Question

11.78 A car is traveling at a constant speed of 54 km/h when its driver sees a child run into the road. The driver applies her brakes until the child returns to the sidewalk and then accelerates to resume her original speed of 54 km/h; the acceleration record of the car is shown in the figure. Assuming x = 0) when t = 0, determine (a) the time t, at which the velocity is again 54 km/h, (b) the position of the car at that time, (c) the average velocity of the car during the interval 1 ssisti. a(m/s) ty t(s) -6 - - - Fig. P11.78

Answer 1

a(m/s) Answer: Given that, A cor is travelling at a cortant speed of su kmhr V = 54 kminy = 54x mis 0 V - 15 mis acceleration record of car shown in figure (a) Determination of time ts:- Agua under at curve upto t=2s is (2-1)x-6 mis (a +22,3-6mls Area under a-t curve uptu t = 4.55 is = * (4.5-2)*(6) (at4.5 {(2,5)(-6) Ca-1)=0.55 = - 7,5 MIS Area under the a-t wive upto tuti (at)t-ti- Į (-4,5) (2) (a-Hitat, = t;-4,5 velocity at t=25 is @t-25 = vt Cathos Sub stitute required valus Nat-25 = 15-6 Netzzs = 9 mis Vet=ti = @t=4.55't (art) t-t (a-t!tzt; = Votzt; - V@t=4.55 Ca-t)tzt = V@tzt; - [v+ Cat)t=u.ss] Subslitute required values tv-4,5 - 15 - [9 +(-7,5] t-4.5 = 15 -1.5 - 13,5 ti = 13.5+ 4,5 = 185 at velocity is against 54 km's the time ti is iss

6) Determination of position of can at that time : By using known values wat vms) curve :- 15 Semi parabola 12 - - -- - - - - - - 124 18 Displacement of the car at t = 185 185 218 = E Area under the v-t cewe tros 218 = [(15* 15m/s) + (15)(1529) m\s]+[(2.5s) (1.5M])+(2,5) (1.5 m/s)] + [(13155) (1.5m/s) + } (13,55)(13,5 m/s)] 18 = [(15+ 4) ]+[Cans)+ 18,15] + [(20.25+ 364 ]" X18 = 5015+12)] + [3,75+6,25] + [20.25 + 121.5]m 218 = 27+10+ 141,75 m X18 = 178.75 m at t = 185 - Position of the car is 178,75 m C) Determination of Average velocity :- Average velocity calculated using below ouation Varg = on there consider intervals for t =!s, t = 185 *** tg - at tals Distance travelled by car = 24 Substitute values at t=185 > Distance touvelled by co X18 Vang - 178.75-15 mis 18-1 Varg = 916324 mis Varg = 218-X1 Convert mis tokmine Norq = 9.6324 x 18 = 34.7 kmlhos Car Average velocity = 34.7 kmlhor