Version PREVIEW Kine Grphs PRACTICE burke (1111) 1 This print-out should he 30 questions. Multiple-choice questions m continue on the next column or pge find ll choices before nswering. Distnce Time Grph 01 001 (prt 1 of 6) 10.0 points Consider the following grph of motion. 50 Distnce (m) 40 30 20 10 0 0 1 2 3 4 5 Time (sec) How fr did the object trel between 2 s nd 4 s? 1. 20 m 2. 30 m 3. 10 m 4. 40 m 5. 50 m Explntion: The prticle moed from 40 m to 20 m, so d = 40 m 20 m = 20 m. 002 (prt 2 of 6) 10.0 points The grph indictes 1. constnt elocit. 2. incresing elocit. 3. no motion. 4. decresing elocit. 5. constnt position. Explntion: The slope of the grph is the sme eerwhere, so the grph indictes constnt positie elocit. 003 (prt 3 of 6) 10.0 points Wht is the speed from 2 s to 4 s? 1. 0 m/s 2. 20 m/s 3. 15 m/s 4. 5 m/s 5. 10 m/s Explntion: = d t = 40 m 20 m 2 s = 10 m/s. 004 (prt 4 of 6) 10.0 points Consider the following grph of motion. Distnce (m) 400 350 300 250 200 150 100 50 0 0 1 2 3 4 5 6 7 8 9 Time (sec) How fr did the object trel between 3 s nd 9 s? 1. 200 m 2. 500 m 3. 250 m 4. 100 m 5. 350 m 6. 400 m 7. 150 m
Version PREVIEW Kine Grphs PRACTICE burke (1111) 2 8. 300 m 9. 50 m 10. 450 m Explntion: The prticle went from 50 m to 400 m, so the distnce d = 400 m 50 m = 350 m. 005 (prt 5 of 6) 10.0 points The grph indictes 1. decresing elocit. 2. constnt elocit. Explntion: = d t = 400 m 50 m 6 s = 58 m/s. Holt SF 02Re 14 007 (prt 1 of 6) 10.0 points The figure shows the position of runner t different times during run. position ( 1000 m) 5 4 3 2 1 3. no motion. 4. incresing elocit. 0 0 10 20 30 40 time (min) 5. constnt position. Explntion: The slopes re steeper s time goes on, so the elocities re incresing. 006 (prt 6 of 6) 10.0 points Wht is the erge speed from 3 s to 9 s? 1. 47 m/s Note: Figure is drwn to scle. For the time interl between 0 min nd 10 min, wht is the runner s displcement? Your nswer must be within ± 150.0 m. Correct nswer: 2600 m. Explntion: 2. 60 m/s 3. 30 m/s 4. 20 m/s Let : x i = 0 m nd x f = 2600 m. 5. 36 m/s 6. 25 m/s 7. 40 m/s 8. 50 m/s 9. 58 m/s x = x f x i = 2600 m 0 m = +2600 m. 008 (prt 2 of 6) 10.0 points For the sme time interl, find the runner s erge elocit. Your nswer must be within ± 0.4 m/s. Correct nswer: 4.33333 m/s.
