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A train, a late passenger, and related velocities - Please help ASAP

A train pulls away from a station with a constant acceleration of 0.37 m/s2. A passenger arrives at a point next to the track 6.1 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 still catch the train?

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1 Answer

An interesting problem in that there is more than one velocity the passenger can take up to his v(max), but what is the minimum velocity necessary to catch the train with a head start?

s = displacement

s(train) = (1/2)a(train)t^2 + v (train)t + 13.77m,  (displacement at 6.1 sec)

s(person) = v(person)t

(1/2)a(train)t^2 + v(train)t + 13.77m = v(person)t

(1/2)a(train)t^2 -  (v(train)-v(person))t  + 13.77m = 0

solve for t

t  =  (v(person) -v (train)  ±  ((v train)-v(person))^2  -  4(.37m/sec)(1/2)(13.77m))^(1/2))/(.37m/sec)
 
Solve the discriminant for positive value.
 
V(train)- v(person) = ± 3.19 m/sec
 
v (person) = v(train) + 3.19 m/sec = 2.26 m/sec + 3.19m/sec = 5.45 m/sec
 
v(person)> 5.45m/sec

Comments

I think that the displacement of train after 6.1 seconds will be (0.5(.37)(6.1)^2 = 6.88385 m

I tried to solve this problem using Microsoft Excell. I found out that if the person travels at a constant speed of 5 m/s he/she will overtake the train after about 13 seconds. When I used the method of George up to the point where you solve the discrimant for positive values of V(perspon)I obtained a minimum value of 4.83 m/s. Note that I did not later add the velocity of the train, V(train) as was done by George after solving the discriminant for positive V(person). This result of 4.83 corresponds with the spreadsheet result of 5.0 where only intgral values for the velocity of the person was used. For the spreadsheet I used 1 m/s, 2 m/s, 3 m/s, 4 m/s, 5 m/s and 6 m/s for the velocity of the person. For the spreedsheet I also used the following columns (A to J): Time-(natural number)--A Time Squared---B Distance travelled by train--C = 0.5*.37*B Velocity of train------D = 0.37*A Distance trav by person(@ V=1)---E =(A-6.1)*1 DistanceTravel by Person(@ V=2)---F=(A-6.1)*2 And so on up to DistanceTravel by Person(@ v=6)---J=(A-6.1)*6 Later I check the spreedsheet for each velocity of the person to see at what time,if any, the distance travelled by the person is greater that the distance travelled by the train. This happens only when the velocity of the person is 5 m/s or greater.

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