is the person moving fastest? B – the greatest slope

Which velocity plot corresponds to the position plot to the right?
CLICKER! Which velocity plot corresponds to the position plot to the right? x vs t graph: Constant positive slope, then zero slope (zero v)

CLICKER! A ball is tossed into the air. Assume the +y direction to be upwards. Which plot best represents the VELOCITY versus time of this graph? Velocity starts upward (+), then transitions to downward (-).

CLICKER! The force due to gravity does not change directions.
A ball is tossed into the air. Think about the direction of the acceleration as the ball is traveling upward and as it is traveling downward. acceleration is upward when ball traveling up; downward when traveling down acceleration is downward when ball traveling up; downward when traveling down acceleration is upward when ball traveling up; upward when traveling down acceleration is downward when ball traveling up; upward when traveling down The force due to gravity does not change directions.

CLICKER! A ball is tossed into the air. Think about the velocity and acceleration at the point when the ball reaches maximum height. the velocity is zero and the acceleration is zero the velocity is non-zero and the acceleration is zero the velocity is zero and the acceleration is non-zero the velocity is non-zero and the acceleration is non-zero Even though velocity is zero right at max height, the gravity doesn't 'turn off' for split second.

y = -25.9 m or 25.9m below release point
Ex. 7 You drop a stone into a well. It takes 2.3s for the stone to reach the bottom. How deep is the well? y = ? v0 = 0m/s v = ? a = 9.8 m/s2 down t = 2.3 s y = v0t + (1/2)g t2 y = 0*(2.3s) + (0.5)(-9.8m/s2)(2.3s)2 y = m or 25.9m below release point

y = -60m v0 = -35m/s v = ? v2 = v02 + 2 g y g = -9.8 m/s2
Ex. 8 You throw a rock from a bridge at a speed of 35 m/s straight down. How fast is it traveling when it hits the water 60m below and how much time does it take to travel the 60m? y = -60m v0 = -35m/s v = ? g = -9.8 m/s2 t = ? v2 = v g y v2 = (-35)2 + 2(-9.8)(-60m) v = 49 m/s v = v0 + g t -49m/s = -35m/s + (-9.8)*t t= 1.43 s

C B A C = A + B B -1 q = tan A 1 ft = 12 in 1 m = 3.281 ft
Unit Conversion QUIZ 1 Equations 1 ft = 12 in 1 m = ft 1 in = 2.54 cm 1 mi = km 1 mi = 5280 ft Acceleration due to gravity g = m/s2 C B (1) V = V0 + at (2) Dx = ½ (V0 + V) t (3) Dx = V0t + ½ at2 (4) V2 = V a Dx q A 2 2 C = A + B B -1 q = tan A

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