On a 2 lane roads, A is travelling with speed of 36km/h. 2 cars B & C approach A in opposite directions with speed of 54km/h each. B is travelling in same direction as A while C is travelling opposite to direction of A. At a certain instant, when distance AB is equal to AC, both being 1km, B decides to overtake A before C does. What minimum acceleration of B is required to avoid an accident?
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Your answer is 1. Apply formula of relative velocity.
Hope it helps you!!!!
Hope it helps you!!!!
Answered by
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From the above diagram we can say that velocity of B w.r.t. A is 18km/hr while C w.r.t. A is 90km/hr.
At the given instant of equal distances time taken by C to overtake A is 1km/(90km/hr) =1/90 hr
So assuming 'a' as required acceleration for B, it will take and applying second equation of motion
1= 18*(1/90) + (1/2)(a)(1/90)^2
upon solving a=12960km.hr^(-2)
At the given instant of equal distances time taken by C to overtake A is 1km/(90km/hr) =1/90 hr
So assuming 'a' as required acceleration for B, it will take and applying second equation of motion
1= 18*(1/90) + (1/2)(a)(1/90)^2
upon solving a=12960km.hr^(-2)
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