# Subway

Subway train went between two stations that gradually accelerated for 26 seconds and reached a speed of 72 km/h. At this rate, went 56 seconds. Then 16 seconds slowed to a stop. What was the distance between the stations?

Correct result:

s =  1.5 km

#### Solution:

$\ \\ s = s_1+s_2 + s_3 \ \\ s = \dfrac{1}{2} a_1 t_1^2 + v\cdot t_2 + \dfrac{1}{2} a_3 t_3^2 \ \\ s = \dfrac{1}{2} \dfrac{v-0}{t_1} t_1^2 + v\cdot t_2 + \dfrac{1}{2} \dfrac{v-0}{t_3} t_3^2 \ \\ s = \dfrac{1}{2} v t_1 + v t_2 + \dfrac{1}{2} v t_3 \ \\ s = v (\dfrac{t_1}{2} + t_2 + \dfrac{t_3}{2} ) \ \\ s = \dfrac{ 72}{3600}( \dfrac{ 26}{2} + 56 + \dfrac{ 16}{2} ) = 1.5 \ \text{km}$ We would be very happy if you find an error in the example, spelling mistakes, or inaccuracies, and please send it to us. We thank you! Tips to related online calculators
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