1. Motion in One Dimension (Velocity vs. Time) Chapter 5.2

2. What is instantaneous velocity?

3. What effect does an increase in velocity have on displacement? d1d1 d2d2 d3d3 d4d4 d5d5 d6d6

4. Instead of position vs. time, consider velocity vs. time. High acceleration Relatively constant velocity

5. How can displacement be determined from a v vs. t graph? Measure the area under the curve. d = v*t Where t is the x component v is the y component Time A2A2 A1A1 A 1 = d 1 = ½  v 1 *  t 1 A 2 = d 2 = v 2 *  t 2 d total = d 1 + d 2

6. Measuring displacement from a velocity vs. time graph. A = ½ b x h A = ½ (2.36)(11.7) A = 13.8 m A = b x h A = (7.37)(11.7) A = 86.2 m

7. What information does the slope of the velocity vs. time curve provide? A.Positively sloped curve = increasing velocity (Speeding up). B.Negatively sloped curve = decreasing velocity (Slowing down). C.Horizontally sloped curve = constant velocity. Time Positive Acceleration A Time Zero Acceleration C Time Negative Acceleration B

8. What is the significance of the slope of the velocity vs. time curve? Since velocity is on the y-axis and time is on the x-axis, it follows that the slope of the line would be: Therefore, slope must equal acceleration. Time

9. Acceleration determined from the slope of the curve. rise run v f – v i t f – t i 8.4m/s-0m/s 1.7s-0.00s m = 4.9 m/s 2 Since m = a: a = 4.9 m/s 2 m = Slope = m = What is the acceleration from t = 0 to t = 1.7 seconds?

10. Determining velocity from acceleration If acceleration is considered constant: a =  v/  t = (v f – v i) /(t f – t i ) Since t i is normally set to 0, this term can be eliminated. Rearranging terms to solve for v f results in: v f = v i + at Time Positive Acceleration

11. Position, velocity and acceleration when t is unknown. d = d i + ½ (v f + v i )*t(1) v f = v i + at(2) Solve (2) for t: t = (v f – v i )/a and substitute back into (1) d f = d i + ½ (v f + v i )(v f – v i )/a By rearranging: v f 2 = v i 2 + 2a*(d f – d i )(3)

12. Alternatively, If time and acceleration are known, but the final velocity is not: d f = d i + ½ (v f + v i )*t(1) v f = v i + at(2) Substitute (2) into (1) for v f d f = d i + ½ (v i + at + v i )*t d f = d i + v i t + ½ at 2 (4)

13. Formulas for Motion of Objects Equations to use when an accelerating object has an initial velocity. Form to use when accelerating object starts from rest (v i = 0).  d = ½ (v i + v f )  t  d = ½ v f  t v f = v i + a  tv f = a  t  d = v i  t + ½ a(  t) 2  d = ½ a(  t) 2 v f 2 = v i 2 + 2a  dv f 2 = 2a  d

14. Acceleration due to Gravity All falling bodies accelerate at the same rate when the effects of friction due to water, air, etc. can be ignored. Acceleration due to gravity is caused by the influences of Earth’s gravity on objects. The acceleration due to gravity is given the special symbol g. The acceleration of gravity is a constant close to the surface of the earth. g = 9.81 m/s 2

15. Example 1: Calculating Distance A stone is dropped from the top of a tall building. After 3.00 seconds of free-fall, what is the displacement, y of the stone? Data y ? a = g -9.81 m/s 2 vfvf n/a vivi 0 m/s t 3.00 s

16. Example 1: Calculating Distance Since v i = 0 we will substitute g for a and get: y = ½ gt 2 y = ½ (-9.81 m/s 2 )(3.00 s) 2 y = -44.1 m

17. Example 2: Calculating Final Velocity What will the final velocity of the stone be? Data y -44.1 m a = g -9.81 m/s 2 vfvf ? vivi 0 m/s t 3.00 s

18. Example 2: Calculating Final Velocity Again, since v i = 0 we will substitute g for a and get: v f = gt v f = (-9.81 m/s 2 )(3.00 s) v f = -29.4 m/s Or, we can also solve the problem with: v f 2 = v i 2 + 2a  d, where v i = 0 v f = [(2(-9.81 m/s 2 )(44.1 m)] 1/2 v f = -29.4 m/s

19. Example 3: Determining the Maximum Height How high will the coin go? Data y ? a = g -9.81 m/s 2 vfvf 0 m/s vivi 5.00 m/s t ?

20. Example 3: Determining the Maximum Height Since we know the initial and final velocity as well as the rate of acceleration we can use: v f 2 = v i 2 + 2a  d Since Δd = Δy we can algebraically rearrange the terms to solve for Δy.

21. Example 4: Determining the Total Time in the Air How long will the coin be in the air? Data y 1.28 m a = g -9.81 m/s 2 vfvf 0 m/s vivi 5.00 m/s t ?

22. Example 4: Determining the Total Time in the Air Since we know the initial and final velocity as well as the rate of acceleration we can use: v f = v i + aΔt, where a = g Solving for t gives us: Since the coin travels both up and down, this value must be doubled to get a total time of 1.02s

23. Key Ideas Instantaneous velocity is equal to the slope of a line tangent to a position vs. time graph. Slope of a velocity vs. time graphs provides an objects acceleration. The area under the curve of a velocity vs. time graph provides the objects displacement. Acceleration due to gravity is the same for all objects when the effects of friction due to wind, water, etc can be ignored.

24. Important equations to know for uniform acceleration. d f = d i + ½ (v i + v f )*t d f = d i + v i t + ½ at 2 v f 2 = v i 2 + 2a*(d f – d i ) v f = v i +at a = Δv/Δt = (v f – v i )/(t f – t i )

25. Displacement when acceleration is constant. Displacement = area under the curve. Δd = v i t + ½ (v f – v i )*t Simplifying: Δd = ½ (v f + v i )*t If the initial position, d i, is not 0, then: d f = d i + ½ (v f + v i )*t By substituting v f = v i + at d f = d i + ½ (v i + at + v i )*t Simplifying: d f = d i + v i t + ½ at 2 d = v i t d = ½ (v f -v i )t vfvf vivi t