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Name: Pebble ims | Date: | Blas Instructor: 4 hea lE eee ee ~ ————- ~-- — Physics 1016 Module 3 Homework OpenStax College Physics, Chapter 2 Problems 1) A well-thrown ball is ae ae in a well-padded mitt. If the deceleration of the ball is 2.10 * 10% m/s*, and 1.85 ms (1 ms = 10s) elapses from the time the ball first touches the mitt until it as what was the initial velocity of the ball? Oe A=~Z.loxiotms * _ 2% } V, =/ ~ ~at= O-(C- -Ulow jot misr 85 X 10 %¢ Jee amie ~3 a. mes] We? #505 03 = Jase} mM 2) A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20 * 10° m/s’for 8.10 * 10*s. What is its muzzle velocity (that is, its { final velocity)? A=$.2 lo? mis” t= 9.10 x Ww S 5 4 = > V=V, tat=0 +(6.20x lo m/s? JP. \ox10's) = sak = 0 Dz M¢ gokm wen Lh = 22,%am/ 3) A light-rail commuter train accelerates at a rate of 1.35 m/s’. lh Xx 60S £5 a. How long does it take to reach its top speed ve 80.0 km/hr, starting from ne / mM gi" 22.274 -0 = 1.35 Se hd” a sO" vf =Broken nas 35 m/¢2 >? b. The same train ordinarily decelerates at a rate of 1.65 m/s*. How long does it take to come to a stop from its top speed? . Q=~1-65 m/z t= Q- 22.22 mé$ - 145s VF=O rer. 22 WS ~libS m/gr c. In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/hr in 8.30 s. What is its emergency deceleration inm/s?? VF>0 W= 22.27 m/f C= 2.306 = VE~-Vo - Q= 22: 2e als — | m/c A= < Ary 3,30 6 2.685 Mi | 4) Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. a. Make a sketch of the situation. | 7 30, 0fm /5 b. List the knowns in this problem. X=0 X=1.800m io =0 y 20,0 CM/S t,=0 c. How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. 2? X=X+VE & 1 GA PerANS % folw for 2 ~ X=% - Ao%o_— .BOCM-00OM— = 1 O. 1206 Vv Vor yy = (2 tenis) d. Is the answer reasonable when compared with the time for a heartbeat? (An entire heartbeat cycle takes about one second.) Yes, it id*readonahle durazin of a heyt pulge. 9) Ina slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33 * 10s, calculate the distance over which the puck accelerates. V, = %.00m /$ Vez to.0m /§ € = 3.33 xlo“s x=? ~ Vis Hy —_ 4D,Om/St 3 K=XtV b= x 1S je = 04 (=e el (5,33 x 19-2 < 2 x10 s)— |! A= ~4.8 mig” 8) Calculate the displacement and velocity at times of the following for a ball chrown. n straight up with an ne velocity of 15.0 m/s. Take the point of release tobeyp>=Om. Vy =!5.0 msg =0.5 005 a. 0.500 5s _ iat alt = (1S.om/5) C0. Soos)t 2 3 (-4.8 mis Jo. 005) Ib. 23m | ViVi tat = 1S.0m/5 + (-4: & m/s*)(b. £065) = |10. | mis | b. 1.00 s X = (Igo mss) 1.008) +3 (- 4.8 m/s*)( 1.005) = |10. | m | Vx 19.0 m5 + (~48 m/s*)( 1.005) =|5, 20m /S | C.1.505 X =(15,0.m/5)(505) + (= 48 m/s*JC1,505) = | | 5 m| Va 15.0m/5+ (-4.8 m/c?) (1905) = | 0.300 MIS | d.2.00s X = (1S Om)(2,005) +4(-468 m/5*) (2, 005)° =| lO. ¢m] = 15,0 m/S + (-4.8 m/s*) (2.008) =|- #.b0 m/s | J) Calculate the displacement and velocity at the following times for a rock thrown straight cdown with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in an York City. The roadway of this bridge is 70.0 m above the water Vj =\tiOmsg JorO Y=~F0.0m < a. 0.500 s yevet yA-b gt = Pas ~$442 = (-\ 4,015) (0.5005) ~HA.8 m/s") (od, s00s)"=|- 8,23 m V=V,- #1 =( ~IF,0m75) ~| 4.8 m/s*) (95005) =|-18,4 m/s b. 1.00 s Y=(~\4, omss) Choos) - $ (4.8 mss*) (1,005)" = 18.4 | Y= (“1F.0018) ~ C4. Big?) (1, 005) _ \- 23,8 nis | re = = = 3 : re 2 ) Continued: E1505 . om y= (-1410m/5) (1.505) ~5 048 mis\(1.505) =|~ 24° V= (-14.0m/5) — (48 m5“) (|,90 5) =| -28.4 M/S | d. 2.00 s |= (-14.0m/5) (2.005) -L(4.8 m/¢*) (2.005) © = |- 44.bm™ | Vz (~4.0m/5) = (4.6 m/5°) (2-005) = ss 33, 6 m/s e.2.50s \J=(-14-0m/5 VC 2,5 0s) = (4.8 m/s 2) (2,505) = E 65, b m) Vaz (~It.0mi5) ~ (4.8 mig") (2.505) = - 33.9 M/S | / Ct O.. ; v= 10) A rock takes 2.35 s to hit the ground after being thrown straight up off a cliff with an initial (oo ab*s velocity of 8.00 m/s. V,=3.00m/5 L=Z35S5 Az ~Adms fy a. Calculate the height of the cliff. ———_———./=0 Yon - Ve -4 at™=~(g,o0mis2.35)~ 4 (-4.3 mis*)(2,355)"= [8.2om | AL“46440=0 b. How long would it take to reach the ground if it is thrown straight down with the same speed? % hy =V, Ly sat - ? ~3,2b m =(~3,00m/S)4 + $4 mityt ~ Bb = ~Z,DOb ~4.44°= 4,44 48,0 -2 2420 Gy" a -R7EA ~"“RNe*~- L= 3,00 * 4(8.00)" -4(4.4)( -8.26) ~ 3.007 422 5.84b_ t= 4 4h C TS Cr ZA 4:8 04,4) ~3. 002 15,02483t (0. t\ts $ -2,356 ZE DE capt Ct ~8:26 4.9 ~ . be — 13) Using approximate values, calculate the slope of the curve in the graph below to verify that the velocity att = 10.0 s is 0.208 m/s. Assume all values are known to be 3 significant figures. Position ys. Time Position {m) 0 i’ 20 30 30) x ot "0 SE) Time (5) bl = 10s -~ 20S = FOS At= 10.95 ~ $0.06 = 40.08% 14.) The graphs below show the position data and velocity data vs. time of a jogger. At bw th tte Position (m) i ee 10 15 Ww 2 Velocity vs. Time Velocity (m/s) Time (s) a. Take the slope of the curve in the position vs. time graph to find the jogger’s velocity att = 2.5 s. This value must be consistent with the velocity vs. time graph. _ Xe -X Z 5.0m —Im y2Ax a oie, ~ iad Z.4-m/6 at t¢-t, “4,05 0.08 b. Repeat at 7.5 s. This value must be consistent with the velocity vs. time graph. 
