Question

a person is approaching a plane mirror with speed 10 cm per sec if the initial distance between person and mirror is 2m then distance between person and his imagd after 2.5 sec will be

Answer 1

Answer:

1.5 m is the correct ans

Explanation:

initial distance between person and his image in plane mirror is 2m.

decrease in seperation after approach is = 2*v*t = 2*10*2.5 = 50 cm

so the seperation now between the person and the image will be == 2 - 0.5 == 1.5m

Answer 2

Answer:

The distance between the person and his image after 2.5 sec will be 3.5m.

Explanation:

The distance between the person and his image is double the distance between person and mirror. So the initial distance between the person and his image:

                                  $d=2\times 2=4m$   (1)

When a person is approaching a plane mirror with speed V his image also approaches a plane mirror with a speed V. So the approach speed between the person and his image is:

                     $V_a=V+V=2V$

                     $V_a=2\times 10\ cm/s=20\ cm/s=0.2m/s$   (2)

The distance between the person and his image after 2.5sec will be

                               $d'=d-V_a\times t$

                              $d'=4-0.2\times 2.5$

                             $d'=4-0.5=3.5m$     (3)

Hence, the distance between the person and his image after 2.5 sec will be 3.5m.

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