Question

15. A man is running to cateh a bus at his top speed of 6 m/s. The man is 25 m behind the b starts to move forward with an acceleration of 1 m/s e man catch the bus? If yes, then fiud out when and where it happens. If no, then determine how w an accurate position-versus-time graph of this scenario for both the man and the bus on the same (c) Repeat Parts (a) and (b) assuming that the bus moves forward, instead, with an acceleration of 0.5 m/ he got to the bus and the minimum speed he would have needed in order to catch it graph. (with all other pieces of information remaining the same)

Answer 1

(a) let us assume that man catches the bus after time �t� so distance covered by man is S_m = ut = 6t During this time, distance covered by bus is S_b = 1/2 at^2 = 0.5 t^2 Because S_m = S_b + 25 implies 6t = 0.5t^2 + 25 implies 0.5t^2 - 6t + 25 = 0 So t = b plusminus squareroot 6^2 - 4 (0.5) 25/2 (0.5) = complex roots So Man never catches the bus with 6 m/s \r\n Now to find, (delta S0 min i.e. (S_b + 25) - S_m delta S = (S_b + 25) - (S_m) delta S = 0.5t^2 + 25 - 6t thatis (1) to find (delta S) min, differentiate with �t� & equate to zero d/dt (delta S) = 2 (0.5) t - 6 0 = t - 6 implies t = 6 s Substitute in (1), delta S = 0.5 (6)^2 + 25 - 6 (6) = 7 m