Version PREVIEW Kine Grphs PRACTICE burke (1111) 3 Explntion: Let : t 1 = 10 min. g = x 1 = 2600 m min 1 t 1 10 min 60 s = +4.33333 m/s. 009 (prt 3 of 6) 10.0 points For the time interl between 10 min nd 20 min, wht is the runner s displcement? Correct nswer: 1000 m. Explntion: Let : x i = 2600 m nd x f = 3900 m. 012 (prt 6 of 6) 10.0 points Find the erge elocit for the entire run. Your nswer must be within ± 0.4 m/s. Correct nswer: 2.33333 m/s. Explntion: t 3 = 30 min 20 min = 10 min, t tot = 10 min + 10 min + 10 min = 30 min, nd g = x tot = 4200 m t tot 30 min 1 min 60 s = +2.33333 m/s. so x 2 = x f x i = 3600 m 2600 m = 1000 m. 010 (prt 4 of 6) 10.0 points For the sme time interl, find the runner s erge elocit. Your nswer must be within ± 0.5 m/s. Correct nswer: 1.66667 m/s. Explntion: t 2 = 20 min 10 min = 10 min, g = x 2 = 1000 m t 2 10 min 1 min 60 s = 1.66667 m/s. 011 (prt 5 of 6) 10.0 points Wht is the runner s totl displcement? Your nswer must be within ± 150.0 m. Correct nswer: 4200 m. Explntion: x 3 = 4200 m 3600 m = +600 m, so x tot = x 1 + x 2 + x 3 = 2600 m + 1000 m + 600 m = +4200 m. so Displcement s Time 02 013 (prt 1 of 5) 10.0 points Consider the displcement cure OABC Displcement s Time x (m) 2 1 0 1 O A 2 0 1 2 t (s) Wht is the erge elocit from point O to A? 1. OA = 3 m/s 2. OA = + 3 m/s 3. OA = +2 m/s 4. OA = 2 m/s 5. OA = 0 m/s B C
Version PREVIEW Kine Grphs PRACTICE burke (1111) 4 Explntion: The displcement x is on the erticl xis nd the time t is on the horizontl xis. Velocit requires net displcement: OA = x A x O = 2 0 = +2 m/s. t A t O 1 0 014 (prt 2 of 5) 10.0 points Wht is the erge elocit for the motion from point O to point B? 1. OB = + 3 m/s 2. OB = 3 m/s 3. OB = 0 m/s 4. OB = +2 m/s 5. OB = 2 m/s Explntion: OB = x B x O t B t O = 0 0 2 0 = 0 m/s. 015 (prt 3 of 5) 10.0 points Wht is the erge speed for the motion from point O to point B? 1. s OB = 0 m/s 2. s OB = + 3 m/s 3. s OB = 3 m/s 4. s OB = 2 m/s 5. s OB = +2 m/s Explntion: s OB = x A x O + x B x A t B t O = 2 + 2 = +2 m/s. 2 0 016 (prt 4 of 5) 10.0 points Wht is the instntneous elocit t point B? 1. B = 2 m/s 2. B = 0 m/s 3. B = + 3 m/s 4. B = +2 m/s 5. B = 3 m/s Explntion: The instntneous elocit t point B cn be obtined b first finding n expression for the position of the moing object nd tking its deritie eluted t time t B. Since the grph ner B is liner, it is simpler to clculte the slope of the line oer the interl from A to B. The result l describes the instntneous elocit t B becuse the deritie of stright line is constnt t ll points on the line nd cn be obtined for our cse b B = x B x A = 0 2 = 2 m/s. t B t A 2 1 017 (prt 5 of 5) 10.0 points Wht is the instntneous speed t point B? 1. s B = 0 2. s B = 3 m/s 3. s B = + 3 m/s 4. s B = 2 m/s 5. s B = +2 m/s Explntion: Instntneous speed is simpl the mgnitude of the elocit: s B = B = 2 = +2 m/s. Velocit s Time 11
Version PREVIEW Kine Grphs PRACTICE burke (1111) 5 018 (prt 1 of 3) 10.0 points The scle on the horizontl xis is 4 s per diision nd on the erticl xis 4 m/s per diision. The prticle hs n initil position t 3 m. elocit ( 4 m/s) 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 time ( 4 s) Wht is the position of the prticle fter the first 8 s? Correct nswer: 131 m. Explntion: Chnge in position is the re under the elocit s time grph. p = t = 4 (4 m/s) t, so the position fter the first (8 s) seconds is p = p o + p = (3 m) + 4 (4 m/s) (8 s) = 131 m. 019 (prt 2 of 3) 10.0 points Wht is the elocit of the prticle fter the first 8 s? Correct nswer: 16 m/s. Explntion: Red the elocit corresponding to the fourth tic mrk on the erticl xis = 4 (4 m/s) = 16 m/s. 020 (prt 3 of 3) 10.0 points Wht is the ccelertion of the prticle fter the first 8 s? Correct nswer: 0 m/s 2. Explntion: The ccelertion is the slope of the elocit s time grph; i.e., the ccelertion is zero. Velocit s Time 17 021 (prt 1 of 3) 10.0 points Consider the elocit cure of one dimensionl motion long the x-xis. The initil position is x = 10 m. The scle on the horizontl xis is 2 s per diision nd on the erticl xis 2 m/s per diision. elocit (2 m/s) per diision b time (2 s) per diision Wht is the position t t = 4 s? 1. x = 22 m 2. x = 16 m 3. x = 14 m 4. x = 30 m 5. x = 20 m 6. x = 18 m 7. x = 10 m 8. x = 28 m 9. x = 24 m c
