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## Chapter 2

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**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 asx • x = xf - xi • SI units are meters(m),xcan 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**Average Velocity**• The average velocity is the rate at which the displacement occurs • The dimensions arelength / time[L/T] • TheSI units arem/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**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 thex vst curve • This would be thegreenline • The blue lines show that astgets 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**Average Acceleration**• Acceleration is the rate of change of the velocity • Dimensions areL/T2 • SI units arem/s²**Instantaneous Acceleration**• The instantaneous acceleration is the limit of the average acceleration astapproaches 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**1D motion with constant acceleration**tf – ti = t**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**1D motion with constant acceleration**(1) Using and than substituting into equation for final position yields (1) (2) (2) Equations (1) and (2) are the basic kinematics equations**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:**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/s2from 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/s2from an initial velocity of 60 km/h? • Using we rearrange to solve for t: • Vf = 0.0 km/h, vi=60 km/h and a= -2.0 m/s2.**A car is approaching a hill at 30.0 m/swhen its engine**suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of –2.00 m/s2while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking x = 0at the bottom of the hill, where vi = 30.0 m/s. (b) Determine the maximum distance the car rolls up the hill.**(a) Take at the bottom of the hill where**xi=0, vi=30m/s, a=-2m/s2. Use these values in the general equation**(a) Take at the bottom of the hill where**xi=0, vi=30m/s, a=-2m/s2. Use these values in the general equation**The distance of travel, xf,**becomes a maximum, when (turning point in the motion). Use the expressions found in part (a) for whent=15sec.**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 isg = 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 withay = 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)**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)**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-st ball: 2-nd ball:**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:**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 Its original height was then:**Motion Equations from Calculus**• Displacement equals the area under the velocity – time curve • The limit of the sum is a definite integral