This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PART 1 MECHANICS CHAPTER 2 MOTION IN A STRAIGHT LINE ActivPhysics can help with these problems: Activities 1.2–1.9 Section 21: Distance, Time, Speed, and Velocity Problem 1. In 1996 Donovan Bailey of Canada set a world record in the 100m dash, with a time of 9.84 s. What was his average speed? Solution Bailey’s average speed was (Equation 21) ¯ v = Δ x/ Δ t = 100 m / 9 . 84 s = 10 . 16 m/s . (One can assume that the race distance was known to more than four significant figures.) Problem 2. When races in a track meet are timed manually, timers start their watches when they see smoke from the starting gun, rather than when they hear the gun. How much error is introduced in timing a 200m dash over a straight track if the watch is started on the sound rather than the smoke? The speed of sound is about 340 m/s. Solution Suppose the timer stands at the finish line, 200 m from where the starting gun is fired. The time required for a signal to travel this distance at speed v is Δ t = Δ r/v = 200 m /v . Because the speed of light is so great, the puff of smoke is seen after a negligible delay, as far as conventional watches are concerned, Δ t = 200 m / 3 × 10 8 m/s = 0 . 67 μ s . The travel time for a sound signal, however, Δ t = 200 m / 340 m/s = . 59 s, would introduce a significant error, if times are recorded to the nearest hundredth of a second. (For a manually operated timing device, the error due to human reaction time is about 0.2 s.) Problem 3. In 1996, Josia Thugwame of South Africa won the Olympic Marathon, completing the 26mi, 385yd course in 2 h 12 min 36 s. What was Thugwame’s average speed, in meters per second? Solution ¯ v = Δ r Δ t = (26 + 385 / 1760) mi (2 + 756 / 3600) h = 11 . 9 mi h ≈ 5 . 30 m s , or a little over half the speed of Bailey’s 100 m dash in Problem 1. (Runners usually compute their average pace, 1 / ¯ v, which in this case was 5 min 3.4 s per mile. See Appendix C for the appropriate conversion factors.) Problem 4. Human nerve impulses travel at about 10 2 m/s. Estimate the minimum time that must elapse between the time you perceive a stalled car in front of you and the time you can activate the muscles in your leg to brake your car. (Your actual “reaction time” is much longer than this estimate.) Moving at 90 km/h, how far would your car travel in this time? Solution Suppose the neural path length from the brain to the leg muscles (the quadriceps) is about 1 m long. The travel time for nerve impulses is about 1 m / 10 2 m/s = 10 − 2 s . A car moving at 90 km/h would travel Δ r = v Δ t = (90 × 10 3 m / 3600 s) × (10 − 2 s) = 25 . 0 cm during this interval. Problem 5. Starting from home, you bicycle 24 km north in 2.5 h, then turn around and pedal straight home in 1.5 h. What are your (a) displacement at the end of the first 2.5 h, (b) average velocity over the first 2.5 h, (c) average velocity for the homeward leg of the trip, (d) displacement for the entire trip, and (e) average velocity for the entire trip?(e) average velocity for the entire trip?...
View
Full
Document
This note was uploaded on 07/10/2008 for the course PHYS 2A taught by Professor Hicks during the Summer '07 term at UCSD.
 Summer '07
 Hicks
 mechanics

Click to edit the document details