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 By: geethavasavi (offline)  Tuesday, March 29 2016 @ 03:28 AM CDT (Read 1468 times)
geethavasavi

Hi Edwin,

At noon, a van driver left Town C for Town D travelling at a constant
speed. 2 hours later, a motorist also left Town C for Town D.
The motorist overtook the van driver at 5pm. The speed of the motorist
was 28km/h faster than that of the van driver.

(a)Find the speed of the van driver.
(b)Find the distance between Town C and Town D if the motorist
was 80km away from Town D at 5pm

Thanks & Regards,
Geetha

Newbie

Registered: 03/17/11
Posts: 10

 By: echeewh (offline)  Tuesday, March 29 2016 @ 01:12 PM CDT
echeewh

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,

= 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

Active Member

Registered: 04/21/11
Posts: 627

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