Question #2801a

Question

125 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-27T14:32:23

Let the speed of running of the man be #v# # m"/"s# and he just catches the bus after #t# s of his start.

So the distance covered by the man is #vt# m.

Starting from rest with uniform acceleration #2ms^-2# the bus covers distance #s=0*t+1/2*2t^2=t^2#

By the problem

#vt-t^2=6#

#=>t^2-vt+6=0#

For real values of #t # of these quadratic equation ,the discriminant #>=0#

#v^2-4*1*6>=0#

#=>v>=2sqrt6# s

So minimum speed #v_"min"=2sqrt6# #m"/"s#