A particle goes from A to B with a speed of 40 km/h and B to C with a speed of 60 km/h . If AB = 6BC , the average speed in km/h between A and C is

Correct answer: 42 V _ av = distance travelled time taken ⇒ V _ av = AB + BC AB 4 0 + BC 6 0 ⇒ V _ av = 6BC + BC 6BC 4 0 + BC 6 0 V _ av = (7 × 2 4 0)/(3 6 + 4) ⇒ V _ av = (7 × 2 4 0)/(4 0) ⇒ V _ av = 4 2 km/h

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

A particle goes from A to B with a speed of 40 km/h and B to C with...

A particle goes from

\text{A}

\text{B}

with a speed of

\text{40 km/h}

and

\text{B}

\text{C}

with a speed of

\text{60 km/h}

. If

\text{AB} = 6\text{BC}

, the average speed in

\text{km/h}

between

\text{A}

and

\text{C}

Correct answer

Step-by-step solution

{\text{V}}_{\text{av}} = \dfrac{\text{distance travelled}}{\text{time taken}}

⇒ {\text{V}}_{\text{av}} = \dfrac{\text{AB} + \text{BC}}{\dfrac{\text{AB}}{4 0} + \dfrac{\text{BC}}{6 0}}

{⇒ \text{V}}_{\text{av}} = \dfrac{\text{6BC} + \text{BC}}{\dfrac{\text{6BC}}{4 0} + \dfrac{\text{BC}}{6 0}}

{\text{V}}_{\text{av}} = \dfrac{7 \times 2 4 0}{3 6 + 4} ⇒ {\text{V}}_{\text{av}} = \dfrac{7 \times 2 4 0}{4 0}

⇒ {\text{V}}_{\text{av}} = 4 2 \text{km/h}

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