Chapter 3: Accelerated Motion

1. Chapter 3:Accelerated Motion Honors Physics Glencoe Science, 2005

2. 3.1 Acceleration • Change in time- ending time minus initial time • Dt = tf - ti • Change in velocity- final velocity minus initial velocity • Dv = vf - vi • Average acceleration- change in velocity between two distinct time intervals • ā = Dv/Dt = (vf - vi)/(tf - ti) (m/s2) • Instantaneous acceleration- change in velocity at an instant in time • Only found by determining the tangent of a curve on a velocity-time graph

3. Motion Diagrams Shows an object slowing down, speeding up, without motion, or at constant motion

4. Velocity-Time Graphs Represents motion graphically Plots the velocity versus the time of the object

5. 3.2 Motion with Constant Acceleration • Sometimes we have an initial velocity • Like when we pull through a stop light as it turns green • Our acceleration formula can be rearranged • Dv= ā Dtvf - vi = ā Dt • If we are looking for the final velocity, then we multiply the acceleration by the time and add the initial velocity • vf= ā Dt + vi

6. Practice Problem 3.2.19 • A bus that is traveling at 30.0km/h speeds up at a constant rate of 3.5m/s2. What velocity does it reach 6.8s later? • Convert all to terms of m and s: • (30.0km/h)(1000m/km)(1h/3600s)=vi • Define known & unknown: • a=3.5m/s2 vi=8.33m/s • Dt=6.8s vf=? • Choose the appropriate equation • vf=āDt + vi • Rearrange if necessary (it is not in this case) • Plug & Chug • vf=(3.5m/s2)(6.8s)+(8.33m/s) • vf=32.13m/s

7. Analyzing Graphs • Looking at a position-time graph, we can find: • Figure 3-9 • Velocity (slope) • Specific positions at specific times • From our original velocity equation, v=Dd/Dt, we can find that Dd=vDt • The area under the line in a velocity-time graph is vDt, which is the displacement! • Figure 3-10 • Slope is v/Dt, which is acceleration!

8. Position with Constant Acceleration • Position (df) of an object under acceleration can be found with: • df=½at2 (m/s2)(s2)=m • If there is an initial distance that we need to add, then: • df=di+½at2 (m)+(m/s2)(s2)=m • If there is an initial velocity, then we can also include that term! • df=di+vit+½at2 (m)+(m/s)(s)+(m/s2)(s2)=m • This equation is only useful if time of travel is known

9. Position with Constant Acceleration • Unlike the prior equation, sometimes time is not known, so we need to relate velocity to distance traveled • vf2=vi2+2a(df-di) (m/s)2=(m/s)2+(m/s2) (m-m)

10. Equations Summary

11. Practice Problem 3.2.27a • A race car travels on a racetrack at 44m/s and slows at a constant rate to a velocity of 22m/s over 11s. How far does it move during this time? • Define known & unknown: • vi=44m/s vf=22m/s • Dt=11s Dd=? • a=? • Choose the appropriate equation • We need to find ā first • vf=āDt + vi • Rearrange if necessary • ā=(-vi+vf)/Dt • Plug & Chug • ā=(-44m/s+22m/s)/(11s) • ā=-2m/s2

12. Practice Problem 3.2.27b • Now that we have ā, we can solve for df • Define known & unknown: • vi=44m/s vf=22m/s • Dt=11s a=-2m/s2 • Dd=? • Choose the appropriate equation • df=di+vit+½at2 • Rearrange if necessary • Plug & Chug • df=(0m)+(44m/s)(11s)+½(-2m/s2)(11s)2 • df=363m

13. Practice Problem 3.2.28a • A car accelerates at a constant rate from 15m/s to 25m/s while it travels a distance of 125m. How long does it take to achieve this speed? • Define known & unknown: • vi=15m/s vf=25m/s • Dt=? a=? • Dd=125m • Choose the appropriate equation • We need to find ā first, but don’t know time • vf2=vi2+2a(df-di) • Rearrange if necessary • a=(vf2-vi2)/(2Dd) • Plug & Chug • a=((25m/s)2-(15m/s)2)/((2)(125m)) • a=1.6m/s2

14. Practice Problem 3.2.28b • Now that we have ā, we can solve for Dt • Define known & unknown: • vi=15m/s vf=25m/s • Dt=? a=1.6m/s2 • Dd=125m • Choose the appropriate equation • vf = ā Dt + vi • Rearrange if necessary • (vf-vi)/ā=Dt • Plug & Chug • (25m/s-15m/s)/(1.6m/s2)=Dt • Dt=6.3s

15. 3.3 Free Fall • Free Fall- an object falling only under the influence of gravity • Acceleration due to gravity- an object speeds up due to the Earth’s gravitational pull • g=9.8m/s2 • Gravity is a specific kind of acceleration: like a quarter is a specific kind of money • Gravity always points toward the center of the Earth

16. Free Fall Equations As gravity (g) is a kind of acceleration (a), we can replace all of the “a”’s with “g” This can only be done if the object is in free fall

17. Practice Problem 3.3.42a • A construction worker accidentally drops a brick from a high scaffold. What is the velocity of the brick after 4.0s? • Define known & unknown: • vi=0m/s vf=? • Dt=4.0s g=9.8m/s2 • Dd=? • Choose the appropriate equation • vf =gDt + vi • Rearrange if necessary (not necessary) • Plug & Chug • vf =(9.8m/s2)(4.0s)+(0m/s) • vf =39.2m/s

18. Practice Problem 3.3.42b • A construction worker accidentally drops a brick from a high scaffold. How far does the brick fall during this time? • Define known & unknown: • vi=0m/s vf=39.2m/s • Dt=4.0s g=9.8m/s2 • Dd=? • Choose the appropriate equation • df=di+vit+½gt2 • Rearrange if necessary (not necessary) • Plug & Chug • df=(0m)+(0m/s)(4.0s)+½(9.8m/s2)(4.0s)2 • df=78.4m

19. Practice Problem 3.3.45a • A tennis ball is thrown straight up with an initial speed of 22.5m/s. It is caught at the same distance above the ground. How high does the ball rise? • Define known & unknown: • vi=22.5m/s vf=0m/s • Dt=? g=-9.8m/s2 • Dd=? • Choose the appropriate equation • vf2=vi2+2g(df-di) • Rearrange if necessary • Dd=(vf2-vi2)/(2g) • Plug & Chug • Dd=((0m/s)2-(22.5m/s)2)/(2(-9.8m/s2)) • Dd=25.82m

20. Practice Problem 3.3.45b • A tennis ball is thrown straight up with an initial speed of 22.5m/s. It is caught at the same distance above the ground. How long does the ball remain in the air? • Define known & unknown: • vi=22.5m/s vf=0m/s • Dt=? g=-9.8m/s2 • Dd=25.82m • Choose the appropriate equation • vf =gDt + vi • Rearrange if necessary • Dt=(vf-vi)/g • Plug & Chug • Dt=(0m/s-22.5m/s)/(-9.8m/s2) • Dt=2.30s

21. Time Dilation As an objects’ speed approaches 3.0x108m/s (c), the time as observed from the outside of the ship changes So if you are travelling very fast,