Answer to Question #218345 in Quantum Mechanics for clay

Question

The speed of a train is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m. How much farther will the train travel before coming to rest, provided the acceleration remains constant? The acceleration is -9.81 m/s2

Answer 1

$v^2+u^2=2as \\ u= 15 \;m/s \\ v=7 \;m/s \\ a=90 \;m \\ 15^2 + 7^2 = 2a \times 90 \\ \frac{176}{180} = a \\ a=0.98 \;m/s^2$

Now the final speed $v_f=0$

$u=7\;m/s \\ v_f^2+u^2=2as \\ 0 -7^2 = 2 \times 0.98 s \\ s = \frac{49}{1.96} \\ s= 25 \;m$