Chapter 3 Acceleration and Newton’s Second Law of Motion

Mechanics (Classical/Newtonian Mechanics): Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object. Kinematics – Mechanics that describes how objects move, no reference to the force or mass. Dynamics– Mechanics that deals with force: why objects move.

Describing Motion Involves: Reference frames Reference Direction Displacement Velocity Acceleration

Reference frames and coordinates When we say a car is moving at 45 km/hr, we usually mean “with respect to the earth” although is is not explicitly stated. An escalator is moving at 3 m/s relative to the ground. A lady walks on the escalator at 2 m/s relative to the escalator. The lady’s speed relative to the ground is 5 m/s.

Reference Direction Can use compass directions: North, East, South, West. In physics, we use coordinate axes 1-D coordinate system: Specify the origin. Show where x axis is pointing: to the right is positive direction of x. Label the axis with the relevant units.

Reference Direction: 2-D coordinate system: Specify the origin. Show x and y axes. To the right is positive direction of x. Vertically upward is positive y. Label the axis with the relevant units. 0 x y

Distance Just a length. A Scalar quantity. SI unit = meter (m).

Position  A vector quantity describing where you are relative to an “origin”.  Point A is located at x =3, y =1 or (3,1). Point B is located at (-1,-2). 10 y x A B

 A vector quantity.  Change in position relative to the starting point.  SI unit = meter (m).   r = r f – r i The displacement from A to B is x-direction: -1 – 3 = -4 y-direction: -2 – 1 = -3  r = (-4, -3), |  r| =  ( ) = 5 y 13 x A B Displacement (  r)

Example You drive 40 km north, then turn east and drive 30 km east. What is your net displacement? 40 km N 30 km E

2. A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of its displacement after this movement? A.8 km B.5.7 km C.16 km D.332 km E.0 km

Average Speed and Average Velocity Average Speed = (distance traveled)/time taken = d/t Speed: - specified only by magnitude. It is a scalar quantity. SI unit = m/s - always a positive number Average velocity = (displacement)/time taken = (x 2 – x 1 )/(t 2 – t 1 ) =  x/  t Velocity: - specified by both magnitude and direction. - It is a vector quantity. - units – m/s - Positive/ negative sign used to indicate direction

Charles walks 120 m due north in 40s. He then walks another 60 m still due north in another 60s. Average speed? Average velocity? Magnitude of average speed and average velocity: not always equal. Average speed = Total distance/time Average velocity = Displacement/time

Charles walks 120 m due north. He then walks another 60 m due south. He took a total of 100 s for the journey. Average speed? Average velocity? Average speed = Total distance/time Average velocity = Displacement/time

Charles walks 120 m due north. He then walks another 120 m due south. He took a total of 100 s for the journey. Average speed? Average velocity?

The average velocity is displacement divided by the change in time. Instantaneous velocity is limit of average velocity as  t gets small. It is the slope of the x(t) versus t graph. v = lim  t  0  x/  t. t x(t) tt xx t 23 Instantaneous Velocity

Magnitude of instantaneous speed and instantaneous velocity are equal. If velocity is uniform (constant) in a motion, then magnitude of Instantaneous velocity = Average speed. time (s) velocity (m/s) uniform (constant) velocity t1t1 t2t2

Determine the velocity of the car at times A, B, C and D. APositiveZeroNegative BPositiveZeroNegative CPositiveZeroNegative DPositiveZeroNegative x(t) t A BCD

Uniform Velocity An object moving with uniform velocity – means velocity stays uniform (constant) throughout the motion. Its magnitude and direction stays the same. Its displacement changes uniformly. Moving along a straight line with constant speed. x(t) t v = slope t v(t)

A car moves at a constant velocity of magnitude 20 m/s. At time t = 0, its position is 50 m from a reference point. What is its position at (i) t = 2s? (ii) t = 4s (iii) t = 10s?

