Answer:
(a) See graph.
(b) t₁ = 30 s, and t₂ = 60 s
(c) d = 900 m
Explanation:
Acceleration (a) is the change in velocity (v) over time (t). An object with constant acceleration will have a linear increase in velocity over time, where the slope of the line is the acceleration. The displacement of the object is equal to the area under the velocity-time curve.
Part A: Velocity-Time Graphs
Train A has a constant acceleration of a = 0.5 m/s² and an initial velocity of u = 0 m/s. Its velocity as a function of time is therefore:
v = u + at
v = (0 m/s) + (0.5 m/s²) t
v = 0.5t
Train B has a constant velocity of 15 m/s. Its velocity as a function of time is therefore:
v = 15
Part B: Equal Velocity and Displacement
When the velocities are equal, the time is:
0.5t₁ = 15
t₁ = 30 s
The displacement is the area under the curve. For train A, the area is a triangle, with width t₂ and height 0.5t₂. For train B, the area is a rectangle, with width t₂ and height 15. When the displacements are equal, the time is:
½ (t₂) (0.5t₂) = (t₂) (15)
0.25 t₂ = 15
t₂ = 60 s
Part C: Distance from X to Y
The distance between the two stations is equal to the displacement at t₂.
d = (60) (15)
d = 900 m
