Question

please show all work.

3. Object A is stationary and is passed by object B traveling at a constant speed of 72 mi/hr. If object A sets off in pursuit of B 1.5 seconds later with a constant acceleration of 3.2 m/s2, how long will it take object A to catch up to object B?

Answer 1

After 1.5 sec. Ane. VASO that meame object 'B' is mowing with constat speed of 32.18 ms. Now after 15 sec distance Bavelo by A object B is K 48.277 S= 32.18x1.5 - 48.27 m. B B so z А B >1 48.277 s QA= 3.2m/s2 Now on so this V8=72 milhy = 32:18 mis VB-32-18 mis given that at t= o(initially) after 1.5 sec VA=0 VA= ? E UB=72milelhr = 32.18 15 QA= 0m/s2 VB= 32.18 m/s² Let consider that at point d' object A and B meets after ag=0m/s2 time t then if distance blw obied 'n and point s' is AB=0m/s2 's meter then distance ble object 'B' and 'x' is After 15 sec. the position of A and B uneill be look like '5-48.27 m. 'A' Newton's law of motion for objet writing PA=3.2 ms 48=0 S= ut+ ļot2 at here on Cinitiae) zo le S=4*3.2x+2= 1-672 Conversion of velocity som milho mis writing Newton's law of motion for object 'B' Imi= 1.609 km = 1609m S-48.27 = (32:18)+ rox+2 = 32.184 - lol. 72 rilar = 115.848 km/ho = 115.848 x 1000 3600 On putting the lalue of epho in m 32:18 mis 1.6+2-48.27= 32.18 t + 1.642-32-180-418.27 00 B 482 32.18 m/s2 А Now on writing CScanned with CamScanner

on soleing above quardatic equation its sooth are -bt √b2cuac here a=1.6 bs-32-12 CE-48.27 20 32.18 |(32•19) 2-U13) (1.6) (-48.27) 2(1.6) 32:18+ 1035.55+ 308.92 3.2 32.18+ 36.66 32.187 1344.47 3.2 3. g the label of t av 21.51 and -1.4, ; Negative time is not acceptable so t= 21:51 sec. so It will take 21:51 sec object À to catch objet B'. CScanned with CamScanner