wach
  • wach
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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wach
  • wach
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?
kropot72
  • kropot72
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 \[v ^{2}=7.84\] \[v=\sqrt{7.84}\ m/s\]
wach
  • wach
Thank you SO much.

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kropot72
  • kropot72
You're welcome :)

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