Chapter 2.ppt

# Chapter 2.ppt - Chapter 2 Motion in One Dimension 2.1...

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Unformatted text preview: Chapter 2 Motion in One Dimension 2.1 position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.4 Freely Falling Objects 2.5 Kinematic Equations Derived from Calculus. Kinematics Kinematics describes motion while ignoring the agents that caused the motion For now, we will consider motion in one dimension Along a straight line We will use the particle model A particle is a point-like object, that has mass but infinitesimal size Position Position is defined in terms of a frame of reference For one dimension the motion is generally along the x- or y-axis The object’s position is its location with respect to the frame of reference Position-Time Graph The position-time graph shows the motion of the particle (car) The smooth curve is a guess as to what happened between the data points Displacement Displacement is defined as the change in position during some time interval Represented as x x = xf - xi SI units are meters (m), x can be positive or negative Displacement is different than distance. Distance is the length of a path followed by a particle. Vectors and Scalars Vector quantities that 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 x fi vaverg tt Average Velocity The average velocity is the rate at which the displacement occurs The dimensions are length / time [L/T] The SI units are m/s Is also the slope of the line in the position – time graph Average Speed Speed is a scalar quantity same units as velocity total distance / total time The average speed is not (necessarily) the magnitude of the average velocity Instantaneous Velocity Instantaneous velocity is 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 x d vx litm 0t Instantaneous Velocity The general equation for instantaneous velocity is The instantaneous velocity can be positive, negative, or zero Instantaneous Velocity The instantaneous velocity is the slope of the line tangent to the x vs t curve This would be the green line The blue lines show that as t gets smaller, they approach the green line Instantaneous Speed The instantaneous speed is the magnitude of the instantaneous velocity Remember that the average speed is not the magnitude of the average velocity v x f x i x ax tt Average Acceleration Acceleration is the rate of change of the velocity Dimensions are L/T2 SI units are m/s² v d v d x a lim tt 2 x x x t 0 Instantaneous Acceleration The instantaneous acceleration is the limit of the average acceleration as t approaches 0 Instantaneous Acceleration The slope of the velocity vs. time graph is the acceleration The green line represents the instantaneous acceleration The blue line is the average acceleration Acceleration and Velocity When an object’s velocity and acceleration are in the same direction, the object is speeding up When an object’s velocity and acceleration are in the opposite direction, the object is slowing down Acceleration and Velocity The car is moving with constant positive velocity (shown by red arrows maintaining the same size) Acceleration equals zero Acceleration and Velocity Velocity and acceleration are in the same direction Acceleration is uniform (blue arrows maintain the same length) Velocity is increasing (red arrows are getting longer) This shows positive acceleration and positive velocity Acceleration and Velocity Acceleration and velocity are in opposite directions Acceleration is uniform (blue arrows maintain the same length) Velocity is decreasing (red arrows are getting shorter) Positive velocity and negative acceleration a v f i tvf a i 1D motion with constant acceleration tf – t i = t x f i vavgx tfi vavgvavg 1 (2v0 f) 1D motion with constant acceleration In a similar manner we can rewrite equation for average velocity: and than solve it for xf Rearranging, and assuming vxff tv0t a ii 12at2 1D motion with constant acceleration (1) Using (1) and than substituting into equation for final position yields (2) (2) Equations (1) and (2) are the basic kinematics equations vx ( 2 x12(v a ))t 2ffi200ffi 1D motion with constant acceleration These two equations can be combined to yield additional equations. We can eliminate t to obtain Second, we can eliminate the acceleration a to produce an equation in which acceleration does not appear: tΔ v a fv i v f 2 1 2 x if2 2 a x Kinematics with constant acceleration - Summary Kinematic Equations - summary Kinematic Equations The kinematic equations may be used to solve any problem involving one-dimensional motion with a constant acceleration You may need to use two of the equations to solve one problem Many times there is more than one way to solve a problem Kinematics - Example 1 How long does it take for a train to come to rest if it decelerates at 2.0m/s2 from an initial velocity of 60 km/h? Kinematics - Example 1 How long does it take for a train to come to rest if it decelerates at 2.0m/s2 from an initial velocity of 60 km/h? km/h Using to solve for t: v vwe arearrange t f i vf vi t a Vf = 0.0 km/h, vi=60 km/h and a= -2.0 m/s2. 