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MOTION An object is in motion if its position changes. The mathematical description of motion is called kinematics. The simplest kind of motion an object can experience is uniform motion in a straight line. The object experiences translational motion if it is moving without rotating.

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One-dimensional motion means that objects are only free to move back and forth along a single line. As a coordinate system for one-dimensional motion, choose this line to be an x-axis together with a specified origin and positive and negative directions.

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**DISTANCE AND DISPLACEMENT**

Distance is the length between any two points in the path of an object. Displacement is the length and direction of the change in position measured from the starting point. The primary difference between the two is that the distance an object travels tells you nothing about the direction of travel, while displacement tells you precisely how far, and in what direction, from its initial position an object is located. Distance is the total length of travel and displacement is the net length of travel accounting for direction.

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**Distance traveled = AC + CB + BC = 400 m.**

Displacement = xf - xi = 200 m

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**PROBLEM SOLVING GUIDELINES**

• Make a list of given quantities (Data) • Convert units if needed • Make a sketch if necessary • Identify what is to be determined • Always solve the equation for the unknown • Be consistent with units • Check that the answer seems reasonable • Physics problems take practice – the more you do the easier it will be!

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**distance = xtotal = xo+ xgs+ xgs = 0 + 4.83 + 4.83 = 9.66 km**

2.1 You leave your home and drive 4.83 km east on Mt Vernon Hwy to go to the grocery store. After shopping, you go back home by traveling west on Mt. Vernon Hwy. a. What distance do you travel during this trip? xo = 0 km xgs = 4.83 km distance = xtotal = xo+ xgs+ xgs = = 9.66 km b. What is your displacement? xo = 0 km xf = 0 km displacement = Δx = xf - xo = 0 - 0 = 0 km

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SPEED If an object takes a time interval t to travel a distance x, then the average speed of the object is given by: Units: m/s 2.2 A ship steams at an average speed of 30 km/h. a. What is the speed in m/s? v = 30 km/h = 8.33 m/s

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**= 30(24) = 720 km = 16.67 h b. How far in km does it travel in a day?**

v = 30 km/h t = 24 h = 30(24) = 720 km c. How long in hours does it take to travel 500 km? x = 500 km = h

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AVERAGE VELOCITY Average velocity is the displacement divided by the amount of time it took to undergo that displacement. The difference between average speed and average velocity is that average speed relates to the distance traveled while average velocity relates to the displacement.

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**= 86 km/h Total distance traveled Total time x = v t**

2.3 A car travels at 100 km/h for 2 h, at 75 km/h for the next 2 h, and finally at 80 km/h for 1 h. What is the car’s average velocity for the entire journey? v1 = 100 km/h t1 = 2 h v2 = 75 km/h t2 = 2 h v3 = 80 km/h t3 = 1 h Total distance traveled x = v t x1 = v1t1 = 100 (2) = 200 km x2 = 75 (2) = 150 km x3 = 80 (1) = 80 km xT = = 430 km Total time tT = = 5 hours = 86 km/h

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**The run ends at the starting point so the displacement is zero then**

2.4 A runner makes one lap around a 200 m track in a time of 25 s. What were the runner's a. average speed and b. average velocity? x = 200 m t = 25 s a. = 8 m/s The run ends at the starting point so the displacement is zero then v = 0 m/s b.

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ACCELERATION Acceleration is the rate of change of velocity. The change in velocity Δv is the final velocity vf minus the initial velocity vo. Acceleration happens when: An object's velocity increases An object's velocity decreases An object changes direction

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**The acceleration of an object is given by:**

Units: m/s2 A positive acceleration means an increase in velocity; a negative acceleration usually means a decrease in velocity (deceleration) unless the object is traveling in the negative direction of the velocity.

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**EQUATIONS FOR MOTION UNDER CONSTANT ACCELERATION:**

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**2.5 An object starts from rest with a constant acceleration of 8 m/s2 along a straight line.**

a. Find the speed at the end of 5 s vo = 0 m/s a = 8 m/s2 t = 5 s = 0 + 8(5) = 40 m/s b. Find the average speed for the 5 s interval = 20 m/s

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**c. Find the distance traveled in the 5 s**

= 0 + ½(8)(5)2 = 100 m or: = 20(5) = 100 m

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**2.6 A truck's speed increases uniformly from 15 km/h to 60 km/h in 20 s.**

a. Determine the average speed vo = 15/3.6 = 4.17 m/s vf = 60/3.6 = 16.7 m/s t = 20 s = 10.4 m/s b. Determine the acceleration = 0.63 m/s2

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**c. Determine the distance traveled, all in units of meters and seconds.**

= 10.4(20) = 208 m

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**2. 7 A skier starts from rest and slides 9. 0 m down a slope in 3. 0 s**

2.7 A skier starts from rest and slides 9.0 m down a slope in 3.0 s. In what time after starting will the skier acquire a speed of 24 m/s? Assume that the acceleration is constant. vo = 0 m/s, x = 9 m t = 3 s, vf = 24 m/s = 2 m/s2 = 12 s

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2.8 A bus moving at a speed of 20 m/s begin to slow at a constant rate of 3.0 m/s each second. Find how far it goes before stopping. vo = 20 m/s a = - 3 m/s/s = 67 m

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**2.9 A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time**

of 5.0 s. a. Determine the acceleration of the car vo = 30 m/s vf = 10 m/s t = 5 s = - 4 m/s2 b. Determine the distance it moves in the third second = 30(3-2)+ ½(-4)(32-22) = 20 m

