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Two cars are traveling on a straight section of dual carriageway in parallel, opposite directions. The perpendicular distance between the cars is a constant $4$ meters. Initially, the horizontal distance between the cars is $1 km$ .

Given that each car is traveling at $20 {ms}^{- 1}$ , find the time taken until the actual distance between the cars is first equal to $500 m$ .

Give your answer correct to one decimal place.

Hint

After $t$ seconds have passed, notice that each car has traveled $20 t$ meters. Let the remaining horizontal distance be called $x$ , and the actual distance be $ℓ$ :

We want to find the $t$ such that $ℓ = 500$ .

Hint

Express $x$ in terms of $t$ . Notice that $x, ℓ$ and $4 m$ form a right angled triangle so that

ℓ^{2} = x^{2} + 4^{2}

Solution

After $t$ seconds have passed, notice that each car has traveled $20 t$ meters. Let the remaining horizontal distance be called $x$ , and the actual distance be $ℓ$ :

We want to find the $t$ such that $ℓ = 500$ . Notice that

x = 1000 - 40 t

By Pythagoras, we have

\begin{aligned} x^{2} & = ℓ^{2} - 16 \\ (1000 - 40 t)^{2} & = 500^{2} - 16 \\ 1000 - 40 t & = \pm \sqrt{249984} \\ 40 t & = 1000 \pm 24 \sqrt{434} \\ t & = 25 \pm \frac{3}{5} \sqrt{69} \end{aligned}

We want the first time $ℓ = 50$ , so we need the smaller of the two solution, which is $12.5 (1 d.p.)$