PHYS Summer Professor Caillault Homework Solutions. Chapter 2


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1 PHYS Summer Professor Cillult Homework Solutions Chpter 2
2 5. Picture the Problem: The runner moves long the ovl trck. Strtegy: The distnce is the totl length of trvel, nd the displcement is the net chnge in position. Solution: 1. () Add the lengths: ( 15 m) + ( 100 m) + ( 15 m) 130 m 2. Subtrct x i from x f to find the displcement. Δ x x f x i m 100 m 3. (b) Add the lengths m 260 m 4. Subtrct x i from x f to find the displcement. Δ x x f x i 0 0 m 0 m Insight: The distnce trveled is lwys positive, but the displcement cn be negtive. The displcement is lwys zero for complete circuit, s in this cse. 17. Picture the Problem: The dog continuously runs bck nd forth s the owners close the distnce between ech other. Strtegy: First find the time tht will elpse before the owners meet ech other. Then determine the distnce the dog will cover if it continues running t constnt speed over tht time intervl. Solution: 1. Find the time it tkes ech owner to wlk 5.00 m before meeting ech other: Δt d s v 5.00 m 1.3 m/s 3.8 s 2. Find the distnce the dog runs: d sδt ( 3.0 m/s) ( 3.8 s) 11 m Insight: The dog will ctully run shorter distnce thn this, becuse it is impossible for it to mintin the sme 3.0 m/s s it turns round to run to the other owner. It must first slow down to zero speed nd then ccelerte gin.
3 27. Picture the Problem: The given position function indictes the prticle begins trveling in the negtive direction but is ccelerting in the positive direction. Strtegy: Crete the xversust plot using spredsheet, or clculte individul vlues by hnd nd sketch the curve using grph pper. Use the known x nd t informtion to determine the verge speed nd velocity. Solution: 1. () Use spredsheet to crete the plot: 2. (b) Find the verge velocity from t to t s: v v Δx Δt ( )( s) 3 ( ) + ( 3 m/s ) 3 ( s) 3 ( 2 m/s) ( s) + 3 m/s 3 ( 2 m/s) s s 1.63 m/s 3. (c) Find the verge velocity from t to t s: v v Δx Δt ( )( s) 3 ( ) + ( 3 m/s ) 3 ( s) 3 ( 2 m/s) ( s) + 3 m/s 3 ( 2 m/s) s s 1.64 m/s 4. (d) The instntneous speed t t s will be closer to 1.64 m/s. As the time intervl becomes smller the verge velocity is pproching 1.64 m/s, so we conclude the verge speed over n infinitesimlly smll time intervl will be very close to tht vlue. Insight: Note tht the instntneous velocity t s is equl to the slope of stright line drwn tngent to the curve t tht point. Since it is difficult to ccurtely drw tngent line, we usully resort to mthemticl methods like those illustrted bove to determine the instntneous velocity. 35. Picture the Problem: The cr trvels in stright line in the positive direction while ccelerting in the negtive direction (slowing down). Strtegy: Use the constnt ccelertion eqution of motion to determine the time elpsed for the specified chnge in velocity. Solution: 1. () The time required to come to stop is the chnge in velocity divided by the ccelertion. In both cses the finl velocity is zero, so the chnge in velocity doubles when you double the initil velocity. Therefore the stopping time will increse by fctor of two when you double your driving speed. 2. (b) Solve eqution 27 for time: t v v m/s 4.2 m/s s
4 3. (c) Solve eqution 27 for time: t v v m/s 4.2 m/s 7.6 s 2 Insight: Note tht the decelertion is treted s negtive ccelertion in this problem nd elsewhere in the text. 43. Picture the Problem: The cheeth runs in stright line with constnt positive ccelertion. Strtegy: The verge velocity is simply hlf the sum of the initil nd finl velocities becuse the ccelertion is uniform. The distnce trveled is the verge velocity multiplied by the time elpsed. Solution: 1. () Clculte hlf the sum of the velocities: v v 1 v 2 ( 0 + v) m/s 2 ( ) 12.5 m/s 2. Use the verge velocity to find the distnce: d v v t ( 12.5 m/s) ( 6.22 s) 77.8 m 3. (b) For constnt ccelertion the velocity vries linerly with time. Therefore we expect the velocity to be equl to 12.5 m/s fter hlf the time (3.11 s) hs elpsed. 