Physics Lec 2 PDF

Physics Prof. Dr. E.. Abo Zeid 10/5/2024  Chapter 2 Motion in One Dimension Topics: 2.1 Position, Velocity and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.5 One-dimensional motion with constant acceleration 2.6 Vec...

Physics Prof. Dr. E.. Abo Zeid 10/5/2024  Chapter 2 Motion in One Dimension Topics: 2.1 Position, Velocity and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.5 One-dimensional motion with constant acceleration 2.6 Vector analysis Physics Prof. Dr. E. F. Abo Zeid 10/5/2024 2.1 Displacement, Velocity and Speed Displacement: Change of the position of a particle. Displacement is a vector quantity. Displacement AX = Xy — X; i — 5¥ A B lfff B )_C)i = AX = +ve lfff < )_(zi = AJ_C) = —ve Note: Displacement # Distance travelled AX, ;= NS 3. Distance travelled = AC + CB Displacement = AB Physics Prof. Dr. E. F. Abo Zeid 10/5/2024  Average Velocity Average velocity The motion of a particle is known if its position at all times are known (position -time graph) A particle is moved from point A to point B Point A:position = ¥;, time = t; Point B: position = X, time = t; i DisplacementAX = X — %; —-S. Time intervalAt = tf — t; A Average velocity: the ratio of its displacement Ax and the time interval At _ AY XX V,=—= < Dimension [\Z ]: L4 Attt T / SI t Unitsof " ° system ftls Br.Eng. system Physics Prof. Dr. E. F. Abo Zeid 10/5/2024 Geometrical meaning of v ¥ 1s the slope of the straight line joining initial (A) and final (B) points on the position-time graph. x(m) 60 Slope of AB = ar V. At Ax At Average Speed Average Speed = Total dlstanct.: travelled 0 T( sec) Total time Average speed is a scalar quantity (has no direction ). Physics Prof. Dr. E. F. Abo Zeid 10/5,/2024  Ex.2.1: A car is moving along the x-axis from pointA to point F. x(m) Point A:X, =30m, 7, =0sec Point B: X =52m,7 =10sec Point F:X, =-53m,7, =50sec Find (i) Displacement G (ii) Average velocity (iii) Average speed ! v N v “. (i) DisplacementAX = X —X; = =53 —30=—83 m AX —83 —83 —83 m (ii)Average velocity = = —1.7m/s At (G, —t) 50-0 50s Note that: The displacement and the velocity are negative because the car is moving to left Total distance travelled (AB) + (BF) (22) + (105) iii) Average Speed = = 2.5m/s Total time - 50 B 50 Physics Prof. Dr. E. F. Abo Zeid 10/5/2024  2.2 Instantaneous velocity and speed Instantaneous velocity The limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero 60 Particlemotion fromAtoB v, = % = slope of AB ! 40 : : 'S Mv ’ Particlemotion from A toB’ v, =—= = slope of AB A ®.. "o— M} Particlemotion from A toB"” v,: = Vi :, slope of AB " At; )(5.00 5)=[5.00 my/s |. ) ) In the first ten seconds, £ || P xp=x+ v,-t+%atz = 0+0+%(2.(IJ m/s? )(10.0 s): =100 m. Over the next five seconds the position changes to xp=x +v,t+%ntz =100 m+(20.0 m/s)(5.00 5)+0 =200 m. And at f=20.0s, x,=x,+u,t+%at2=zoom+(zo.o m/s)(5.00 s)+%(—3.00 m/s2)(5.00 5)* =[262m |. Physics Prof. Dr. E. F. Abo Zeid 10/5/2024  21. An object moving with uniform acceleration has a velocity of 72.0 c¢m/s in the positive x direction when its x coordinate is 3.00 cm. If its X coordinate 2.00 s later is —5.00 cm, what is its acceleration? Given v; =12.0 cm/s when x; =3.00 cm(f=0),and at {=2.00 s, x; =—5.00 cm, xp—x, = v,~t+%atz: ~5.00—3.00 =12.0(2.00)+%a(2.00)2 —8.00=240+2a a=—&2°=‘ -160 cm/s” |. Physics Prof. Dr. E.. Abo Zeid 10/5/2024 25. A particle moves along the x axis. Its position is given by the equation x =2 + 3t — 4 with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at z = 0. (a) Compare the position equation x = 2.00+3.00¢ —4.00¢ to the general form Xp=2x; +ui!+%nf1 to recognize that x, =2.00 m, v; =3.00 m/s, and a=—8.00 m/s?. The velocity equation, vy =v; +at,is then v;=3.00 m/s—(8.00 m/s?)t. The particle changes direction when v, =0, which occurs at ¢ =% s. The position at this time is: x=2.00 m+(3.00 m/sn(% s)—(.Loo u\/sz(% s)z =[256m]. ®) From x; =x; +v,f+%nt2, observe that when x; = x,, the time is given by t=—2ni. Thus, when the particle returns to its initial position, the time is _— —2(3.00 n\/s)=% - —8.00 m/s> and the velocity is v, = 3.00 m/s —(8.00 m/s> ;(% s |00 mv=] Physics Prof. Dr. E.. Abo Zeid 10/5/2024  Thanks for your attention Prof. Dr. E. F. Abo Zeid 10/5/2024