Non-uniform Velocity An object moving with non-uniform velocity – means velocity does not stays uniform (constant) during the motion. Either its magnitude or direction stays the same. Moving along a straight line with varying speed, or moving in a curved path. x(t) tt v(t)

Area Under velocity-time Graph Area under the velocity-time graph = magnitude of the displacement over the time interval. time (s) velocity (m/s) uniform (constant) velocity t1t1 t2t2 Area = v(t 2 -t 1 ) = (x 2 – x 1 ) =  x time (s) velocity (m/s)

A stray dog ran 3 km due north and then 4 km due east. What is the magnitude of its average velocity if it took 45 min?

The average acceleration is the change in velocity divided by the change in time. SI unit = m/s 2 Slope of velocity-time graph. t v(t) tt vv A C E B D Acceleration (a)

Instantaneous acceleration is limit of average acceleration as  t gets small. It is the slope of the v(t) versus t graph. a = lim  t  0  v/  t. v(t) t Acceleration (a)

Uniform Acceleration An object moving with uniform acceleration – means acceleration stays uniform (constant) throughout the motion. Its magnitude and direction stays the same. Its velocity changes uniformly. Moving along a straight line. a = slope t a(t) v(t) t vv tt

A car moves at a constant acceleration of magnitude 5 m/s 2. At time t = 0, the magnitude of its velocity is 8 m/s. What is the magnitude of its velocity at (i) t = 2s? (ii) t = 4s? (iii) t = 10s?

Average acceleration =  v/  t T/F? If the acceleration of an object is zero, it must be at rest. T/F? If an object is at rest, its acceleration must be zero. Acceleration (a)

Example Train A moves due east along a straight line with a velocity 8 m/s. Within 7 seconds, it’s velocity increases to 22 m/s. What is its average acceleration? Train B is moves due east along a straight line at a velocity of 18 m/s. Within 10 s, its velocity drops to 3 m/s. What is its average acceleration? When an object slows down, we say it is decelerating. In that case, the direction of the acceleration will be opposite to that of the velocity.

Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration? 1 - Yes 2 - No 37 “Yes, because an object can be going forward but at the same time slowing down which would give it a negative acceleration.” Example

Graphical Representation Position, x (m) Time, t (s) Slope of the line = average velocity If the slope is zero, the object is at rest

Graphical Representation velocity, v (m/s) Time, t (s) Slope of the line = average acceleration O If the slope is zero, the object is moving with zero acceleration (constant velocity)

Slope of the line = average acceleration =  v/  t velocity, v (m/s) Time, t (s) O A B CD E 1. When is a = 0? 2. When is a < 0? 3. When is a = maximum? (A) OA (B) AB (C) BC (D) CD (E) DE

1. The area under a velocity-time graph represents A.Average velocity B.Average acceleration C.Total displacement D.Instantaneous velocity

2. A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of its displacement after this movement? A.8 km B.5.7 km C.16 km D.332 km E.0 km

3. Which car has a southward acceleration? A car traveling A.Southward at constant speed B.Northward at constant speed C.Southward and slowing down D.Northward and speeding up E.Northward and slowing down

4. A train moves due north along a straight path with a uniform acceleration of 0.18 m/s 2. If its velocity is 2.4 m/s, what will its velocity be after 1 minute? A.13.2 m/s north B.2.58 m/s north C.10.8 m/s north D.144 m/s north E.None of these

5. A bird flew 6.00 km north, then turned around and flew 1.50 km west. If it took 25.0 min, what was the average speed of the bird? A.5.00 m/s B m/s C.300 m/s D.4.12 m/s E.247 m/s

6. A bird flew 6.00 km north, then turned around and flew 1.50 km west. If it took 25.0 min, what was magnitude of the average velocity of the bird? A.5.00 m/s B m/s C.300 m/s D.4.12 m/s E.247 m/s

7. A train moving along a straight path due north has a velocity of 20 m/s. Within 5.0 seconds, its velocity became 5.0 m/s. What was the train's average acceleration? A.15 m/s 2 north B.15 m/s 2 south C.3.0 m/s 2 north D.3.0 m/s 2 south E.5.0 m/s 2 north F.5.0 m/s 2 south

VECTORS Vector A tail tip Equal Vectors: A = B only if they have equal magnitudes and same directions. A B Displacing a vector parallel to itself does not change it +y +x

The negative of a vector: Two vectors with equal magnitude but opposite directions are negatives of each other A -A Vectors can be multiplied by a scalar: A 2A ½ A

