Two trucks

Two trucks left cities A and B against each other and met after an hour. The first car came to B 27 minutes later than the second car to A. Calculate the car speed if the distance between cities A and B is 90 km.

Final Answer:

v₁ =  40 km/h
v₂ =  50 km/h

Step-by-step explanation:

$$\begin{gathered} s=90\text{km} \\ {t}_1=1\text{h} \\ {t}_2=27\text{min}\to \text{h}=27:60\text{h}=0.45\text{h} \\ {s}_1+s_2=s \\ {v}_1\cdot t_1+v_2\cdot t_1=s \\ {v}_1\cdot (t+t_2)=s \\ {v}_2\cdot t=s \\ 1\cdot v_1+1\cdot v_2=90 \\ {v}_1\cdot (t+27/60)=90 \\ {v}_2\cdot t=90 \\ {v}_1+v_2=90 \\ {v}_1\cdot t+v_1\cdot 27/60=90 \\ t=90/v_2 \\ {v}_1+v_2=90 \\ {v}_1\cdot 90/v_2+v_1\cdot 27/60=90 \\ {v}_1\cdot 90+v_2\cdot v_1\cdot 27/60=90\cdot v_2 \\ (90-v_2)\cdot 90+v_2\cdot (90-v_2)\cdot 27/60=90\cdot v_2 \\ (90-x)\cdot 90\cdot 60+x\cdot (90-x)\cdot 27=60\cdot 90\cdot x \\ (90-x)\cdot 90\cdot 60+x\cdot (90-x)\cdot 27=60\cdot 90\cdot x \\ -27x^2-8370x+486000=0 \\ 27x^2+8370x-486000=0 \\ 27=3^3 \\ 8370=2\cdot 3^3\cdot 5\cdot 31 \\ 486000=2^4\cdot 3^5\cdot 5^3 \\ \text{GCD}(27,8370,486000)=3^3=27 \\ {x}^2+310x-18000=0 \\ a=1;b=310;c=-18000 \\ D=b^2-4ac=310^2-4\cdot 1\cdot (-18000)=168100 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-310\pm \sqrt{168100}}{2} \\ {x}_{1,2}=\frac{-310\pm 410}{2} \\ {x}_{1,2}=-155\pm 205 \\ {x}_1=50 \\ {x}_2=-360 \\ {v}_2=x_1=50 \\ {v}_1=90-v_2=90-50=40\text{km/h} \end{gathered}$$

Our quadratic equation calculator calculates it.

$v_2=50=50\text{km/h}$

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