Two cars X and Y are positioned as shown at time t = 0. They are travelling in the same direction.

Q#27 (Past Exam Paper – June 2016 Paper 11 Q8)

Two cars X and Y are positioned as shown at time = 0.

They are travelling in the same direction.

X is 50 m behind Y and has a constant velocity of 30 m s-1. Y has a constant velocity of 20 m s-1.

What is the value of when X is level with Y?
1.0 s                         1.7 s                         2.5 s                         5.0 s



Solution:
Answer: D.


Both cars X and Y are travelling at the constant speed. Since car X is travelling faster, it would, at one point, be level with car Y.

Let the time at which the cars are level be t.

Speed = Distance / time

Distance = Speed × Time

Distance travelled by car Y in this time t = 20t

Initially, car Y was 50 m in front of car X.

Position of car Y at this time t = 50 + 20t


The cars would be level at the same time (t).

Distance travelled by car X = 30t


When the cars at level, they would be at the same position.

Position of car X = Position of car Y

30t = 50 + 20t

10t = 50

Time t = 50 / 10 = 5.0 s

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