Answer:
Δt=6.43h
Δt1 = 5.38h
Explanation:
We know that the train has to stop at the destination, so in the interval Δt2:
Vf = Vo + a2*Δt2 where Vo is the final velocity in the intervat Δt1.
Vf = a1*Δt1 + a2*Δt2 = 0 Solving for Δt2:
[tex]\Delta t2 = \frac{a1*\Delta t1}{a2}[/tex] Let this be eq1.
The total distance is 1.48km, and this is the sum of the displacements during Δt1 and Δt2:
[tex]\Delta X = \Delta X1 + \Delta X2 = \frac{a1*\Delta t1^2}{2} + (a1*\Delta t1)*\Delta t2+\frac{a2*\Delta t2^2}{2}[/tex]
Replacing our previous value for Δt2 in this equation we can solve for Δt1:
Δt1 = 19351.36s ≅ 5.38h Using this value in eq1, we get Δt2:
Δt2 = 3794.39s ≅ 1.054h Therefore, total time Δt= 23145.75s ≅ 6.43h