t ro E tryrfqlo20 trioq .nim 0.8 L loot qiu 9dT .m 0.Et Yd b9269Tni 1E3 19d no 1
ID: 1533704 • Letter: T
Question
t ro E tryrfqlo20 trioq .nim 0.8 L loot qiu 9dT .m 0.Et Yd b9269Tni 1E3 19d no 10srnobo srt, ,ut, baton brs ornorl Tad moil Vialovinu srt, ot ovob nabuta A tbaoqa apST9VE 19rl aGw 16/fW (6) ' lov sesnv6 1arl esw s tw taco To thuoa oo.es nothaib 6 n mi t 01 ai Vinev inu 9rt, ot smsod d mort 93nsteib sni tiplatte adiu nothaib 6 ni mA E.01 2i Y1ianovinu 9rt1 of omorlisd mot sonsteib snil-Meiste sti 11 (d) (3 to 2 oes) rlma Sqht stitn9 oril 10t Y1bolav bn6 bsoqa 90619v6 10d 019w ,sdw J1ol orla 1stls nim OC rlrteq ormea orvd smorl bonudst oria tr (3) booga oQE19V6 rlVmA YIbolov op619V6 rNmAExplanation / Answer
Distance travelled = 13 km
Displacement = 10.3 km
Time taken for travel = 18 min = 18/60 hr = 0.3 hr
A) avg speed = distance/time = 13/0.3 = 43.33 km/hr
B) avg velocity = displacement / time = 10.3/0.3 = 34.33 km/hr
C) total time from start of his journey back to his home = 7 hr 30 min = 7.5 hr
Then average speed = 13×2 / 7.5 = 3.466 m/s (forward and backward journey distance = 13×2 )
Average velocity = 10.3 ×2/7.5 =2.7466m/s ( forward and return journey total displacement = 10.3×2)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.