Coming in hot, coming out cold?

The amount of time taken by the train to cross the silo is calculated in this problem . The expression for velocity at any time $t$ is also calculated in the problem .

$v = \frac{m_0 v_0 }{m_0 + \alpha _m t}$

Using the formula for velocity and substituting the values gives $v \approx \boxed{37.671 \text{m/s}} ~~~~ _\square$

Momentum conservation gives us (for the component of momentum parallel to the velocity of the train) $p(0)=p_0=m_0v_0=m(t)v(t)=p(t).$ That means $v(t) = \frac{m_0v_0}{m(t)}= \frac{m_0v_0}{m_0 + \alpha_m t}.$ The second equality comes from the fact that the train gains mass in time as $m(t)=m_0+\alpha_m t$ .

To get how far the train has traveled at time $T$ we integrate: $s(T) = \int_0^Tv(t)dt = v_0 \int_0^T\frac{m_0}{m_0 + \alpha_m t}dt = ... = \ln \left(\frac{m_0 + \alpha_m T}{m_0}\right)$ Now we solve the above equation for $T=t_l$ where $l=s(t_l)$ . That gives us $t_l = \frac{m_0}{\alpha_m}\left(e^{\frac{\alpha_m l}{v_0 m_0}}-1\right)$

and by computing $v(t_l)$ we get: $v(t_l) = v_0 e^{-\frac{\alpha_m l}{v_0 m_0}} = 45 \times e^{\frac{20 \times 200}{45 \times 500}} \mathrm{m/s} \approx 37.6708 \mathrm{m/s} .$