Acceleration 7124

The train passes 700m, braking with an acceleration of -0.15 m/s². How long does it break, and what is the final speed of the train if the initial was 55 km/h?

Correct answer:

t =  69.5956 s
v₂ =  4.8384 m/s

Step-by-step explanation:

$$\begin{gathered} s=700\text{m} \\ a=0.15{\text{m/s}}^2 \\ {v}_1=55\text{km/h}\to \text{m/s}=55:3.6\text{m/s}=15.27778\text{m/s} \\ s=v_1\cdot t-\frac{1}{2}\cdot a\cdot t^2 \\ 700=15.277777777\cdot t-0.5\cdot 0.15\cdot t^2 \\ 700=15.277777777\cdot t-0.5\cdot 0.15\cdot t^2 \\ 0.075000000000045t^2-15.278t+700=0 \\ a=0.075;b=-15.278;c=700 \\ D=b^2-4ac=15.278^2-4\cdot 0.075\cdot 700=23.4104938033 \\ D>0 \\ {t}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{15.28\pm \sqrt{23.41}}{0.15} \\ {t}_{1,2}=101.851852\pm 32.256261 \\ {t}_1=134.108113119 \\ {t}_2=69.595590574 \\ t>0 \\ {v}_2>0 \\ t=t_2=69.5956=69.5956\text{s} \end{gathered}$$

Our quadratic equation calculator calculates it.

$v_2=v_1-a\cdot t=15.2778-0.15\cdot 69.5956=4.8384\text{m/s}$

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