0 (60x 1000 / 3600) t 8.3 s 2.0 A car is approaching a hill at 30.0 m/s when its engine suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of –2.00 m/s2 while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking x = 0 at the bottom of the hill, where vi = 30.0 m/s. (b) Determine the maximum distance the car rolls up the hill. ti 0 (a) Take at the bottom of the hill where xi=0, =0 vi=30m/s, =30m/s a=-2m/s2. Use these values in the general equation 1 2 x f xi vi t at 2 1 x f 0 30.0t m s 2.00 m s 2 t 2 2 x f 30.0t t 2 m (a) Take ti 0 at the bottom of the hill where xi=0, vi=30m/s, a=-2m/s2. Use these values in the general equation 2 v f vi at 30.0 m s 2.00 m s t v f 30.0 2.00t m s The distance of travel, xf, becomes a maximum, xmax when when v f 0 (turning point in the motion). Use the expressions found in part (a) for v f 30.00 2.00t m / s v f 0 when t=15sec. xmax 30.0t t 2 m 30.0 15.0 15.0 225 m 2 Graphical Look at Motion: displacement-time curve The slope of the curve is the velocity The curved line indicates the velocity is changing Therefore, there is an acceleration Graphical Look at Motion: velocity-time curve The slope gives the acceleration The straight line indicates a constant acceleration Graphical Look at Motion: acceleration-time curve The zero slope indicates a constant acceleration Freely Falling Objects A freely falling object is any object moving freely under the influence of gravity alone. It does not depend upon the initial motion of the object Dropped – released from rest Thrown downward Thrown upward Acceleration of Freely Falling Object The acceleration of an object in free fall is directed downward, regardless of the initial motion The magnitude of free fall acceleration is g = 9.80 m/s2 g decreases with increasing altitude g varies with latitude 9.80 m/s2 is the average at the Earth’s surface Acceleration of Free Fall We will neglect air resistance Free fall motion is constantly accelerated motion in one dimension Let upward be positive Use the kinematic equations with ay = g = -9.80 m/s2 Free Fall Example Initial velocity at A is upward (+) and acceleration is g (-9.8 m/s2) At B, the velocity is 0 and the acceleration is g (-9.8 m/s2) At C, the velocity has the same magnitude as at A, but is in the opposite direction A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? (a) 1 2 y f yi vi t at 2 4.00 1.50 vi 4.90 1.50 vi 10.0 m s upward 2 A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? (b) v f vi at 10.0 9.80 1.50 4.68 m s v f 4.68 m s downward A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. 1 2 1-st ball: y1 h gt 2 2-nd ball: 1 2 y2 vi t gt 2 h y1 2 h 1 2 h gt 2 2 h h 1 h vi g 2 g 2 g t vi gh h g A freely falling object requires 1.50 s to travel the last 30.0 m before it hits the ground. From what height above the ground did it fall? A freely falling object requires 1.50 s to travel the last 30.0 m before it hits the ground. From what height above the ground did it fall? Consider the last 30 m of fall. We find its speed 30 m above the ground: 1 y f yi v yi t 2 ay t 2 1 2 2 0 30 m vyi 1.5 s 9.8 m s 1.5 s 2 30 m 11.0 m vyi 12.6 m s. 1.5 s A freely falling object requires 1.50 s to travel the last 30.0 m before it hits the ground. From what height above the ground did it fall? Now consider the portion of its fall above the 30 m point. We assume it starts from rest v 2 v 2 2a y y yf 12.6 m s 2 0 2 9.8 m s y y Its original height was then: yi y f i 2 160 m 2 s 2 19.6 m s 2 8.16 m. 30 m 8.16 m 38.2 m Motion Equations from Calculus Displacement equals the area under the velocity – time curve lim tn 0 v n tf xn tn v x (t )dt ti The limit of the sum is a definite integral Kinematic Equations – General Calculus Form dv x ax dt t v xf v xi a x dt 0 dx vx dt t x f xi v x dt 0 vxxfixfxi avixt 12xt2 Kinematic Equations – Calculus Form with Constant Acceleration The integration form of vf – vi gives The integration form of xf – xi gives The height of a helicopter above the ground is