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Question

From the same place at 7 am 'A' started walking in the north at the speed of 5km/hr. After 1 hour B started cycling in the east at a speed of 16km/hr. At what time they will be at distance of 52 km apart from each other?

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Solution

Let A start walking from the point O in the north at 7 am at a speed of 5 km/hr and B also start cycling from the point O in the east at 8 am at a speed of 16 km/hr.
Now, let they be 52 km apart from each other after t hrs.
Then, the distance covered by A in t hrs is 5 × 1 + 5 × t = 5 + 5t in the northern direction and the distance covered by B in t hrs is 16 × t = 16t in the eastern direction.
Now, we draw a figure of their motion as shown below.


From the figure, we observe that ΔAOB is a right-angled triangle. Using the Pythagoras' Theorem, we get:
OA2 + OB2 = AB2
(5 + 5t)2 + (16t)2 = (52)2
25 + 50t + 25t2 + 256t2 = 2704
281t2 + 50t – 2679 = 0
On comparing this equation with at2 + bt + c = 0, we get:
a = 281, b = 50, c = –2679
For the quadratic equation at2 + bt + c = 0, we know that the

Quadratic formula is,t = -b±b2-4ac2a

On substituting a = 281, b = 50, c = –2679, we get:


t=-50±502-4281-26792281 =-50±2500+3011196562 =-50±3013696562 =-50±1736562 =-1786562,1686562 =-3.177,3

Since t is time, which cannot be negative, t = 3
Thus, they will be 52 km apart from each other after 3 hours and the time will be 7 am + 1 hr + 3 hr = 11 am.

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