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Two trains (A and B) are travelling on two parallel rails to the same direction. Train A
which was at rest starts to travel with an acceleration of 0.5ms² from the station X. Train
B passes A with a velocity of 15 ms when A starts to travel.
(a) Draw velocity-time graphs using the same axis to the motion of the trains.
(b) If both the trains are having same velocity at t₁ and displacement at t2, find t; and
12.
(c) If both the trains meet at the station Y, what is the distance from X to Y?


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Sagot :

MathPhysidn

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

View image MathPhys
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