From the same place at 7am 'A' started walking in north at the speed of 5 km/hr. After 1 hour B started cycling in the east at the speed of 16 km/hr. At what time will they be at a distance of 52 km apart from each other?
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Step-by-step explanation:
Let t= travel time of bicycle
then (t+1)= Walking time
Distance = Speed × Time
This is a Pythagoras problem
a
2
+b
2
=c
2
Where a=8t, bicycle distance
b=4(t+1), Walking distance
c=20, distance apart in t hrs from 8 a.m.
∴(8t)
2
+[4(t+1)]
2
=(20)
2
64t
2
+16(t
2
+1+2t)=400
64t
2
+16t
2
+32t+16=400
80t
2
+32t−384=0
16(5t
2
+2t−24)=0
5t
2
+2t−24=0
5t
2
+(12−10)t−24=0
5t
2
+12t−10t−24=0
5t
2
−10t+12t−24=0
5t(t−2)+12(t−2)=0
(t−2)(5t+12)=0
t=2
and t=−
15
12
which is not possible.
Positive solution t=2 hrs.
At 10 a.m. they will be 20 km apart
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