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

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

2. What is instantaneous velocity?

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

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

5. Velocity A2 A1 Time 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 A1 = d1 = ½ v1*t1 A2 = d2 = v2*t2 dtotal = d1 + d2

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

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

8. Velocity Time 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.

9. rise run vf – vi tf – ti 8.4m/s-0m/s 1.7s-0.00s m = 4.9 m/s2 Since m = a: a = 4.9 m/s2 Slope = m = m = Acceleration determined from the slope of the curve. What is the acceleration from t = 0 to t = 1.7 seconds?

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

11. Position, velocity and acceleration when t is unknown. d = di + ½ (vf + vi)*t (1) vf = vi + at (2) Solve (2) for t: t = (vf – vi)/a and substitute back into (1) df = di + ½ (vf + vi)(vf – vi)/a By rearranging: vf2 = vi2 + 2a*(df – di) (3)

12. Alternatively, If time and acceleration are known, but the final velocity is not: df = di + ½ (vf + vi)*t (1) vf = vi + at (2) Substitute (2) into (1) for vf df = di + ½ (vi + at + vi)*t df = di + vit + ½ at2 (4)

13. Formulas for Motion of Objects

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/s2

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?

16. Example 1: Calculating Distance • From your reference table: d = vit + ½ at2 • Since vi = 0 we will substitute g for a and y for d to get: y = ½ gt2 y = ½ (-9.81 m/s2)(3.00 s)2 y = -44.1 m

17. Example 2: Calculating Final Velocity • What will the final velocity of the stone be?

18. Example 2: Calculating Final Velocity • Using your reference table: vf = vi + at • Again, since vi = 0 and substituting g for a, we get: vf = gt vf = (-9.81 m/s2)(3.00 s) vf = -29.4 m/s • Or, we can also solve the problem with: vf2 = vi2 + 2ad, where vi = 0 vf = [(2(-9.81 m/s2)(-44.1 m)]1/2 vf = -29.4 m/s

19. Example 3: Determining the Maximum Height • How high will the coin go?

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: vf2 = vi2 + 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?

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: vf = vi+ 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. • df = di + ½ (vi + vf)*t • df = di + vit + ½ at2 • vf2 = vi2 + 2a*(df – di) • vf = vi +at • a = Δv/Δt = (vf – vi)/(tf – ti)

25. vf d = ½ (vf-vi)t vi d = vit t Displacement when acceleration is constant. Displacement = area under the curve. Δd = vit + ½ (vf – vi)*t Simplifying: Δd = ½ (vf + vi)*t If the initial position, di, is not 0, then: df = di + ½ (vf + vi)*t By substituting vf = vi + at df = di + ½ (vi + at + vi)*t Simplifying: df = di + vit + ½ at2