Question

A train pulls away from a station with a constant acceleration of 0.7 m/s
A passenger arrives at a position on the track 4 s after the end of the train left that position.
What is the slowest constant speed at which
she can run and catch the train?
Answer in units of m/s

Answer 1

Can anyone help with me with this? I think I'm supposed to find the position function and then derive for the velocity equations, setting equal .. But I'm not really sure how to attack the question. How would I go about this?

Answer 2

Let the distance travelled by the train (including the distance travelled before the woman starts running) and by the woman be x meters.
For the train
\[x=\frac{at ^{2}}{2}+\frac{at ^{2}}{2}=\frac{0.7\times4^{2}}{2}+\frac{0.7t ^{2}}{2}=5.6+\frac{0.7t ^{2}}{2}\ .....(1)\]
For the woman
x=vt ...........................(2)
Equating equations (1) and (2) gives:
\[vt=5.6+\frac{0.7t ^{2}}{2}\ ...............(3)\]
From equation (3) we can form the quadratic
\[0.7t ^{2}-2vt+11.2=0\]
For the solution of the quadratic to be real, the value of the disciminant must be zero or positive. A zero value will give the slowest value of v.
\[(-2v)^{2}-(4\times0.7\times11.2)=0\]
from which we get
\[v ^{2}=7.84\]
\[v=\sqrt{7.84}\ m/s\]

Answer 3

Thank you SO much.

Answer 4

You're welcome :)