Question

2) Shown below is the velocity graph of a lg coffee filter released from rest. (Note the break in the time axis.) Four distinct intervals are shown on the graph. a) Speed downward increasing from rest, Os <t<0.02s. b) Speed downward increasing, 0.02s sis 0.09. c) Speed downward is constant, 0.098 <134.99s. d) Landing on the floor, 4.995 <155.00s. Draw an acceleration graph for the same time intervals above. You may use the tangent lines drawn on the velocity graph to calculate the average slopes of the velocity curve during the first two intervals. +0.2 +0.1 (s/w] A 0.0 t[sec] 5.50 0.05 0.10 0.15 4.90 4.95 -0.1 -0.2 (b) -0.3 d) -0.4 -0.51

Answer 1

* Answers Given that: will help Find the slope of the velocity Curve finding acceleration in that interval. interval (oss + 30.025) Y₂ - Y, Acceleration slope A-2, here (2., 4, (0,0) (0.08 ,-0.9) - 0.0 -0 .. aceleration 0.02-0 - 10 m/s acceleraton is decelerating : Negative sigh means in nature. 6 In interval (0.02s < t < 0.095) Here (using graph] Y2 = -0.5 12 = 0.09 acceleration -0.5 - (-0.9) 3, = -0.9 21 = 0.02 0.09 -0.02 - 0.5 +0.2 0.07

-0.3 0.07 =-4.9857 m/s © to In Interval 0.ogs <ta4.995 When Speed downward is constant. So v is constant [sance, as a dy dt zero Hence, acceleration (a) is In interval 4.995 < t < 500S When landing on the floor. object have landed Speed (-V) is zero. since, So, zero acceleration (@) is * Acceleration graph:- 0.02 0.06 0.09 0.10 0.15 5.00 o -01 tises) V (m/s) -10 Acceleratim