Question

As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0 m/s to catch the bus, but the door closes when she's still 6.0 m behind the door, and the bus leaves the stop at a constant acceleration of 2.0 m/s2 . She has missed her bus, but as a physics exercise she keeps running at 6.0 m/s until she draws even with the bus door. She keeps running at a constant 6.0 m/s after drawing even with the bus door and pulls ahead for a while, but the accelerating bus soon overtakes her. By what maximum distance does she get ahead of the door?

Answer 1

velocity of the physics instructor with respect to bus

$v = 6 m/s$

acceleration of the bus is given as

$a = 2 m/s^2$

acceleration of instructor with respect to bus is given as

$a = -2 m/s^2$

now the maximum distance that instructor will move with respect to bus is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 6^2 = 2(-2)(d)$

$-36 = - 4 d$

$d = 9 m$

so the position of the instructor with respect to door is exceed by

$\delta x = 9 - 6 = 3 m$

so it will be moved maximum by 3 m distance