Question

A very long train is rolling at 4 m/s along a straight track. An observer is standing on the ground very dangerously close to the train as moves past notices a person standing on top of a car. When the person on the car is 300 m from the observer, the person begins running toward the observer at 6 m/s.
A. Please write down the velocity vector equation for the person with respect to the ground.
B. How much time does the person take to reach the observer?

Answer 1

Answer:

A. $\vec{r}=(6\frac{m}{s})t\ \ \hat{i}$

B.  t = 50 s

Explanation:

A. The vectorial equation of the person who is getting closer to the other person is:

$\vec{r}=\vec{v}t$

r: position vector

v: speed vector = 6m/s i  (if you consider the motion as a horizontal motion)

Then, you replace and obtain:

$\vec{r}=(6\frac{m}{s})t\ \ \hat{i}$

B. The time is:

$t=\frac{d}{v}$

d: distance to the observer = 300m

v: speed of the person on the car = 6.00 m/s

$t=\frac{300m}{6m/s}=50s$