Question

2) a ) A train passes a point, already traveling at 2.0 m/s and accelerating at 0.6 m/s / s . Write down the equation that describes the train’s position. A girl on her bicycle reaches that very point 4 seconds later and begins to ride at a constant velocity v in order to catch the train. Write down the equation that describes her position. b) Show that she will never catch the train if her velocity is 6 m/s . c) Show that she will actually catch the train if her velocity is 12 m/s . In this case calculate when and where she will first catch up with the train and the velocity of the train.

Answer 1

a] Equation of trains position with respect to the given point,

x = ut+ 0.5at^2

x₁  = 2t + 0.5*0.6t^2

Equation of the girl :

x = v*(t-4)

x₂ = 6*(t-4)

b] in order to catch the train, there should be a t>4s, for which x1 = x2

   2t + 0.5*0.6t^2 = 6*(t-4)

has no real solution, so she will not catch the train.

If her velocity is 12 m/s, her position equation will be : x2=12*(t-4)

Then 2t + 0.5*0.6t^2 = 12*(t-4)

She will catch the train at t= 5.81 s after the train passed the point.

At a distance d =   12*(5.81-4) = 21.8 m from the point