Version PREVIEW Kine Grphs PRACTICE burke (1111) 6 10. x = 12 m Explntion: the re of tringle is 1 bse height. 2 Chnge in position is the re under the elocit s time grph, so x = x + x o = 1 2 (10 m/s)(4 s) + (1 s)(10 m/s) = 30 m. 022 (prt 2 of 3) 10.0 points Wht is the erge elocit between 0 s nd 6 s? 1. 4 m/s < 5 m/s 2. 6 m/s < 7 m/s 3. 8 m/s < 9 m/s 4. 0 m/s < 1 m/s 5. 1 m/s < 2 m/s 6. 7 m/s < 8 m/s 7. 2 m/s < 3 m/s 8. 5 m/s < 6 m/s 9. 3 m/s < 4 m/s 10. 9 m/s < 10 m/s Explntion: During the time interl 0 to 6 s, x = 1 (10 m/s) (4 s) + (2 s) (10 m/s) 2 = 40 m, so the erge elocit between 0 to 6 s is = x t = 40 m 6.667 m/s. 6 s Wht is the erge ccelertion between 14 s nd 16 s? 1. 1 m/s 2 < 2 m/s 2 2. 4 m/s 2 < 3 m/s 2 3. 3 m/s 2 < 4 m/s 2 4. 2 m/s 2 < 3 m/s 2 5. 5 m/s 2 < 4 m/s 2 6. 2 m/s 2 < 1 m/s 2 7. 4 m/s 2 < 5 m/s 2 8. 0 m/s 2 < 1 m/s 2 9. 1 m/s 2 < 0 m/s 2 10. 3 m/s 2 < 2 m/s 2 Explntion: The ccelertion is the slope of the cure in the elocit s time grph. The ccelertion is constnt from 12 s to 18 s, so the ccelertion is = t = 8 m/s 0 m/s 18 s 12 s 1.333 m/s 2. Velocit s Time 18 024 10.0 points An object ws suspended in fixed plce nd then llowed to drop in free fll. Tking down s the positie erticl direction, which grph l represents its motion s erticl elocit s time? 1. t 023 (prt 3 of 3) 10.0 points
Version PREVIEW Kine Grphs PRACTICE burke (1111) 7 ccelertion, so the grph is t Velocit s Time 025 10.0 points An object is thrown erticll upwrd. Disregrding ir resistnce, which grph represents the elocit of the object s function of time t? 1. t 5. t 7. t 8. t Explntion: The object is undergoing constnt positie grittionl ccelertion g. The slope of elocit s time cure represents the Explntion: The object undergoes constnt downwrd grittionl ccelertion g. It hs decresing positie elocit on the w up, brief instnt of zero elocit, then negtie elocit on the w down (which continues to decrese in lue: 3 < 1).
Version PREVIEW Kine Grphs PRACTICE burke (1111) 8 = 0 g t This is stright line with negtie slope nd positie initil elocit. t 5. t Accelertion Time Grph 01 026 (prt 1 of 5) 10.0 points Consider to cr which cn moe to the right (positie direction) or left on horizontl surfce long stright line. cr O + Wht is the ccelertion-time grph if the cr moes towrd the right (w from the origin), speeding up t sted rte? 1. None of these grphs is. 7. t 8. t Explntion: Since the cr speeds up t sted rte, the ccelertion is constnt. 027 (prt 2 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moes towrd the right, slowing down t sted rte? 1. t
Version PREVIEW Kine Grphs PRACTICE burke (1111) 9 5. t 5. t 7. None of these grphs is. 8. t Explntion: Since the cr slows down, the ccelertion is in the opposite direction. 028 (prt 3 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moes towrds the left (towrd the origin) t constnt elocit? 1. t 7. t 8. None of these grphs is. Explntion: Since the cr moes t constnt elocit, the ccelertion is zero. 029 (prt 4 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moes towrd the left, speeding up t sted rte?
Version PREVIEW Kine Grphs PRACTICE burke (1111) 10 1. t 030 (prt 5 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moes towrd the right t constnt elocit? 1. None of these grphs is. 5. None of these grphs is. 7. t 8. t 5. t 7. t Explntion: The sme reson s Prt 1.
Version PREVIEW Kine Grphs PRACTICE burke (1111) 11 8. t Explntion: The sme reson s Prt 3.