# Pls solve the first one

Average speed=total distance travelled/total time taken.

Let the distance between A and B is‘d’.

From A to B:

Let time taken = t

_{1}

Then, ${t}_{1}=\frac{d}{60}$ (in hr)

From B to A:

Let time taken = t

_{2}

Then, ${t}_{2}=\frac{d}{80}$ (in hr)

Now, total time taken,

$t={t}_{1}+{t}_{2}=\frac{d}{60}+\frac{d}{80}=\frac{7d}{240}hr$

Total distance travelled =

*d + d*= 2

*d*

Then, Average speed is $=\frac{totaldis\mathrm{tan}ce}{totaltime}=\frac{2d}{{\displaystyle \raisebox{1ex}{$7d$}\!\left/ \!\raisebox{-1ex}{$240$}\right.}}=\frac{480}{7}=68.57km/hr$

$averagevelocity=\frac{totaldisplacement}{totaltime}=\frac{0}{totaltime}=0km/hr$

(displcament is zero as the initial point = final point)

hence,

Average speed=68.57 km/hr

Average velocity =0

Regards,

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