GRAPH Part B. Acceleration versus sin ? What is the curve's slope? What is this value equal to? Answer to Question 1. What quantity does the slope of velocity-time grapph represent? 2. From the data obtained from part 2 how is time t related to the inclination of the track? Explain why. Analysis ConclusionPART 2 x0 = 25 am L= 100 am 1 = 75 cam AX = X = JOOM Trial Height of Track a she= sine Time = 1 47 am/s' 0.05 940 com / s 1.465 2 10 CAMM 99 cm 6 2 0.10 790 cm/ s' 1.015 SOLUTION : ( FUR TRAL | 2 2 ) trial 1: Sin Of L 9 = 0 Time = J2 x sint Sin e = JU 9 = 47 cm /sz time : 2 ( 50 0m ) 47 cmrs 2 100 chy 10.05 time = Sin 0 = 0.05 9 = 940 cm / 32 47 cox / sz trial 2 : Hime = 1. 46 5 9 = a Sin O = + Sing I time = ) z x Time = J 1. 952 Sin D= 10 0xx g = 99 am /5 2 0 . 10 time = Time = 1. 005s sin # = 0-10 9 = 9 9 0 m / s z time = ! 1.015Trial 3 15 6m 4in 9 - 0.15 138em / 9 = 920 Gm/ 0. 15 Time : V3/ Time 1( 50am) 13 5am4" . lime ' Do. Instant . Time = VO. TZ. Time : 0.85, Treal 4 Ain 0 - 0.20 -1 09 am 100am /5 20.10 8 : 1000 m/ Time = is . Time = 2150.) Tome =meem . lime - Voss , Time = 0.7/, Trial 5 Aim . 0.25 109um 9 : 214 um il 1 : 976 am/' . 0.15 Time = Vy , Time . :( 50um) 111 am4 . Time . florum , Time : Vons, Time : 0.69, average ] 2 940 um/ + 940 m/ : 965. 2 6 m /1! 1 910 ult % erron 6.". - a.Y . 100 % on = _ |9 6 5. 1am'- 99. - 94/4m A1 . 100 + 476 wit a.v 4 1 14/ % eres = 0.0151 . 100 , Town : 1.51%