Kindly answer this with full explanation.

#ThankYou.

Dear student,

As shown in the figure,

distance travelled between 4:00-4:20 pm($d_1$) = 100m
distance travelled between 4:20-4:30pm ($d_2$) = 50m
distance travelled between 4:30-5:00 pm($d_3$) = 450m
total time taken ($t$) = 1 h= $1\times 60\times 60 seconds= 3600 s$

Therefore, average speed = $\frac{Total dis\tan ce travelled}{Total time taken}$
                                            = $\frac{d_{1 }+ d_{2 }+d_3}{t}$
                                           = $\frac{100+50+450}{3600}$
                                          = $0.167 m/s$
Hence, option (D) is the correct answer.

Regards