Question

Example 2. A bus 'A' is moving with velocity of 10 ms-1. Another bus 'B' is coming behind the bus A. The direction of motion of A and B are along
same direction. A bus C is coming opposite to A.Velocity of B and C each are 15 ms-1. When the distance AB and AC each are equal to 500 m, the
driver of bus B think to overtake the bus A before
the bus C. To make it possible what should be the
minimum acceleration by bus B?​

Answer 1

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• Velocity of bus A = 10m/s

• Velocity of bus B = 15m/s

• Velocity of bus C = -15m/s

Here negative sign shows opposite direction.

Relative velocity of bus B wrt bus A is given by

➝ V(BA) = V(B) - V(A)

➝ V(BA) = 15 - 10 = 5m/s

Relative velocity of bus C wrt bus A is given by

➝ V(CA) = V(C) - V(A)

➝ V(CA) = -15 - 10 = -25m/s

At a certain instant, both buses B and C are at the same distance from bus A.

• AB = BC = 500m

Time taken by bus C to cover 500m to reach bus A is given by

➝ t = BC/V(BC)

➝ t = 500/25

➝ t = 20s

In order to avoid an accident, the bus B accelerates such that it overtakes bus A in less than 20s.

Let the minimum required acceleration be a. Then

• u = 5m/s
• t = 20s
• S = 500m

➝ S = ut + 1/2 at²

➝ 500 = (5 × 20) + 1/2 a(20)²

➝ 400 × 2 = 400a

➝ a = 2 m/s²