phy2048-ch2_new - MECHANICS Kinematics Chapter 2 Motion...

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Unformatted text preview: MECHANICS Kinematics Chapter 2 - Motion along a straight line I. Position and displacement II. Velocity III. Acceleration IV. Motion in one dimension with constant acceleration V. Free fall Particle: point-like object that has a mass but infinitesimal size. I. Position and displacement Position: Defined in terms of a frame of reference: x or y axis in 1D. - The object’s position is its location with respect to the frame of reference. Position-Time graph: shows the motion of the particle (car). The smooth curve is a guess as to what happened between the data points. 1 I. Position and displacement Displacement: Change from position x1 to x2 during a time interval. ∆x = x2-x1 (2.1) - Vector quantity: Magnitude (absolute value) and direction (sign). - Coordinate (position) ≠ Displacement x ≠ ∆x x x Coordinate system x1=x2 x2 x1 t ∆x >0 t ∆x = 0 Only the initial and final coordinates influence the displacement different motions between x1and x2 give the same displacement. many Distance: length of a path followed by a particle. - Scalar quantity Displacement ≠ Distance Example: round trip house-work-house distance traveled = 10 km displacement = 0 Review: - Vector quantities need both magnitude (size or numerical value) and direction to completely describe them. - We will use + and – signs to indicate vector directions. - Scalar quantities are completely described by magnitude only. 2 II. Velocity Average velocity: Ratio of the displacement ∆x that occurs during a particular time interval ∆t to that interval. v avg = ∆x x 2 − x 1 = ∆t t 2 − t1 Motion along x-axis (2.2) -Vector quantity indicates not just how fast an object is moving but also in which direction it is moving. - SI Units: m/s - Dimensions: Length/Time [L]/[T] - The slope of a straight line connecting 2 points on an x-versus-t plot is equal to the average velocity during that time interval. Average speed: Total distance covered in a time interval. Savg = Total distance ∆t (2.3) Savg ≠ magnitude Vavg Savg always >0 Scalar quantity Same units as velocity Example: A person drives 4 mi at 30 mi/h and 4 mi and 50 mi/h average speed >,<,= 40 mi/h ? <40 mi/h t1= 4 mi/(30 mi/h)=0.13 h ; t2= 4 mi/(50 mi/h)=0.08 h Is the ttot= 0.213 h Savg= 8 mi/0.213h = 37.5mi/h 3 Instantaneous velocity: How fast a particle is moving at a given instant. vx = lim ∆t → 0 ∆x dx = ∆t dt (2.4) - Vector quantity - The limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero. - The instantaneous velocity indicates what is happening at every point of time. - Can be positive, negative, or zero. x(t) - The instantaneous velocity is the slope of the line tangent to the x vs. t curve (green line). t Instantaneous velocity: Position Slope of the particle’s position-time curve at a given instant of time. V is tangent to x(t) when ∆t 0 Time When the velocity is constant, the average velocity over any time interval is equal to the instantaneous velocity at any time. Instantaneous speed: Magnitude of the instantaneous velocity. Example: car speedometer. - Scalar quantity Average velocity (or average acceleration) always refers to an specific time interval. Instantaneous velocity (acceleration) refers to an specific instant of time. 4 III. Acceleration Average acceleration: Ratio of a change in velocity ∆v to the time interval ∆t in which the change occurs. aavg = v2 − v1 ∆v = t 2 − t1 ∆t (2.5) - Vector quantity V t - Dimensions [L]/[T]2, Units: m/s2 - The average acceleration in a “v-t” plot is the slope of a straight line connecting points corresponding to two different times. t t Instantaneous acceleration: Limit of the average acceleration as ∆t approaches zero. - Vector quantity ∆v dv d 2 x = = ∆t → 0 ∆t dt dt 2 a = lim (2.6) - The instantaneous acceleration is the slope of the tangent line (v-t plot) at a particular time. (green line in