Question

12. A train starts from a station with a constant acceleration of at = 0.40 m/s2.

A passenger arrives at the track time t = 6.0s after the end of the train left the very same point.

What is the slowest constant speed at which she can run and catch the train? Sketch curves for the motion of the pasenger and the train as functions of time.

Answer 1

Answer:

If the passenger instantly gets to this minimal speed to catch the train, his speed would be 4.8m/s

If we consider our initial time when the passenger start running, we can describe the motion equations:

$X_{pas}=v_0t$

$X_{train}=\frac{1}{2} 0.4\frac{m}{s^2}(t+6s)^2$

The minimum passenger speed will be the one that makes these 2 curves intersect in a single point. In other words, the motion curve of the passenger must be the tangent curve of the train motion curve that crosses for the point (0,0).

The tangent curve of a function f(t) is defined:

$T(t)=f'(a)(t-a)+f(a)$ where t=a is the time at which both curves cross.

In this case:

$T(0)=X_{pas}(0)=0=f'(a)(-a)+f(a)$

$0=f'(a) \cdot(-a)+f(a)$ and $f(t)=X_{train}(t)$

$X_{train}(a)=\frac{1}{2} 0.4\frac{m}{s^2}(a+6s)^2\\X_{train}'(a)=\frac{1}{2} 0.4\frac{m}{s^2}(2a+12s)\\-\frac{1}{2} 0.4\frac{m}{s^2}(2a+12s)a+\frac{1}{2} 0.4\frac{m}{s^2}(a+6s)^2 \leftrightarrow a=6s$.

Therefore:

$X_{pas}=v_0t=f'(6s)(t-6s)+f(6s)=4.8\frac{m}{s}t$