Two places

The distance to the places is 60 km. From place A a pedestrian came out at a speed of 4 km/h and at the same time a car drove against him from place B. What was the speed of the car if the pedestrian met him in 90 minutes?

Correct result:

v =  36 km/h

Solution:

$$\begin{gathered} s=60\text{ km } \\ {v}_1=4\text{ km/h } \\ t=90\text{ min}\to \text{ h}=90\mathrm{/}60\text{ h}=1.5\text{ h } \\ s=v_1\cdot t+v\cdot t \\ 60=4\cdot 1.5+v\cdot 1.5 \\ 1.5v=54 \\ v=36=36\text{ km/h} \end{gathered}$$

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