B) - Average acceleration: blue line. - When an object’s velocity and acceleration are in the same direction (same sign), the object is speeding up. - When an object’s velocity and acceleration are in the opposite direction, the object is slowing down. 5 - Positive acceleration does not necessarily imply speeding up, and negative acceleration slowing down. Example (1): v1= -25m/s ; v2= 0m/s in 5s particle slows down, aavg= 5m/s2 - An object can have simultaneously v=0 and a≠0 Example (2): x(t)=At2 v(t)=2At a(t)=2A ; At t=0s, v(0)=0 but a(0)=2A Example (3): - The car is moving with constant positive velocity (red arrows maintaining same size) Acceleration equals zero. Example (4): + acceleration + velocity - Velocity and acceleration are in the same direction, “a” is uniform (blue arrows of same length) Velocity is increasing (red arrows are getting longer). Example (5): - acceleration + velocity - Acceleration and velocity are in opposite directions. - Acceleration is uniform (blue arrows same length). - Velocity is decreasing (red arrows are getting shorter). 6 IV. Motion in one dimension with constant acceleration - Average acceleration and instantaneous acceleration are equal. a = aavg = v − v0 t −0 - Equations for motion with constant acceleration: v = v 0 + at v avg v avg ( 2 .7 ) t x − x0 = → x = x 0 + v avg t ( 2 .8 ) t v +v at =0 and ( 2 . 7 ) → v avg = v 0 + 2 2 ( 2 . 8 ), ( 2 . 9 ) → x a =x 0 = v − t + − a = v0 v 0 avg t−0 at 2 ( 2 .9 ) 2 ( 2 .10 ) 2 2 ( 2 . 7 ), ( 2 . 10 ) → v 2 = v 0 + a 2 t 2 + 2 a ( v 0 t ) = v 0 + a 2 t 2 + 2 a ( x − x 0 − 2 → v 2 = v0 + 2 a ( x − x0 ) at 2 ) 2 t ( 2 . 11 ) t missing t PROBLEMS - Chapter 2 P1. A red car and a green car move toward each other in adjacent lanes and parallel to The x-axis. At time t=0, the red car is at x=0 and the green car at x=220 m. If the red car has a constant velocity of 20km/h, the cars pass each other at x=44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x=76.6m. What are (a) the initial velocity, and (b) the acceleration of the green car? vr2=40km/h vr1=20km/h 3 km 1h 10 m = 11.11m / s 40 ⋅ ⋅ h 3600s 1km Xr2=76.6m x O Xr1=44.5 m d=220 m 44.5m x = v t →t = = 8s r1 r1 1 1 5.55m / s 76.6m x = v t → t2 = = 6.9s r2 r2 2 11.11m / s Xg=220m x = x +v t r r0 r 1 x = x + v t + at 2 g g0 g0 2 (1) (2) x − x g = −v t − 0.5 ⋅ a g t2 2 → 76.6 − 220 = −v ⋅ (6.9s) − 0.5 ⋅ (6.9s) 2 a g r2 g0 2 g0 x − x g = −v t − 0.5 ⋅ ag t12 → 44.5 − 220 = −v ⋅ (8s) − 0.5 ⋅ (8s) 2 a g r1 g0 1 g0 The car moves to the left (-) in my reference system a<0, v<0 ag = 2.1 m/s2 v0g = 13.55 m/sc 7 P2: At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s2. At the same instant, a truck, traveling with constant speed of 9.5 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant? ac = 2.2 m/s2, vc0=0 m/s t = 0s t = 0s x (m) Car Truck x (m) x=0 x=d ? vt = 9.5 m/s xT = d = vT t = 9.5 t → (1) (a) 9.5 ⋅ t = 1.1⋅ t 2 → t = 8.63 s → d = (9.5m / s )(8.63s) ≈ 82m Truck 1 xc = d = v t + act 2 → d = 0 + 0.5 ⋅ (2.2m / s 2 ) ⋅ t 2 = 1.1t 2 (2) Car C0 2 2 (b) v 2 = v0 + 2 ⋅ ac ⋅ d = 2 ⋅ (2.2m / s 2 ) ⋅ (82m) → v f = 19m / s f P3: A proton moves along the x-axis according to the equation: x = 50t+10t2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3s of its motion. v avg = x (3) − x (0) (50)(3) + (10)( 3) 2 − 0 = = 80 m / s. ∆t 3 (b) Instantaneous velocity of the proton at t = 3s. v (t ) = (c) Instantaneous acceleration of the proton at t = 3s. dx = 50 + 20 t → v(3s ) = 50 + 20 ⋅ 3 = 110 m / s dt a (t ) = dv = 20 m / s 2 = a(3s ) dt (d) Graph x versus t and indicate how the answer to (a) (average velocity) can be obtained from the plot. (e) Indicate the answer to (b) (instantaneous velocity) on the graph. (f) Plot v versus t and indicate on it the answer to (c). x = 50t + 10t2 v = 50 + 20t P4. An electron moving along the x-axis has a position given by: x = 16t·exp(-t) m, where t is in seconds. How far is the electron from the origin when it momentarily stops? x(t) when