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eltavo809
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Homework Statement
A train pulls away from a station with a constant
acceleration of 0.40 m/s2. A passenger arrives at the
track 6.0 s after the end of the train has passed the
very same point. What is the slowest constant speed at
which she can run and catch the train?
Homework Equations
X = X0 + V0 * t + 0.5 * a * t2
V = V0 + a * t
The Attempt at a Solution
The passenger arrieves 6 sec. latter, so: tp = tt - 6sec
Passenger
X = V0 * ( t - 6 sec)
Train
X = 0.5 * 0.4 m/s2 * t2
Both, train and passenger, have to be at the same place at the same time in order to achieve an encounter, so:
xtrain = x passenger
V0 * ( t - 6 sec) = 0.2 m/s2 * t2So here is the problem. I know I have to find out V0 , but I have 3 variables and 2 equations. Any ideas? This is supose to be easy... :(
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