Question

75 To stop a car, first you require a certain reaction time to be 2The psition of a particle moving along the x sxis depends on gin braking: then the car slows at a constant rate. Suppose that the the time according to the equationxbP, where a is in me total distance moved by your car during these two phases is 56.7 m lers and rin secondh What are the units of (a) constant e and (b) con- when its initial speed i·M3 kmh, and 24.4 m when its initial speed an' b? Let their numerical values be 30 and 20, respectively (e) Al 4x3 kmh. What are (a) your reaction time and (b) the magni what time does the particle reach its maximum positive x position? 00s to r-4.0(d) what distance does the particle move tude of the acceleration? (g) 2.0%,(b) 30 (k)20x1, 30%,and (m)40 and (i) 4D s Find its acceleration at times G, 1.0 "51 As a runaway scientific bal loon ascends at 19.6 m/s one of its instrument packages breaks free of a The indicated acceleradions are What is the outward speed of the songue at aceeleration phase? -49 nW How far docs the run- ner whose welocity-time graph is shoun in Fig 2-4) travel in 16s7 The Sigure's vertical scaling is set bys,80ms harness and free-falls. Figure 2-34 o N6T'ω gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground. (a) What maximum height above the break-free point does it Figure 2-34 Problem 51. rise? (b) How high is the break-free point above the ground? Figun 2-48 Problem t9 OAir 1.5sit ases thetop of t lower nd 1.0s later desits maximum height Wha isthe height ofthetower? 11You are to drive 300 km to an interview The interview is at 11:15 A.M. You plan to drive at 100 km/h, so you leave at 8:00 A.M. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to skow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview

Answer 1

V_1 = 80.5 km/n = 80.5 times 1000/3600 m/s = 22.36m/s V_2 = 48.3 km/n = 48.3 times 1000/3600 m/s = 13.42 m/s S_1 = 56.7 m S_2 = 24.4 m S_1 = V_1t_1 + 1/2 a_1 t_1^2 rightdoublearrow 56.7 = 22.36 t + 1/2 at^2 - (1) S_2 = V_2t_2 + 1/2 a_2t_2^2 rightdoublearrow 24.4 = 13.42t + 1/2 at^2 - (2) subtracting (2) from (1) 56.7 = 22.36t + 1/2 at^2 24.4 = 13.42 + 1/2 at^2/32.3 = 8.94t rightdoublearrow t = 3.61S putting value of t in (1) 56.7 = 22.36 times 3.61 + 1/2 a times (3.61)^2 rightdoublearrow a = -3.69 m/s^2 (a) t = 3.61S (b) a = 3.69 m/s^2