4. (c) Clculte hlf the sum of the velocities: v v,1 1 2 v 0 + v 5. Clculte hlf the sum of the velocities: v v,2 1 2 v 0 + v ( ) m/s 2 ( ) 6.25 m/s ( ) m/s 2 ( ) 18.8 m/s 6. (d) Use the verge velocity to find the d v t 1 v,1 ( 6.25 m/s) ( 3.11 s) 19.4 m distnce: 7. Use the verge velocity to find the distnce: d 2 v v,2 t ( 18.8 m/s) ( 3.11 s) 58.5 m Insight: The distnce trveled is lwys the verge velocity multiplied by the time. This stems from the definition of verge velocity. 53. Picture the Problem: You drive in stright line nd then slow down to stop. Strtegy: Use the constnt ccelertion eqution of motion (eqution 27) to find the time. Once the time is known, we cn use the sme eqution to find the speed. In this cse, the ccelertion is negtive becuse the cr is slowing down. Solution: 1. () Solve eqution 27 for t: t v v m/s 3.2 m/s s 2. (b) Since the velocity vries linerly with time for constnt ccelertion, the velocity will be hlf the initil velocity when you hve brked for hlf the time. Therefore the speed fter brking 2.5 s will be equl to 8.0 m/s. 3. Evlute eqution 27 directly: v v 0 + t 16 m/s + ( 3.2 m/s ) 2 ( 2.5 s) 8.0 m/s 4. (c) The totl distnce trveled is the distnce the cr trvels t 16 m/s before you hit the brkes ( time intervl given by your rection time) plus the distnce covered s Δx v 0 t rect + v v t stop t rect Δx v v t stop v 0 ( )( 5.0 s) 55 m 8.0 m/s 16 m/s 0.94 s
5 the cr stops. Insight: For constnt ccelertion, the velocity chnges linerly with time, but nonlinerly with distnce. 77. Picture the Problem: The glove rises stright up, momentrily comes to rest, nd then flls stright bck down. Strtegy: The glove will lnd with the sme speed it ws relesed, neglecting ny ir friction, so the finl velocity v 6.0 m/s. We cn use the eqution of motion for velocity s function of time to find the time of flight. Solution: 1. () Solve eqution 27 for t: t v v 0 ( ) ( 6.0) m/s s 9.81 m/s 2 2. (b) The time to rech mximum height: t v v m/s 9.81 m/s s Insight: Throwing the glove upwrd with twice the speed will double the time of flight but the mximum height ttined by the glove (3.66 m for 6.0 m/s initil speed) will increse by only fctor of 2.
6 95. Picture the Problem: The person is thrown stright upwrd, slows down, nd momentrily comes to rest before flling stright downwrd gin. Strtegy: Find the time of flight by exploiting the symmetry of the sitution. If it tkes time t for grvity to slow the person down from her initil speed v 0 to zero, it will tke the sme mount of time to ccelerte her bck to the sme speed. It therefore tkes the sme mount of time for her to rise to the pek of her flight thn it does for her to return to the blnket. Use this fct together with eqution 211 with v 0 0 (corresponding to the second hlf of her flight, from the pek bck down to the blnket) to find the time of flight. The time bove nd below 14.0 ft cn be found using the sme eqution. Solution: 1. () The time of flight cn be found from eqution 211: t 2 t down 2 2Δx g 2( 28.0 ft m/ft) s 9.81 m/s 2 2. (b) The person s verge speed is less during the upper hlf of her trjectory, so the time she spends in tht portion of her flight is more thn the time she spends in the lower hlf of her flight. 3. (c) The time she spends bove 14.0 ft is the sme time of her flight if her mximum height were 14.0 ft: 4. The time spent below 14.0 ft is the remining portion of the totl time of flight: t bove 2 2Δx g 2( 14.0 ft m/ft) s 9.81 m/s 2 t below t totl t bove s 0.77 s Insight: The symmetry of the motion of freely flling object cn often be useful tool for solving problems quickly.
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