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


 s=s1+s2+s3 s=12a1t12+vt2+12a3t32 s=12v0t1t12+vt2+12v0t3t32 s=12vt1+vt2+12vt3 s=v(t12+t2+t32) s=723600(262+56+162)=1.5 km \ \\ 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}

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