Hey <Geetha>

Following is my worked solution :

*** - for alignment purpose

C ********************************************** D

|---------------------------------------------------------|

|-> ********* |-> ************** |->

V1 ********* V2 ************** V3

t=12 ******* t=2 ************** t=5

s(V)=s

|-> *************************** |->

M1 ************************** M2

t=2 ************************** t=5

s(M)=(s + 28)

In order for M to catch up and overtake V at 5pm (t=5), M would have to travel an additional distance since V started 2 hours earlier than M. Such additional distance is covered by M's additional speed of 28 km/h.

Given that M overtook V 3 hours after it started,

Additional distance (by M) = (Additional speed) × time taken

= 28 × 3 = 84 km

This additional distance travelled by M means that at the time M started its journey from C, V was ahead by 84 km (V2), i.e. d(V1V2) = 84.

(a)

s(V) = d(V1V2) ÷ t(V1V2) = 84 ÷ 2 = **42 km/h**

(b)

Given that motorist (M) was 80km away from Town D at 5pm,

d(M2D) = 80

d(V2V3) = s(V) × t(V2V3) = 42 × 3 = 126

Hence, total distance (CD) = d(V1V2) + d(V2V3) + d(M2D)

= 84 + 126 + 80 = **290 km**

=========

Trust this helps.

Do let me know again if this is different from the <anskey> or if there's further clarification.

Cheers,

Edward