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Physics Subject Area Test MECHANICS: KINEMATICS.

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1 Physics Subject Area Test MECHANICS: KINEMATICS


3  To simplify the concept of motion, we will first consider motion that takes place in one direction.  One example is the motion of a commuter train on a straight track.  To measure motion, you must first choose a frame of reference. A frame of reference is a system for specifying the precise location of objects in space and time.  In the train example, any station along the route.

4  Displacement is a change in position.  Displacement is not always equal to the distance traveled.  The SI unit of displacement is the meter, m. ∆ x = x f -x i Displacement – final position – initial position

5 Displacement is not always equal to the distance traveled. Example: If a gecko starts at an initial position of 20 cm and moves to the 80 cm mark, then retreats back to the 50 cm mark as its final position, How far has the gecko traveled? What is its displacement? The gecko traveled 90 cm, but its displacement is 30 cm.

6  In general, right (east) is positive as well as upward (north) and left (west) is negative as well as downward (south).

7 Average velocity is the total displacement divided by the time interval during which the displacement occurred. In SI, the unit of velocity is meters per second abbreviated as m/s.

8  Consider a trip to a friend’s house 370 km to the west (negative direction) along a straight highway. If you left at 10 AM and arrived at 3 PM, what is your average velocity? This is your average. You did not travel at 74 km/h at every moment.

9  Velocity is not the same as speed.  Velocity describes motion with both direction and a numerical value (magnitude).  Speed has no direction, only magnitude.  Average speed is equal to the total distance traveled divided by the time interval.

10  Consider an object whose position-time graph is not a straight line, but a curve.  We obtain different average velocities depending on the time interval. The instantaneous velocity is the velocity of the object at a specific point in the object’s path The instaneous velocity can be determined by measuring the slope of the line that is tangent to that point on the diatance-vs-time graph.


12  Acceleration – Rate at which velocity changes over time  An object accelerates if its speed, direction or both change.  Acceleration has direction and magnitude.  Acceleration is a vector quantity.

13 Acceleration has the dimensions of length divided by time squared.  SI units are m/s 2  Remember we have (m/s)/s = m/s 2 Acceleration

14 A bus slows down with an average acceleration of -1.8 m/s 2. How long does it take the bus to slow down from 9.0 m/s to a complete stop?

15  Consider a train moving to the right, so that the displacement and velocity are positive.  The slope of the velocity-time graph is the average acceleration. When the velocity in the positive direction is increasing, the acceleration is positive, as at A. When the velocity is constant, there is no acceleration, as at B. When the velocity in the positive direction is decreasing, the acceleration is negative, as at C.


17  When velocity changes by the same amount during each time interval, acceleration is constant.  The relationships between displacement, time, velocity, and constant acceleration are expressed by the equations shown on the next slide.

18 These equations apply to any object moving with constant acceleration. These equations use the following symbols:


20  A racing car reaches a speed of 42 m/s. It begins a uniform negative acceleration, using its parachute and braking system, and comes to a complete rest 5.5 s later. Find the distance that the car travels during braking.


22 * A scalar is only a magnitude (length) (Example: Temperature, time, mass) * A vector has magnitude and direction (Example: displacement = 10 m East, Velocity= 50 mph west)

23 * A vector will be symbolized by the “letter” with an arrow over it. The arrow indicates direction. * Vectors are equal if they have the same units, magnitude, and direction. * A vector can be moved anywhere parallel to itself.

24 * To add vectors they must have the same units. * Tip-to -tail method put them head to tail and connect them so you end up with a triangle. * Parallelogram Method- (put them tail to tail) make vectors parallel and draw a line making 2 triangles


26 * Resultant Vector * The resultant vector is the sum of a given set of vectors

27 * Tip to tail- subtract by putting vector in the opposite direction * If you change the sign of a vector it is not the same vector. It is a new vector. * A – B does not equal B - A

28 * A component is a part * It is useful to use rectangular components * These are the projections of the vector along the x- and y-axes

29 * The x-component of a vector is the projection along the x-axis * The y-component of a vector is the projection along the y-axis * Then,

30 The Pythagorean Theorem can only be used with right triangles!

31 When its not 90 0 R 2 = A 2 + B 2 – 2AB(COSӨ)

32 * Find the magnitude of the sum of a 15 km displacement and a 25 km displacement when the angle between them is 90 0 and when the angle between them is

33 (a) Find the horizontal and vertical components of the 100m displacement of a superhero who flies from the top of a tall building at an angle of (b) (b) Suppose instead the superhero leaps in the other direction along a displacement vector B to the top of a flagpole where the displacement components are given B x = -25m and B Y =10.0m. Find the magnitude and direction of the displacement vector.

34 * A GPS receiver indicates that your home is 15.0 km and 40 0 north of west, but the only path through the woods leads directly north. If you follow the path 5.0 km before it opens into a field, how far, and in what direction, would you have to walk to reach your home? * R= * Ө = 158’

35 Resolving a Vector Into Components +x +y A AxAx AyAy  The horizontal, or x-component, of A is found by A x = A cos  The vertical, or y-component, of A is found by A y = A sin   By the Pythagorean Theorem, A x 2 + A y 2 = A 2 Every vector can be resolved using these formulas, such that A is the magnitude of A, and  is the angle the vector makes with the x- axis. Each component must have the proper “sign” according to the quadrant the vector terminates in.

