2. “Motion in a Straight Line” Two types of kinematics' analysis: Scalar: measures magnitude (size) only Examples: mass, distance, speed, Vector: measures magnitude and direction Examples: displacement, velocity, acceleration, Speed represents the magnitude of velocity in one-dimensional motion.. Kinematics: mathematical description of motion

3. Distance: d T he measure of how far one travels Distance can be represented graphically S.I. Units are meters “m” Scalar quantity The resultant measure of how far one travels from the starting position to the ending position. (ie..as the crow flies) Displacement can be represented graphically S.I. Unit s are meters, direction ie.. 16 m, 30° Vector quantity Displacement: d

4. An arrow represents vectors: tailhead The tail of a vector represents the starting position, The head represents the ending position and direction of the motion. Vector Length indicates magnitude. In one-dimensional motion we add vectors in the same direction and subtract vectors in the opposite directions. A: 32 m, E B: 10 m, E Vector A should be drawn approximately 3 times longer than vector B since the vector length represents the magnitude. Vectors can be added by placing the tail of one vector to the head of another vector.

5. A + B Graphical Representation B: 10 m, E A: 32 m, E 32 m,E + 10 m,E = 42 m, E Math representation When adding vectors tail to head, order doesn’t matter. Draw and compute the adding of vectors C and D. C : 115 m, W D : 80.0 m, E resultant Graphically Mathematically -35 mor35 m, W- 115 m+ 80.0 m = Best answer A + B = B + A

6. Speed and velocity are the same in one-dimensional motion Speed is the total distance per time unit. S = d/t  SI units…m/s  Scalar quantity  Measures magnitude only  Speedometers on cars measure your instantaneous speed.  Constant speed can be achieved by putting on the cruise control  Graphically, a distance or position vs time graph will relate speed to the reader.

7. Speed (m/s) Time (s) Time (s)  A straight line indicates constant velocity or speed o Slope represents average or constant velocity or speed o Linear equation: y = mx + b where - m = slope - b = y-intercept (where the line intersects the y-axis) - x = independent variable - y = dependent variable  A parabolic curve indicates acceleration o Slope on a parabola indicates instantaneous velocity or speed o To straighten out a parabolic curve plot y vs x 2

8. Velocity is displacement per time unit. V = d t  Velocity is a vector quantity  SI units are m/s, direction  Average velocity: v= Δ displacement or final velocity + initial velocity Δ time 2 V ave = Δd = (v f + v i ) Δt 2  Instantaneous velocity is the speedometer reading in a specific direction.  Constant velocity is putting on the cruise control in a specific direction  G raphically speaking a position vs time graph can also represent velocity. However, the direction is not part of the graph unless it is a vector graph with time as a given.

9. Velocity Displacement (m) vs Time (s) A B C D A to B = constant velocity B to C = changing velocity C to D = zero velocity E = instantaneous velocity (x,, y)Read the x and y values off the graph and then calculate d/t. x y Draw a tangent line through the data point.

10. Linear graph: Equation  y = mx + b m = slope b = y-intercept (y-intercept always has coordinates of (0, y)) Linear graphs always use the best fit line. Linear graphs indicate direct proportionality Hyperbolic graphs: Equation  y = k/x Where k = some constant value Hyperbolic graphs indicate inverse proportionality Parabolic graphs: Equation  y = kx 2 or x = ky 2 The y = kx 2 is most commonly used. Again, k = some constant value Parabolic graphs (y = kx 2) indicate the dependent variable increases at a faster rate than the independent variable. Best-fit line Y vs X Speed Distance (m) vs Mass (kg)

11. Acceleration (a): The rate of change in velocity per time unit. + a = acceleration = increasing speed – a = deceleration = decreasing speed Average acceleration = a ave = v f – v i = Δvelocity t f - t i Δ time FYI..Time is generally given already in terms of change, so rarely do you ever need to calculate the Δ time (time final – time initial ) A ave  (a)  if the direction of velocity and acceleration are the same.

12. a = Δv = v f – v i t t Solve for t, v f and v i t = Δv = v f – v i a a v f = at + v i v i = v f - at Acceleration (m/s 2 ) Constant acceleration Calculate slope for acceleration area under the curve defines distance Variable acceleration Instantaneous velocity is read off the graph. area under the curve requires calculus, but also defines distance. (m/s) Time (s)

13. a = v f – v i and Δd = (v f +v i ) Δt Δt 2 Using the two equations above, we have five different variable available to us. They are atvfvf vivi d acceleration, time, final velocity, initial velocity, displacement, Steps to deriving equations 1. Determine which variable is missing from the problem. Example: Nathan accelerates his skateboard along a straight path to a final velocity of 12.5 m/s in 2.5 s. If Nathan accelerates at a rate of 5.0 m/s 2. What is his displacement? Givens are:a = 5.0 m/s 2 t = 2.5 s v f = 12.5 m/s d = ?? V i is missing

14. a = v f – v i and V ave = d = (v f +v i ) t t 2 v f - at = v i and 2d - v f = v i t

15. Example: Nathan accelerates his skateboard along a straight path to a final velocity of 12.5 m/s in 2.5 s. If Nathan accelerates at a rate of 5.0 m/s 2. What is his displacement? Givens are:a = 5.0 m/s 2 t = 2.5 s v i = 0 v f = 12.5 m/s d = ?? v f - at = 2d - v f t d = v f t – at 2 2 or d = v f t – ½ at 2

16. d = v f t – at 2 = 2 Example: Nathan accelerates his skateboard along a straight path to a final velocity of 12.5 m/s in 2.5 s. If Nathan accelerates at a rate of 5.0 m/s 2. What is his displacement? Givens are:a = 5.0 m/s 2 t = 2.5 s v i = 0 v f = 12.5 m/s d = ?? d = 16 m (12.5m/s)(2.5 s) – (5.0 m/s 2 )(2.5s) 2 2

17. Final Velocity = maximum Initial Velocity = 0 Free Fall  In the absence of air resistance, all objects fall to earth at a rate of – 9.81 m/s 2. What goes up, must come down! V i = max V f = 0 Time = ½ total time Distance = max Acceleration = -9.81m/s 2 Distance going up is + Distance coming down is - Terminal velocity should = V i

18. If Jeronimo jumps off a cliff into the river below, how high was the cliff if it took him 2.20 s to reach the river? Herbert jumped out of an airplane that was 1800.0 m above the ground. How long would it take him to hit the ground if his parachute didn’t deploy? How long would hit take him to reach a terminal velocity of 160.0 m/s. 1.Check to see if Herbert reached terminal velocity in the air or on the ground. d = (v f 2 -v i 2 )/2a = (-160.0m/s – 0)/2(-9.81m/s 2)