Question

Two motorcyclists \(A\) and \(B\) leave a place at 12 noon. \(A\) travels north at \(60\) km/hr and \(B\) travels east at \(80\) km/hr. At 2 PM, they are separating at the rate

• A. \(50\) km/hr
• B. \(100\) km/hr
• C. \(75\) km/hr
• D. \(25\) km/hr

Hint:

When two objects move at right angles with speeds \(u\) and \(v\), \[ \text{Rate of separation} = \sqrt{u^2+v^2} \] if both start simultaneously from the same point. Here, \[ \sqrt{60^2+80^2}=100 \]

Answer 1

The Correct Option is B

Concept: If two objects move along perpendicular directions, then the distance between them is \[\begin{aligned} s=\sqrt{x^2+y^2} \end{aligned}\] Differentiating with respect to time, \[\begin{aligned} \frac{ds}{dt} = \frac{x\frac{dx}{dt}+y\frac{dy}{dt}} {\sqrt{x^2+y^2}} \end{aligned}\]

Step 1: Find the distances travelled by \(A\) and \(B\) in 2 hours. \[\begin{aligned} x=80\times2=160\text{ km} \end{aligned}\] \[\begin{aligned} y=60\times2=120\text{ km} \end{aligned}\]

Step 2: Calculate the distance between them at 2 PM. \[\begin{aligned} s &=\sqrt{160^2+120^2}\\ &=\sqrt{25600+14400}\\ &=\sqrt{40000}\\ &=200 \end{aligned}\]

Step 3: Find the rate of separation. \[\begin{aligned} \frac{ds}{dt} &= \frac{160(80)+120(60)}{200}\\ &= \frac{12800+7200}{200}\\ &= \frac{20000}{200}\\ &=100 \end{aligned}\] \[\begin{aligned} \boxed{100\text{ km/hr}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.