Components of a vector A vector can be expressed as a sum of 2 vectors called “components” that are parallel to the x and y axes: AxAx AyAy A +y +x A = A x + A y A x = x- component of A. A y = y-component of A.

 h = hypotenus a = adjacent o = opposite SOH CAH TOA sin  = opposite/hypotenuse = o/h cos  = adjacent/hypotenuse = a/h tan  = opposite/adjacent = o/a Pythagorean theorem: h 2 = a 2 + o 2

 AxAx sin  = o/h = A y /A or y-component:A y = A. sin  cos  = a/h = A x /A or x-component: A x = A. cos  A 2 = A x 2 + A y 2 or magnitude of vector A =  (A x 2 + A y 2 ) tan  = o/a = A y /A x or  = tan -1 (A y /A x ) AyAy A

Addition of vectors Three methods: Component method (analytical). Tip-to-tail (graphical). Parallelogram (graphical).

Component method Add vectors V 1 + V 2 + V 3 : 1.Find x and y components of each vector. {V 1 =V 1x + V 1y }, {V 2 = V 2x + V 2y }, {V 3 = V 3x + V 3y 2. Add x-components and y-components separately: {V x = V 1x + V 2x +V 3x } and {V y = V 1y + V 2y +V 3y } 3. Find the magnitude of the resultant vector using pythagorean theorem: V =  {V 2 x + V 2 y } 4. Find the angle of the resultant measured from the +x axis:  = tan -1 (V y /V x )

Example A B C y x 30 o Three displacement vectors are shown in the figure below. Their magnitudes are A = 20 cm, B = 16 cm and C = 12 cm. Find the magnitude and angle of the resultant vector.

Tip-to-tail Method A B Draw the vectors such that the “tail” of the second vector connects to the tip of the first vector. The resultant is from the tail of the first to the tip of the second. B A C C = A + B -B D = A - B D A

Parallelogram Method A Draw the vectors such that their “tails” are joined to a common origin. Construct a parallelogram with the two vectors as adjacent sides. The resultant vector is the diagonal line of the parallelogram drawn from the common origin. B A B C C = A + B

? A B (A) A – B (B) A + B (C) -A – B (D) – A + B From the diagram shown in the figure below, the unknown vector is

From the diagram shown in the figure below, what are vectors Z 1 and Z 2 in terms of X and Y? Y X Z2Z2 Z1Z1

y 35 o x A In the diagram below, what are the x and y-components of vector A if the magnitude of A is 40 units?

From the diagram shown in the figure below, what are vectors Z 1 and Z 2 in terms of B and A? A B Z1Z1 A B Z2Z2

You are on a train traveling 40 mph North. If you walk 5 mph toward the front of the train, what is your speed relative to the ground? 1) 45 mph2) 40 mph3) 35 mph 40 mph N + 5 mph N = 45 mph N Relative Velocity

You are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground? 1) 45 mph2) 40 mph3) 35 mph 40 mph N - 5 mph N = 35 mph N Relative Velocity

You are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative to the ground? 1) 40 mph 40 mph N + 5 mph W = 41 mph N Relative Motion (Add vector components) Relative Velocity

P2.27: Find the magnitude and direction of the vector with the following components: (a) x = -5.0 cm, y = +8.0 cm (b) F x = +120 N, F y = N (c) v x = m/s, v y = -8.8 m/s (d) a x = 2.3 m/s 2, a y = 6.5 cm/s 2

P2.56: A 3.0 kg block is at rest on a horizontal floor. If you push horizontally on the block with a force of 12.0 N, it just starts to move. (a) What is the coefficient of static friction? (b) A 7.0-kg block is stacked on top of the 3.0-kg block. What is the magnitude F of the force acting horizontally on the 3.0-kg block as before, that is required to make the two blocks start to move?

P3.47: A 2010-kg elevator moves with an upward acceleration of 1.50 m/s 2. What is the tension that supports the elevator? P3.48: A 2010-kg elevator moves with a downward acceleration of 1.50 m/s 2. What is the tension that supports the elevator?

A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of its displacement after this movement?