Describing Motion: Kinematics in One Dimension Chapter 2

MechanicsMechanics: study of the motion of objects and the related concepts of force and energy Kinematics: description of how objects move Dynamics: description of why objects move as they do Translational motion – objects move with out rotating Rectilinear motion – objects move along a straight-line path (one-dimensional)

Frame of Reference – coordinate system used to define motion. – Need a reference point - zero location in a coordinate system or frame of reference.

Position, Velocity, and Speed Position : location of a particle with respect to a chosen reference point Displacement : the change in position in some time interval  x = x f – x i Distance : the length of a path followed by a particle. Distance does not need a ref. pt. Displacement is a VECTOR QUANTITY Distance is a SCALAR QUANTITY

Average Velocity : a particle’s displacement (  x) divided by the time interval (  t) during which that displacement occurs V = xx tt Average Speed : the total distance traveled divided by the total time interval required to travel that distance Avg speed = total distance total time Avg Velocity is a VECTOR QUANTITY Avg Speed is a SCALAR QUANTITY

Units: km/h or m/sec Everyday Speeds in SI Units 40 km/h = 25 mi/h = 11 m/s 80 km/h = 50 mi/h = 22 m/s 100 km/h = 62 mi/h = 28 m/s 120 km/h = 75 mi/h = 33 m/s

Conversion Factor from km/h to m/s:

Example 1 Positiont(s)x(m) A030 B1052 C2038 D300 E40-37 F50-53 Find the displacement, average velocity, and average speed of the object between positions A and F.

POSITION – TIME GRAPHS A graph that show how position depends on time. The slope of a position –time graph -The displacement is the vertical separation between two points. -The time interval is the horizontal separation. SLOPE = rise =  d = average velocity = v run  t

When a position-time graph is a linear relationship, and thus has a constant slope, then the slope = constant  Velocity is uniform or constant Time (s) Position (m) Position vs. time

? Can an object have a varying velocity if its speed is constant? For the special case of uniform velocity, we can rewrite the equation for velocity as: v = x t

Instantaneous Velocity and Speed Instantaneous Velocity : defined as the average velocity over an indefinitely short time interval. Or we can think of it as the limiting value of the ratio  x/  t as  t approaches zero v = xx tt lim  t  0 time position

The slope of the tangent is the instantaneous velocity. If an object moves at uniform velocity over a particular time interval, the the instantaneous velocity at any instant is the same as its average velocity. Instantaneous Speed : the magnitude of the instantaneous velocity

Example 1 A) A ball thrown directly upward rises to a highest point and falls back in the thrower’s hand. B) A race car starts from rest and speeds up to 100 m/s. C) A spacecraft drifts through space at constant velocity. Are there any points at which the instantaneous velocity has the same value as the average velocity over the entire motion ? If so, identify the point(s).

What is the instantaneous velocity at t = 5.0 s ? Answer : Slope of the tangent line drawn at the time in question. Example 2

Lesson 3 : Acceleration Average Acceleration : the change in velocity (  v) divided by the time interval (  t) during which that change occurs a = vv tt = v f – v i t f – t i time V Linear: y = mx + b v = at run rise

Instantaneous Acceleration : the limit of the average acceleration as  t appoaches zero. - Once again we would take the slope of the tangent to the velocity-time graph. a = vv tt lim  t  0

Example 1 What is the instantaneous acceleration at t = 2.0 s ? Answer : Slope of the tangent line drawn at the time in question.

Example 2 The velocity of a particle moving along the x- axis varies in time according to the expression v x = (40 – 5t 2 ) m/s, where t is in seconds. A) Find the average acceleration in the time interval t =0 to t = 2.0 s. B) Determine the acceleration at t = 2.0 s.

Lesson 4 : Motion Diagrams v No acc Motion A v a Motion B v a Motion C

Graphs for each motion : x Motion A t v t a t x Motion B t v t a t x Motion C t v t a t

Example 1 Draw the corresponding x vs. t and a vs. t graphs. A) v t x t a t B) v t x t a t v C) t x t a t

Example 2 Draw the corresponding v vs. t and a vs. t graphs. A) x t v t a t B) x t v t a t x C) t v t a t

FINDING DISPLACEMENT FROM A GRAPH The area under the line on a velocity-time graph is equal to the displacement of the object from its original position to its position at time t. When velocity is constant, displacment increases linearly with time.

What is the displacement of the object from t = 0 to 3 s ? Answer : The area under the graph equals displacement Example 3

Negative Area : Object is moving toward smaller x values and displacement is decreasing

RELATIVE VELOCITY When measuring the position or velocity, you must first define your frame of reference. –e.g. For a viewer on a ship or on the land. For one, the frame of reference is the ship, for the other, it is the land.

Lesson 5 : Kinematic Equations

EQUATIONS OF MOTION FOR UNIFORM ACCELERATION Equation Variables v = v o + at v o v a t x = x o + v o t + ½ at 2 v o x - x o a t v 2 = v a(x – xo) v o x - x o v a v = v + v o 2 v = x - x o a = v - v o t t

Lesson 5 : Kinematic Equations a = vv tt = v f – v i t f – t i v f = v i + a  t v f = v i + at (for constant a) v = v i + v f 2 (for constant a)

 x = x f - x i  x = vt x f – x i = vt x f – x i = ½ (v i + v f )t x f = x i + ½ (v i + v f )t (for constant a)

x f = x i + ½ (v i + v f )t x f = x i + ½ [v i + (v i +at)]t v f = v i + at x f = x i + v i t + ½ at 2 (for constant a) x f = x i + ½ (v i + v f )t x f = x i + ½ (v i + v f ) ( v f - v i a ) v f 2 = v i 2 + 2a (x f – x i ) (for constant a)

Summary v f = v i + at (for constant a) v = v i + v f 2 (for constant a) x f = x i + ½ (v i + v f )t (for constant a) x f = x i + v i t + ½ at 2 (for constant a) v f 2 = v i 2 + 2a (x f – x i ) (for constant a)

Lesson 6 : Freely Falling Objects Neglecting air resistance, all objects dropped near the Earth’s surface fall toward the Earth with the same constant acceleration under the influence of the Earth’s gravity. A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion.

Free-fall acceleration (g) = 9.80 m/s 2 +y-y a y = -g = m/s 2 For making quick estimates, use g = 10 m/s 2

Example 1 A ball is tossed straight up at 25 m/s. Estimate its velocity at 1 s intervals. Time (s) Velocity (m/s)

Example 2 A stone is thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. A)Determine the time at which the stone reaches its maximum height. B)Determine the maximum height.

D)Determine the velocity of the stone at this instant. E)Determine the velocity and position of the stone at t = 5.00 s. C)Determine the time at which the stone returns to the height from which it was thrown.