2003 1. (b)   A man runs at constant speed to catch a bus.
At the instant the man is 40 metres from the bus, it begins to accelerate uniformly from rest away from him.
The man just catches the bus 20 seconds later.

                            (i).            Find the constant speed of the man.

                          (ii).            If the constant speed of the man had instead been 3 m/s, show that the closest he gets to the bus is 17.5 metres.

ANSWER
1 (b) (i)

Say the bus travels s_b metres in the 20 seconds.
The man must travel s_m = 40+ s_b in this time.

Simultaneous equations are needed for the two unknowns.

For the bus:
   no “v”     i.e    i.e.

For the man:
   i.e.    

Since the man just catches the bus, the speed of the man, v, equals the speed of the bus, v_b.

Thus,  v_b = 20a = v

i.e.  implies that  i.e. a = 0.2 ms^-2.

Substituting in  we get  i.e. v = 4 ms^-1.

(ii)

Let x = the distance of the man from the bus at time t.

x = 40 – the distance covered by the man + the distance covered by the bus.

i.e.

i.e.

To find the minimum value of x we differentiate and set

i.e.   thus