Two straight roads cross at right angles at a point O. At a certain instant, car A is 6 km north of O and moving north at 50 km/h, while car B is 8 km east of O and moving east at 30 km/h. The rate at which the distance between the two cars is increasing at that instant is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 54 km/h
Let y be the distance of A from O, x the distance of B from O, and D the distance between the cars. The roads are at right angles, so by Pythagoras
\( D^{2} = x^{2} + y^{2} \)
At that instant \(D = \sqrt{64 + 36} = 10\) km. Differentiating with respect to t:
\( \displaystyle 2D\frac{dD}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \)
With x = 8, \(\frac{dx}{dt} = 30\), y = 6, \(\frac{dy}{dt} = 50\):
\( \displaystyle 10\frac{dD}{dt} = 8 \times 30 + 6 \times 50 = 540 \)
\( \displaystyle \frac{dD}{dt} = 54 \) km/h
Why the other options are incorrect
Option (1) 80 km/h — Incorrect. This is 50 + 30; speeds add like this only when the cars move along the same line in opposite directions.
Option (2) 58 km/h — Incorrect. This is \((6 \times 30 + 8 \times 50) \div 10\): each speed is paired with the other car’s distance.
Option (3) 40 km/h — Incorrect. The equation \(10\frac{dD}{dt} = 540\) gives 54 km/h, not 40 km/h.