A passenger is walking on an escalator at a speed of 6 km / hr relative to escalator. He starts from point P on the ground and reaches point Q on ground. The escalator is moving at 3 km / hr relative

Question

A passenger is walking on an escalator at a speed of $6 km / hr$ relative to escalator. He starts from point P on the ground and reaches point Q on ground. The escalator is moving at $3 km / hr$ relative to ground in the direction of point Q . The distance PQ is 120 m . The time taken by him to reach point Q is :

Options:

Answer:

B

Solution:

$v_{man / ground} = 6 + 3 = 9 km / hr$ .

$= 9 \times \frac{5}{18} = \frac{5}{2} m / s time = \frac{120}{5/2} = 48 sec$

TestHub · Accepted Answer

v man / ground ​ = 6 + 3 = 9 km / hr . ​ = 9 × 18 5 ​ = 2 5 ​ m / s ​ time = 5/2 120 ​ = 48 sec ​

Physics - Kinematics Question with Solution | TestHub

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