Subway

A subway train accelerated uniformly for 26 seconds, reaching a speed of 72 km/h. It then travelled at this speed for 56 seconds, before decelerating uniformly to a stop in 16 seconds. What was the distance between the stations?

Final Answer:

s =  1.5 km

Step-by-step explanation:

$$\begin{gathered} s=s_1+s_2+s_3 \\ s=\frac{1}{2}{a}_1{t}_1^2+v\cdot t_2+\frac{1}{2}{a}_3{t}_3^2 \\ s=\frac{1}{2}\frac{v-0}{t_1}{t}_1^2+v\cdot t_2+\frac{1}{2}\frac{v-0}{t_3}{t}_3^2 \\ s=\frac{1}{2}v{t}_1+v{t}_2+\frac{1}{2}v{t}_3 \\ s=v(\frac{t_1}{2}+t_2+\frac{t_3}{2}) \\ s=\frac{72}{3600}(\frac{26}{2}+56+\frac{16}{2})=1.5\text{km} \end{gathered}$$

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