36 Analytical Method of Vector Addition 1. Find the x- and y-components of each vector. A x = A cos  =A y = A sin  = B x = B cos  = B y = B sin  = C x = C cos  =C y = C sin  = 2. Sum the x-components. This is the x-component of the resultant. Rx =Rx = 3. Sum the y-components. This is the y-component of the resultant. R y = Pythagorean Theorem 4. Use the Pythagorean Theorem to find the magnitude of the resultant vector. R x 2 + R y 2 = R 2

37 * A roller coaster moves 215 ft horizontally and then rises 130 ft at an angle of above the horizontally. Next, it travels 125 ft at an angle of below the horizontal. Find the roller coaster’s displacement from its starting point to the end of this movement.

38 * A quarter back takes the ball from the line of scrimmage, runs backwards for 15.0 yards, then runs sideways parallel to the line of scrimmage for 15.0 yards. At this point, he throws a 60.0 yard forward pass straight downfield, perpendicular to the line of scrimmage. What is the magnitude of the football’s resultant displacement?

39 Vector Multiplication DOT PRODUCT scalar product A ∙ B A ∙ B = AB cosφ The product of the 2 vectors and the cosine of the angle between them

40 A ∙ B = (A x i + A y j) (B x i + B y j) = A x B x i ∙ i + A x B y i ∙ j + A y B x j ∙ i + A y B y j ∙ j i.i = j.j = k.k = 1 and i.j = j.i = i.k = k.i = j.k = k.j = 0 A ∙ B = A x B x i ∙ i + A y B y j ∙ j With 3 dimension: A ∙ B = A x B x i ∙ i + A y B y j ∙ j + A z B z k ∙ k

41 CROSS PRODUCT vector product A x B The product of the 2 vectors and the sine of the angle between them A x B is not the same as B x A … the direction is opposite

42 i x i = j x j = k x k = 0 i x j = k j x k = I k x i = j a x b = (a 2 b 3 – a 3 b 2 ) i + (a 3 b 1 – a 1 b 3 ) j + (a 1 b 2 - a 2 b 1 ) k kij

43 * Two vectors in component forms are written as : In evaluating the product, we make use of the fact that multiplication of the same unit vectors gives the value of 0, while multiplication of two different unit vectors result in remaining vector with appropriate sign. Finally, the vector product evaluates to vector terms :


45 * Moving in the x and y direction * A projectile is an object shot through the air. This occurs in a parabola curve.

46 Object dropped Object thrown up Object thrown at an angle projectile- any object that moves through the air or through space, acted on only by gravity (and air resistance, if any)

47 The vertical acceleration of a projectile is caused by gravity, so a y = -9.8 m/s 2 Parabolic Trajectory

48 * g remains constant (g= -9.8m/s 2) * a in the x direction is 0 because gravity is not acting on it. * Neglect air resistance * Neglect the effects of the earths rotation

49 Projectiles launched horizontally

50 To find how far the ball falls, you use the formula. y =v iy t + 1/2gt 2 1 st second- 5m After 2 seconds- 20m After 3 seconds- 45m The curved path of a projectile produced is a parabola (caused by both horizontal motion and vertical motion. It must accelerate only in the vertical direction)

51 * The projectile will experience two: * Accelerations (a x = o and a Y = -9.8m/s 2 ) * Velocities * Displacements

52 Upwardly Launched Projectiles When a projectile is launched at an upward angle, it follows a curved path and finally hits the ground because of gravity. The Vertical distance a cannonball falls below “imaginary path if no gravity” is the same vertical distance it would fall if it were dropped from rest & had been falling for the same amount of time.


54 * Draw a free body diagram with a coordinate system. * Divide the information into x and y components * Look at your formulas and decided which one(s) to use.

55 Objects that have been thrown will have a horizontal velocity that stays the same (no horizontal acceleration a x = 0m/s 2) So v fx =v ix in the second formula and third formulas under horizontal motion. (X) Horizontal (Y) Vertical x f- x i = v ix t + ½ a x t 2 y f -y i = v iy t + ½ a y t 2 v fx = v ix + a x t v fy = v iy + a y t v fx 2 - v ix 2 = 2a x (x f- x i ) v fy 2 = v iy 2 + 2a y (y f -y i )

56 This equation only works when y f and y i are both the same magnitude a = 2v iy t

57 If a ball is thrown up in the air from a moving truck, where will it land? (Ignore air resistance) In front of the truck, behind the truck, or back in the truck

58 Where will a package land if it is released from a plane? Behind the plane, in front of the plane below the plane

59 What is the horizontal distance covered by an arrow that was shot through the air at a 60 0 angle with a velocity of 55 m/s? Given v = 55m/s v ix =27.5 m/s v iy =47.6m/s a x =0 a y =-9.8m/s 2 t=? d x =? Solve V xi = cos 60(55m/s)=27.5 m/s V iy = sin60(55m/s)=47.6m/s v fy =v iy +a y t d x = V ix t (we need time) m/s dydy V ix d x 0 = 47.6m/s m/s 2 t -47.6m/s = -9.8m/st 4.86 s =t d x = 27.5 m/s(9.7s) d x = m Total time in the air 4.86s x 2 = 9.7s Need to find time first! To find x dist: x = v 0x t

60 * Frames of Reference Observers using different frames of reference may measure different displacements or velocities for an object in motion Relative Velocities the difference between the velocities relative to some common point

61 * Relative Motion: Suppose you are on a train platform as the train rushes through the station without stopping. Someone on board the train is pitching a ball, throwing it has hard as they can towards the back of the train. If the train’s speed is 60 mph and the pitcher is capable of throwing at 60 mph, what is the speed of the ball as you see it from the platform?

62 * A boat heading due north crosses a river with a speed of 10.0 km/h. The water in the river has a speed of 5.0 km/h due east. Moving frame of reference In general we have (a)Determine the velocity of the boat. (b)If the river is 3.0 km wide how long does it take to cross it?

63 Conservation of Linear Momentum Completely Inelastic Collision Velocity of Center of Mass

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