A BUS TRAVELS THE FIRST ONE - THIRD DISTANCE AT A SPEED OF 10 KM H-1 , THE NEXT ONE - THIRD DISTANCE ATA SPEED OF 20 KM H-1 AND THE NEXT ONE - THIRD DISTANCE AT A SPEED OF 30 KM H-1. WHAT IS THE AVERAGE SPEED OF THE BUS?
Answers
Answer:
Step-by-step explanation:
Let total distance is dd kmkm.
Using s=uts=ut \Rightarrow t_1= \dfrac{\dfrac{d}{3}}{10}=\dfrac{d}{30}⇒t
1
=
10
3
d
=
30
d
hours,
\Rightarrow t_2= \dfrac{\dfrac{d}{3}}{20}=\dfrac{d}{60}⇒t
2
=
20
3
d
=
60
d
hours,
\Rightarrow t_3= \dfrac{\dfrac{d}{3}}{30}=\dfrac{d}{90}⇒t
3
=
30
3
d
=
90
d
hours,
\text{Average speed}= \dfrac{\text{Total distance}}{\text{Total Time}}Average speed=
Total Time
Total distance
Average speed= \dfrac {d}{\dfrac{d}{30} + \dfrac{d}{60} +\dfrac{d}{90}}=\dfrac{180}{11} {km}{h}= \dfrac{50}{11} ms^{-1}=
30
d
+
60
d
+
90
d
d
=
11
180
kmh=
11
50
ms
−1
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