v(t)=0?? dx = v = 16e −t − 16te −t = 16e −t (1 − t ) dt v = 0 → (1 − t ) = 0; (e−t > 0) → t = 1s x(1) = 16 / e = 5.9m 8 P5. When a high speed passenger train traveling at 161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered into the track from a siding and is a distance D= 676 m ahead. The locomotive is moving at 29 km/h. The engineer of the high speed train immediately applies the brakes. (a) What must be the magnitude of the resultant deceleration if a collision is to be avoided? (b) Assume that the engineer is at x=0 when at t=0 he first spots the locomotive. Sketch x(t) curves representing the locomotive and high speed train for the situation in which a collision is just avoided and is not quite avoided. Locomotive Train t = 0s x (m) dL x (m) t > 0s x=0 vT=161km/h = 44.72 m/s = vT0 x=D x=D+dL 1D movement with a<0=cte vL=29 km/h = 8.05 m/s is constant d L = vLt = 8.05 t → t = dL 8.05 (1) Locomotive 1 1 d L + D = v t + aT t 2 → d L + 676 = 44.72 t + aT t 2 (2) Train T0 2 2 P5. vTf = vT 0 + aT t = 0 → aT = −44.72m / s (−44.72m / s)(8.05m / s) −360m2 / s 2 = (eq. 1) = = t dL dL v2Tf = v 2T 0 + 2aT ( D + d L ) = 0 → aT = −(44.72m / s)2 2(676m+d L ) (3) (4) (3) = (4) → d L = 380.3m from (1) → t = (1) + (3) → aT = Collision can be avoided Locomotive dL 8.05 = 47.24s −360m2 / s 2 = −0.947m / s 2 380.3m x L = 676 + 8 . 05 t x T = 44 . 72 t + 0 . 5 a T t 2 - Collision can be avoided: Train Collision can not be avoided Slope of x(t) vs. t locomotive at t = 47.24 s (the point were both Lines meet) = v instantaneous locom > Slope of x(t) vs. t train - Collision cannot be avoided: Slope of x(t) vs. t locomotive at t = 47.24 s < Slope of x(t) vs. t train 9 - The motion equations can also be obtained by indefinite integration: dv = a dt → ∫ dv = ∫ a dt → v = at + C ; v = v0 at t = 0 → v0 = (a )(0) + C → v0 = C → v = v0 + at 1 dx = v dt → ∫ dx = ∫ v dt → ∫ dx = ∫ (v0 + at )dt → ∫ dx = v0 ∫ dt + a ∫ t dt → x = v0t + at 2 + C ' ; 2 1 12 x = x0 at t = 0 → x0 = v0 (0) + a(0) + C ' → x0 = C ' → x = x0 + v0t + at 2 2 V. Free fall Motion direction along y-axis ( y >0 upwards) Free fall acceleration: (near Earth’s surface) a= -g = -9.8 m/s2 (in mov. eqs. with constant acceleration) Due to gravity Earth’s center downward on y, directed toward Approximations: - Locally, Earth’s surface essentially flat at slightly different points. free fall “a” has same direction - All objects at the same place have same free fall “a” (neglecting air influence). VI. Graphical integration in motion analysis From a(t) versus t graph integration = area between acceleration curve and time axis, from t0 to t1 v(t) t1 v1 − v0 = ∫ a ⋅dt t0 Similarly, from v(t) versus t graph under curve from t0 to t1 x(t) integration = area t1 x1 − x0 = ∫ v ⋅dt t0 10 P6: A rocket is launched vertically from the ground with an initial velocity of 80m/s. It ascends with a constant acceleration of 4 m/s2 to an altitude of 10 km. Its motors then fail, and the rocket continues upward as a free fall particle and then falls back down. (a) What is the total time elapsed from takeoff until the rocket strikes the ground? (b) What is the maximum altitude reached? 1) Ascent a0= 4m/s2 (c) What is the velocity just before hitting ground? y1 − y0 = v t + 0.5 ⋅ a0t12 → 104 = 80t + 2t12 → t = 53.48s y 01 y2=ymax a1= -g t2 t3=t2 2) Ascent +v1, t1 y1= 10 km a0= 4m/s2 t1 t4 v0= 80m/s a1 = − g = a2= -g 1 a= -9.8 m/s2 0−v1 − 294m / s → t2 = = 29.96s t2 −9.8m / s 2 Total time ascent = t1+t2= 53.48 s+29.96 s= 83.44 s t0=0 v3 1 vv a0 = 1− 0 → v1 = (4m / s 2 ) ⋅ (53.48s) + 80m / s = 294m / s t1 v2=0, t2 x 3) Descent a= -9.8 m/s2 4 0 − y1 = −v1t + 0.5 ⋅ a0t4 2 → −10 = −294t4 − 4.9t4 2 → t = 24.22 s 4 4 ttotal=t1+2t2+t4= 53.48 s + 2·29.96 s + 24.22 s=137.62 s hmax= y2 y2-104 m = v1t2-4.9t22= (294 m/s)(29.96s)-(4.9m/s2)(29.96s)2 = 4410 m a2 = − g = hmax=14.4 km v3−(−v ) 1 → v = − g ⋅ t − v = −531.35m / s 3 41 t4 11 ...
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