Question

Two boy A and B running in the opposite direction in adjacent lanes, at t=0, boy A accelerates at a constant rate of 6.0 ft/s2. Five seconds later, boy B starts and accelerates at a constant rate of 9.7 ft/s2. Knowing that boy B passes original point 18 s after boy A started from there .Determine (a) when and where boy B will overtake boy A. (for example t=1 s x= 1 m )

b) the speed of each boys at that) (time( va=1 m/s, vb31 m/s

a= 9.7 ft/s? a= 6.0 ft/s? Original point

Answer 1

Given Data : 6 ft /s2 Acceleration Acceleration of boy A (AA) of boy B (96) no 9.7 ft/sa Boy A time t=0 sec. Boy B starts accelerating appar 5 sec later (at t = 5 sec.) starts accelerating 6 ft 182 g.7ft 182 LoB original point Boy B passes original point after boy A started from there. (@t- 188ec.) 18 sec solution > Assuming time t the distance that S is = 0 sec. between A & B And time taken by B to travel 's' distance t = 18-5 13 sec. of Motion Hence using equation s = ut + 1/2a +2 since u=0 - s = 12 Opt² hence S = 1/2 x 9.7 X (13) 819.65 ft 1

Now B starts » distance travelled by A, before SA d At + 1 Ata UA 0 ал 6 ft/82 f t = 5 sec. 1 SA 0 + } * 6x(5)2 = 75 ft g.7ft/82 6 ft /s2 744.65 ft * 154+ AVA t=0 t = 5 sec B t=0 to t= 5 sec. velocity of Bed A a t = 5 sec.3 VA UA + Aat UA=0 GA 6 ft 182 & t = 5 ser. 2 NA 6x5 30 ft/sec. AA = 6ft/82 ag=9.7 ft/82 (H t-o VA t=5 1 744.65 ft- X Now using relative velocity & accelerations = VAR VA - VB 30-0 30 ft 182 AAR a A - AB = 6-7-9-1) = 15.7 ft/ge & t time taken by A to reach B.

Now using equation of motion SR VART + I PAR to 144.65 30*t + } *15.7 x70 7-85 t2 + 30t - 744.65=0 after solving t 8.0 144 sec. k 75 sty NA - VBS B t=5 to t=13-014 sec. N is the meeting point of A R B. Distance travelled by A 13.0144 sec. (x) ut & Lat2 QA = 6 ft/sec. UA =0 X = 0 + < x 6 x (13.0144) 508.124 ft 154.88 mtr Boy B Boy H A time t & point will N from overtake distance X: t 13.0144 sec. х 154.88 mtr

B) velocity of a boy A & B meeting point → For As CA 6ft/g2 VA t=13.014 sec t=0 UA=0 using equation of Motion VA UA taat VA O + 6X 13-014 NA 78.084 ft Isec. INA 23.8 mis For B AB= 9.7 ft lg 2 VB N t=13.014 sec. UB=0 t=5 sec. using equation of motion UB + abt Where UB=0 AB 9.7 ft 182 & t = 13.014 = 8.014 sec. NB 0 + 8.014 X 9-7 NB 77,7358 ftisec. IVB = 23.69 misec.