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?

Answer 1

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
$\Rightarrow$(5 + 5t)² + (16t)² = (52)²
$\Rightarrow$ 25 + 50t + 25t² + 256t² = 2704
$\Rightarrow$ 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

$$\begin{gathered} \mathrm{Quadratic}\mathrm{formula}\mathrm{is}, \\ t=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$

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


$$\begin{gathered} t=\frac{-50\pm \sqrt{{\left(50\right)}^2-4\left(281\right)\left(-2679\right)}}{2\left(281\right)} \\ =\frac{-50\pm \sqrt{2500+3011196}}{562} \\ =\frac{-50\pm \sqrt{3013696}}{562} \\ =\frac{-50\pm 1736}{562} \\ =-\frac{1786}{562},\frac{1686}{562} \\ =-3.177,3 \end{gathered}$$

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.