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**2. 10 The speed of a train is reduced uniformly from 15 m/s to 7**

2.10 The speed of a train is reduced uniformly from 15 m/s to 7.0 m/s while traveling a distance of 90 m. a. Calculate the acceleration. vo = 15 m/s vf = 7 m/s x = 90 m = m/s2 b. How much farther will the train travel before coming to rest, provided the acceleration remains constant? = 25 m

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**ACCELERATION DUE TO GRAVITY**

All bodies in free fall near the Earth's surface have the same downward acceleration of: g = 9.8 m/s2 A body falling from rest in a vacuum thus has a velocity of 9.8 m/s at the end of the first second, 19.6 m/s at the end of the next second, and so forth. The farther the body falls, the faster it moves. A body in free fall has the same downward acceleration whether it starts from rest or has an initial velocity in some direction.

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**Galileo is alleged to have performed free-fall experiments by dropping objects off the Tower of Pisa**

Galileo Galilei ( )

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The presence of air affects the motion of falling bodies partly through buoyancy and partly through air resistance. Thus two different objects falling in air from the same height will not, in general, reach the ground at exactly the same time. Because air resistance increases with velocity, eventually a falling body reaches a terminal velocity that depends on its mass, size, shape, and it cannot fall any faster than that.

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FREE FALL When air resistance can be neglected, a falling body has the constant acceleration g, and the equations for uniformly accelerated motion apply. Just substitute a for g. Sign Convention for direction of motion: If the object is thrown downwards then g = 9.8 m/s2 If the object is thrown upwards then g = m/s2

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**2.11 A ball is dropped from rest at a height of 50 m above the ground.**

a. What is its speed just before it hits the ground? vo = 0 m/s y = 50 m g = 9.8 m/s2 = 31.3 m/s b. How long does it take to reach the ground? = 3.19 s

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**At highest point vf = 0 = 19.8 m/s**

2.12 A stone is thrown straight upward and it rises to a height of 20 m. With what speed was it thrown? y = 20 m g = -9.8 m/s2 At highest point vf = 0 = 19.8 m/s

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**2. 13 A stone is thrown straight upward with a speed of 20 m/s**

2.13 A stone is thrown straight upward with a speed of 20 m/s. It is caught on its way down at a point 5.0 m above where it was thrown. a. How fast was it going when it was caught? vo = 20 m/s y = 5 m g = -9.8 m/s2 = m/s (moving down) b. How long did the trip take? = 3.8 s

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**At highest point vf = 0 = 382. 8 m = 21.9 s**

2.14 A baseball is thrown straight upward on the Moon with an initial speed of 35 m/s. (g = 1.60 m/s2) Find a. The maximum height reached by the ball vo = 35 m/s g = -1.6 m/s2 At highest point vf = 0 = m b. The time taken to reach that height = 21.9 s

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**= - 13 m/s = 30 m/s c. Its velocity 30 s after it is thrown t = 30 s**

d. Its velocity when the ball's height is 100 m y = 100 m = 30 m/s

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2.15 A ball that is thrown vertically upward on the Moon returns to its starting point in 4.0 s. Find the ball's original speed. (g = 1.60 m/s2) t = 4 s g = -1.6 m/s2 y = 0 m = 3.2 m/s

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2.16 A rock is thrown vertically upward with a velocity of 20 m/s from the edge of a bridge 42 m above a river. How long does the rock stay in the air? First, find the velocity of the rock at the moment that it hits the river. vo = 20 m/s y = - 42 m g = -9.8 m/s2 = -35 m/s We select the negative answer because the rock will be moving toward the river, in the negative direction. Now, use this value to find t = 5.6 s

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**GRAPHICAL ANALYSIS OF MOTION**

Graphical interpretations for motion along a straight line (the x-axis) are as follows: The instantaneous velocity of an object at a certain time is the slope of the position (displacement) versus time graph at that time. It can be positive, negative, or zero.

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The instantaneous acceleration of an object at a certain time is the slope of the velocity versus time graph at that time. It can be positive, negative, or zero.

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**For constant velocity motion, the x versus t graph is**

a straight line.

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For constant acceleration motion, the v versus t graph is a straight line. The area under the curve gives the displacement.

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**= 0.5 m/s Same for A and B = 0.5 m/s a = 0 m/s2**

2.17 A car moves in a straight line and its odometer readings are plotted against time. a. Find the instantaneous speed of the car at points A and B. = 0.5 m/s B Same for A and B A b. What is the car's average speed? = 0.5 m/s c. What is its acceleration? a = 0 m/s2

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2.18 An object's one-dimensional motion along the x-axis is graphed in the figure. Describe its motion. t = 0-2 s object is at rest t = 2-4 s slope is = 1.5 m/s object moves at constant velocity t = 4-6 s object is at rest t = 6-10 s slope is = -1.3 m/s object moves in opposite direction

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**2. 19 The vertical motion of an object is graphed in the figure**

2.19 The vertical motion of an object is graphed in the figure. Describe its motion qualitatively and find its instantaneous velocity at points A, B and C. Object moves fastest at t = 0 as it slows and stops at B, it begins to fall back downward at increasing speed.

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**vA = 2.3 m/s vB slope = 0 = 0 m/s vC = -1.2 m/s**

Find its instantaneous velocity at points A, B and C. = 2.3 m/s vB slope = 0 = 0 m/s vC = -1.2 m/s

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