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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, thedriver of bus B think to overtake the bus A beforethe bus C. To make it possible what should be theminimum acceleration by bus B? |
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Answer» ity of bus A = 10m/s• Velocity of bus B = 15m/s• Velocity of bus C = -15m/sHere 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/sRelative velocity of bus C wrt bus A is given by➝ V(CA) = V(C) - V(A)➝ V(CA) = -15 - 10 = -25m/sAt a certain instant, both buses B and C are at the same distance from bus A.AB = BC = 500mTime TAKEN by bus C to cover 500m to reach bus A is given by➝ t = BC/V(BC)➝ t = 500/25➝ t = 20sIn 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. Thenu = 5m/st = 20sS = 500m➝ S = UT + 1/2 at²➝ 500 = (5 × 20) + 1/2 a(20)²➝ 400 × 2 = 400a➝ a = 2 m/s² |
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