Answer to Question #161789 in Mechanics | Relativity for Dr. Horus

Question

Two trucks A and B, each

traveling at 40 km/hr, move

towards each other. At the

instance when the two trucks are

at a distance 50 km from one

another, a car moving at 60 lm/hr

overtakes A. How far behind the

car is A at the point when the car

meets B?

Answer 1

Explanations & Calculations

• If we assume the car to be traveled some $\small x$ $\small km$ distance from the beginning of the $\small 50km$ until it meets the truck B, then the truck B would have traveled $\small (50-x)km$.
• Both incidents take place at the same period of time: $\small t(h)$
• And during that time, truck A would travel $\small s=ut\to s=40kmh^{-1}\times t=40t$
• Then the truck A would be $\small (x-40t) km$ distance behind the car.
• $\small t$ and $\small x$ should be found
• To calculate x,

$\qquad\qquad \begin{aligned} \small \frac{x}{60kmh^{-1}}&= \small \frac{50-x}{40kmh^{-1}}\\ \small x&= \small \bold{30\,km} \end{aligned}$

• To calculate time,

$\qquad\qquad \begin{aligned} \small t&= \small \frac{x}{60}\,\,\,\,or\,\,\,\,t=\frac{50-x}{40}\\ &= \small \bold{0.5h=30min} \end{aligned}$

• Then the gap is,

$\qquad\qquad \begin{aligned} &=\small 30km-40kmh^{-1}\times0.5h\\ &=\small \bold{10\,km} \end{aligned}$