given by h = 3.00t3, where h is in meters and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? The height of a helicopter above the ground is given by h = 3.00t3, where h is in meters and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? y 3.00t 3 t 2.00 s dy vy 9.00t 2 36.0 m s dt y 3.00 2.00 24.0 m 3 The height of a helicopter above the ground is given by h = 3.00t3, where h is in meters and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? The equation of motion of the mailbag is: 1 2 1 yb yb ybi vi t gt 24.0 36.0t 9.80 t 2 2 2 2 t 7.96 s 0 24.0 36.0t 4.90t 0 Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that 2 a x2 a xi 2 j (v x v xi ) Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that 2 a x2 a xi 2 j (v x vxi ) (a) da J constant dt a J dt Jt c1 a Jt ai da Jdt a ai when t 0 c1 ai Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that 2 a x2 a xi 2 j (v x v xi ) (a) dv a dt dv adt v v vi adt when t 1 2 Jt ai dt Jt ai t c2 2 1 2 0 c2 vi v Jt ai t vi 2 Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that 2 a x2 a xi 2 j (v x vxi ) dx (a) dt dx vdt v x 1 2 Jt a t v i i dt 2 1 ai t 2 vi t c3 2 vdt 1 Jt 3 6 x xi when x t 0 c3 x i x 1 3 1 Jt ai t 2 vi t xi 6 2 (b) a 2 Jt ai 2 J 2t 2 ai2 2 Jai t a 2 ai2 J 2 t 2 2 Jai t a 2 ai2 1 2J Jt 2 ai t 2 Recall the expression for vv: 1 Jt 2 ai t vi 2 1 2 v vi 2 Jt ai t 2 a 2 ai 2 J v vi The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a = –3.00 v2 for v > 0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to half of its initial value? The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a = –3.00 v2 for v > 0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to half of its initial value? dv a 3.00v 2 dt dv 3.00v 2 dt vi 1.50 m s v v vi v 2 dv 3.00 t dt t 0 1 1 1 1 3.00t or 3.00t . v vi v vi The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a = –3.00 v2 for v > 0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to half of its initial value? dv a 3.00v 2 dt vi v 2 vi 1.50 m s 1 t 0.222 s 3.00vi A test rocket is fired vertically upward from a well. A catapult gives it initial velocity 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point its engines fail and the rocket goes into free fall, with an acceleration of –9.80 m/s2. (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? A test rocket is fired vertically upward from a well. A catapult gives it initial velocity 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point its engines fail and the rocket goes into free fall, with an acceleration of –9.80 m/s2. (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? Let point 0 be at ground level and point 1 be at the end of the engine burn. Let point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table are found for each phase of the rocket’s motion. 2 (0 to 1): v 80.0 2 2 4.00 1 000 f so v f 120 m s 120 80.0 4.00 t t 10.0 s A test rocket is fired vertically upward from a well. A catapult gives it initial velocity 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point its engines fail and the rocket goes into free fall, with an acceleration of –9.80 m/s2. (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (1 to 2): 0 120 2 9.80 x f xi 2 x f xi 735 m 0 120 9.80t t 12.2 s This is the time of maximum height of the rocket. A test rocket is fired vertically upward from a well. A catapult gives it initial velocity 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point its engines fail and the rocket goes into free fall, with an acceleration of –9.80 m/s2. (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? v 2f 0 2 9.80 1 735 (2 to 3): v f 184m / s v f 184 9.80 t (a): (b): (c): t 18.8 s ttotal 10 12.2 18.8 41.0 s x f xi total 1.73 km vfinal 184 m s t 0 Launch x v a 0.0 0 80 +4.00 #1 End Thrust 10.0 1 000 120 +4.00 #2 Rise Upwards 22.2 1 735 0 –9.80 #3 Fall to Earth 41.0 0 –184 –9.80 An inquisitive physics student and mountain climber climbs a 50.0-m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water? inquisitive physics student and mountain climber climbs a 50.0-m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water? An (a) y f vi1t 1 2 1 at 50.0 2.00t 9.80 t 2 2 2 2 4.90t 2.00t 50.0 0 t 2.00 2.00 2 4 4.90 50.0 2 4.90 Only the positive root is physically meaningful: t 3.00 s after the first stone is thrown. (b) 1 2 y f vi 2 t at 2 t 3.00 1.00 2.00 s 1 2 50.0 vi 2 2.00 9.80 2.00 2 vi 2 15.3 m s downward (c) v1 f vi1 at 2.00 9.80 3.00 31.4 m s v2 f vi 2 at 15.3 9.80 2.00 34.8 m s downward downward ...
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