Physics 141MechanicsLecture 2 Kinematics in One Dimension Kinematics is the study of the motion of an object. We begin by studying a single particle in a straight line. A particle is a point-like object such as an electron. For a non-point-like object, if all its parts move together in exactly the same way, its motion can also be described as a particle. In a baseball game, if the pitcher throws the ball without spin, the ball motion can be considered a particle. When he/she throws the ball with spin, the ball motion can not be described precisely as a particle since different part of the ball moves differently.

Coordinate Imagine you are going to dash straight in a playground. Your starting point is the origin. The running track is marked in meters, and any point on the track is of a unique number of meters from the origin. This number is the coordinate. The coordinate of the origin is zero. The straight track, the origin and the meter marks form the coordinate system. We can further define the positive direction as you move forward, and negative direction as you move backward. We usually use a straight horizontal line, the x-axis, with an arrow to the right and with tick marks to represent a one dimensional coordinate system. The arrow point to the positive direction.

Position and Displacement The coordinate of a particle at any given time t gives the particles position,x(t), as a function of time. The position has the unit of length. If you know explicitly the functional form of x(t), you know everything about the motion. The time t is typically represented by the number of seconds after some original event, say the start of the timing clock. We call it time origin, or t = 0 second. t is positive for instants after the time origin, and negative for those before the time origin. When checking the units of an equation, remember t has a unit of time. The displacement of a particle is defined as the difference between the positions of the particles at different times ∆x=x(t 2 )-x(t 1 ) (1) Again the displacement has the unit of length.

Average Velocity The average velocity is defined as the displacement divided by the time interval taken for the displacement (2) If we present x(t) graphically, the average velocity is the slope of the straight line connecting the two points (t 1,x(t 1 )) and (t 2,x(t 2 )). Velocity and speed, used almost interchangeably in daily life, are very different in physics. Speed is defined as distance traveled divided by the time interval. If you drive round-trip from Rochester to New York city in 10 hours, your displacement is zero, and the distance traveled is 700 miles. Your average velocity is therefore zero, while your average speed is 70 mi/hr. We’ll learn later that velocity is a vector and speed is a scalar.

Instantaneous Velocity You drive a new car and one thing you’d like to test is the performance: how long it takes the car to get to 65 mi/hr from rest. In this case, the average velocity is ever changing. How do we now the velocity v(t) at the instant t? Imagine a very small interval [t, t+∆t], and the displacement during the interval is ∆x(t) = x(t+∆t) - x(t). If ∆t is small enough, we get a good approximation v(t)≈∆x/∆t. As ∆t becomes smaller and smaller, the ratio becomes closer and closer to v(t). The instantaneous velocity, is defined as the limit of the ratio when ∆t -> 0 (3)

The above discussion of instantaneous velocity, or simply velocity, is how the concepts of derivative was first formulated and therefore forms the foundation of calculus. The instantaneous speed, or speed, is the magnitude of the velocity. Graphically, the derivative of a curve at any point is the slope of its tangent line at that point. Calculus tells us (4) where c,n are constants. This formula is enough for us now for the time being.

Example: Let the position of particle be described by the function x(t) = a + bt + ct 2,where a,b,c are constants. Its velocity at time t is v(t)=dx/dt=b+2ct(5) Now let’s try something numerical. Suppose that a = 4.0 m, b = 3.0 m/s, and c = 2.0 m/s 2, (check the unit of each term!) and look at the instantaneous velocity and average velocity about t=1.0 s. x 1 =x(t=1.0 s) = 9.0 m, v(t=1.0 s) = 7.0 m/s x 2 =x(t=2.0 s) = 18.0 m, v= 9.0 m/s x 2 =x(t=1.1 s) = 9.72 m, v= 7.2 m/s x 2 =x(t=1.01 s) = m, v= 7.02 m/s

Acceleration Now let’s get back to test your car. The performance of your car is measured by how much it accelerates. This quantity in physics is the acceleration, the measurement of the rate of velocity change. It is defined as the derivative of velocity with respect to time, (6) or, in combination with Eq. 3, the acceleration is the second derivative of position, (7) The unit of acceleration in SI is m/s 2.

Example: If x(t) = x 0 + v 0 t, where x 0 and v 0 are constants, then the velocity is a constant: (8) and the acceleration is zero. (9) x(t) here describes an object moving with constant velocity and therefore no acceleration. Example: If, then (10)

Motion with Constant Acceleration If (11) where x 0, v 0,a are constants, then the velocity is (12) The acceleration is (13) We see that if the acceleration is a constant, the position is a quadratic function of time. What are the physical meanings of x 0, v 0, and a? a is obviously the constant acceleration, x 0 is the initial position, or the position at t = 0, and v 0 is the initial velocity.

Example: The driver of a train traveling at a speed v 0 sees a car trapped in the track in a distance D in front of the train. What must be the acceleration of the train the driver has to make to avoid a collision? Solution: To avoid a collision, the train must stop at the car. We have (14) Eliminate t, (15) Or, (16) The required acceleration is of course negative.

We see from the above example that in motion with constant acceleration, sometimes we do not have to know the time t explicitly. We can find the formula relating among the acceleration a, the initial and final positions x 0, x, and the initial and final velocities, v 0, v by eliminating t from Eqs.11&12. From Eq.11 (17) (18) Or, (19)

Demonstration: Motion of Constant Acceleration We have a air track on which an Al cart move almost frictionless. The air track is tilted a little so the cart will slide downward with constant acceleration. We put two sensors along the track to measure the velocity of the cart when passing the sensors. Let’s now calculate the acceleration from Eq. 19 and check for ourselves if it is in fact motion of constant acceleration. Letx 0 =0 and v 0 =0. Measured values x (in m)v (in m/s) (in m/s 2 )

Free-Fall Acceleration Galeleo made the first precise measurements on falling bodies and discovered that in the absence of friction, all bodies fall at the same acceleration due to gravity. The magnitude of the gravitational acceleration is found to be g=9.80 m/s 2. We usually define the positive direction to be upward, and then the acceleration a = -g. Eqs.11,12&19 become (20) (21) (22)

Example: If you throw a smooth lead ball straight up with initial velocity v 0, what is the maximum height the ball will reach? Solution: At t = 0, x 0 = h (if the origin is at the ground and your height is h). The initial velocity is v 0. and the acceleration a is -g. At the maximum height, x m, v = 0. Using Eq. 22, we have - 2g(x m -h)=0 2 -v 0 2, or x m =h+v 0 2 /2g If you have taken calculus, you see the typical may of finding the maximum of x(t) is to let dx/dt=0 to solve t m, and the maximum x m =x(t=t m ). Check it yourself.