Chapter Motion Along a Straight Line In this chapter we will study how objects moe along a straight line The following parameters will be defined: (1) Displacement () Aerage elocity (3) Aerage speed (4) Instantaneous elocity (5) Aerage and instantaneous acceleration For constant acceleration we will deelop the equations that gie us the elocity and position at any time..
Position ector Its magnitude is the distance between the object and the origin. Its direction is positie when the object is in the positie side of axis, and negatie when the object is in the negatie side.
Position and Displacement. It consists of: the origin (or zero point), a coordinate axis: the direction along it is positie. The other direction is negatie
Scalars and Vectors A scalar quantity can be described by its magnitude only A Vector is described with both its magnitude and direction. A ector can be represented by an arrow: magnitude direction
The elocity is a ector. Displacement and acceleration are also ectors. N W E S
Displacement. If an object moes from position x 1 to position x, the change in position is described by the displacement x x x 1 O. x. 1 x. x-axis Δx motion It is the change of the object s position It points from the initial position to the final position of the object Its magnitude equals the distance between the two positions. SI Unit of Displacement: meter (m)
Note: The actual distance for a trip is irreleant as far as the displacement is concerned.. O x. 1 x. Δx m 13 m x c x-axis Consider as an example the motion of an object from an initial position x 1 = 5 m to x c = 5 m and then back to x = 1 m. Een though the total distance coered is 33 m the displacement then is Δx = +7 m.
. O x. 1 x. x-axis Δx For example if x 1 = 5 m and x = 1 m then Δx = 1 5 = +7 m. The positie sign of Δx indicates that the motion is along the positie x-direction.
. O x x.. 1 x-axis Δx If instead the object moes from x 1 = 5 m and x = 1 m then Δx = 1 5 = - 4 m. The negatie sign of Δx indicates that the motion is along the negatie x-direction. Displacement is a ector quantity that has both magnitude and direction. In this restricted one-dimensional motion the direction is described by the algebraic sign of Δx. Question What is the magnitude of the displacement in this case?
Example : The initial and the final positions of a particle moing along x-axis are X 1 = -73 m, X = 97 m, then its displacement x equals: (A) +4 m (B) +17 m (C) -17 m (D) -34 m
Aerage Velocity x and x 1 are the position ectors at the final and initial times Angle brackets denotes the aerage of a quantity. SI Unit of Aerage Velocity: meter per second (m/s)
1 m/s 4 48-4 48m x4 4x4 4) (t x x 4x ) (t x 1 1 1 1 t t x x t x t at t at ag
Aerage Speed speed: the magnitude of elocity Aerage speed is always positie Aerage elocity could be negatie, positie or zero depending on the direction of the partial elocities
Example A bike traels1km in 9 mins. Its aerage speed is: (A) 8 km/h (B) 18 km/h (C) 8 km/h (D) 48 km/h
3 t 3 min 6 dis tan ce ag t.5 h 15.5 3 km/h t 3 ( t ) 3 m/s
Instantaneous Velocity and Speed It is the time deriatie of the object's position. It is obtained at any instant from the aerage elocity by shrinking the time interal t closer and closer to zero Instantaneous speed (speed) is the magnitude of the instantaneous elocity ector
Example : The position of an object is gien by x = 4t 3 - t +.5t, where x and t are in SI units. What is the instantaneous elocity of the object when t =.5 s. (A) 3.75 m/s (B) 5.45 m/s (C).5 m/s (D) 4.55m/s
Example : The position of a particle moing on an X- axis is gien by x= 4+7t-t² with x in (m) t in (s). The elocity at 3 s is : (A) 4 m/s (B) m/s (C) 1 m/s (D).4 m/s
Aerage Acceleration We define the aerage acceleration a ag between t 1 and t as: a ag 1 t t t 1 Units: m/s Example : a 1 m / s m / s ag t t at at t 1 1 t t 1 3s 8s 8 3 5 4 m/s
Instantaneous Acceleration If we take the limit of a ag as Δt we get the instantaneous acceleration a, which describes how fast the elocity is changing at any time t. Units: m/s
Example : (E)
Motion with Constant Acceleration a = constant We will deelop the equations that describe motion with a constant a (a special case). We assume that initially (at time t = ) the particle is at x o and moes with o, and some time t later (final state) the particle is at x and moe with.
Motion with Constant Acceleration We will deelop the equations that describe motion with a constant a ( a special case ). We assume that initially (at time t= ) the particle is at x and moes with, and some time t later (final state) the particle is at x and moe with.
(1) () (3) (4) (5) The first two equations are originals, but all others are drien from the first two.
االستنتاج الرياضي لمعادالت الحركة ليس مطلوب من طالب.Phys 11 (1) () (3) (4) (5)
? 8.5 s.8 m/s 4m t x x s m t x x a / 6.61.8)8.5 ( 1 4 (4) ) ( 1 ) ( /.448 8.5 6.6.8 (1) ) ( s m a a at b
x x x x 4 m/s a 1 t at 1 4x4 xx4 m/s () 3 m t 4 s x-x (?)
Example 5 a1 a 5m at / s (1)
Example m x x x x x x s t s m h km s m h km 75 6 ) 36 1 36 1 (7 1? 6 / / / / 7
Free Fall Close to the surface of the Earth all objects moe toward the center of the Earth with an acceleration whose magnitude is constant and equal to 9.8 m/s. We use the symbol g to indicate the acceleration of an object in free fall.
Free Fall
its magnitude is g; it is independent of the object's characteristics, such as mass, density, or shape g aries slightly with latitude and with eleation; at the sea leel g=9.8 m/s (or 3 ft/s ) The equations of motion for constant acceleration also apply to free fall near Earth's surface either up or down The directions of motion are now along a ertical y axis: it is +e for upward motion and e for downward motion (a = - g)
y a Free Fall gg
Free Fall Equations
Example 6.83 m/s.84 39. 39..84 39..16 39..16 9.8.4 (3) y y g y y g m? 9.8 m/s.4 y-y g
Example A stone is released from rest from the edge of a building roof 19 m aboe the ground. Neglecting air resistance, the speed of the stone, just before striking the ground, is (A)1 m/s (B)19 m/s (C) 43 m/s (D) 61 m/s 371 61m / s y y 9.819 ( )9.819 g 9.819 y-y g 19 m 9.8 m/s? (3) 374 why