Question

Class‐ 11 th
Subject ‐ Physics

8) A man covers half distance with
speed v1 and remaining half distance with speed V2 and V3 for equal time interval • Find his Average speed .

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Answer 1

Given :-

◉ A man covers half distance with speed v₁ and remaining half distance with speed v₂ and v₃ for equal interval of time.

To Find :-

◉ Average speed of the man

Solution :-

Let the total distance the man travelled be d.

Case 1

Given that half of the distance is covered by the man with a speed of v₁ also we assumed the total distance travelled to be d.

Hence, we have

• Speed = v₁
• Distance travelled = d / 2

We know,

⇒ Time = Distance / Speed

⇒ t₁ = (d/2) / v₁

⇒ t₁ = d / 2v₁ ...(1)

Case 2

Now the remaining distance is d/2

Also, It is given that the man travelled with speed v₂ and v₃ for equal time interval.

⇒ Distance travelled = Distance travelled with v₂ + Distance travelled with v₁

Since, the time taken by the man to travel the remaining distance with speed v₂ and v₃ is equal. so let the time be t

Also, We know

• Distance = Speed × Time

⇒ d / 2 = v₂t + v₃t

⇒ d / 2 = t(v₂ + v₃)

⇒ t = d / 2(v₂ + v₃) ...(2)

Now, For the whole journey,

⇒ Average speed = Total distance / Total time

⇒ Avg. speed = d / (t₁ + t + t)

⇒ Avg. speed = d / { (d / 2v₁) + d/2(v₂ + v₃) + d/2(v₂ + v₃)}

⇒ Avg. speed = d / { d / 2v₁ + 2d / 2(v₂ + v₃) }

⇒ Avg. speed = d / { d(v₂ + v₃) + 2dv₁ / 2v₁(v₂ + v₃) }

⇒ Avg. speed = d / { d( v₂ + v₃ + 2v₁) / 2v₁(v₂ + v₃) }

⇒ Avg. speed = 1 / { (2v₁ + v₂ + v₃) / 2v₁(v₂ + v₃) }

⇒ Avg. speed = 2v₁(v₂ + v₃) / (2v₁ + v₂ + v₃)

Answer 2

Answer:

$\boxed{\mathfrak{Average \ speed \ (v_{avg} )= \dfrac{2v_{1}(v_{2} + v_{3}) }{2v_{1} + v_{2} + v_{3} }}}$

Explanation:

Let the total distance covered by man be 'd'

Velocity to cover first half distance is given as $\rm v_1$

Let time taken to cover first half distance be $\rm t_1$

$\rm \implies t_1 = \dfrac{d}{2v_1}$

Velocity to cover second half distance is given as $\rm v_2$ & $\rm v_3$.

Time interval for $\rm v_2$ & $\rm v_3$ are equal let it be 't'

Let the distance travelled by $\rm v_2$ be $\rm d_2$ & $\rm v_3$ be $\rm d_3$

So,

$\rm \implies t = \dfrac{ d_2}{v_2} \: \: \: \: \: ...eq_1 \\ \rm \implies t = \dfrac{d_3}{v_ 3} \: \: \: \: \: ...eq_2$

$\rm d_2$ & $\rm d_3$ is equal to second half distance i.e.

$\rm d_2 + d_3 = \dfrac{d}{2} \: \: \: \: \: ...eq_3$

From $\rm eq_1$ & $\rm eq_2$ we can find $\rm d_2$ & $\rm d_3$ and substitute it in $\rm eq_3$

$\sf \implies tv_2 + tv_3 = \frac{d}{2} \\ \\ \sf \implies t(v_2 + v_3) = \frac{d}{2} \\ \\ \sf \implies t= \frac{d}{2(v_2 + v_3) }$

Total time = $\rm t_1 + t + t$

$\rm Average \: speed \: ( v_{avg})= \dfrac{Total \: distance \: travelled}{Total \: time}$

$\rm \implies v_{avg} = \frac{d}{t_1 + t + t} \\ \\ \rm \implies v_{avg} = \dfrac{d}{ \dfrac{d}{2v_{1}} + \dfrac{d}{2(v_{2} + v_{3}) } + \dfrac{d}{2(v_{2} + v_{3}) } } \\ \\ \rm \implies v_{avg} = \dfrac{ \cancel{d}}{ \dfrac{ \cancel{d}}{2v_{1}} + \dfrac{ \cancel{2d}}{ \cancel{2}(v_{2} + v_{3}) }} \\ \\ \rm \implies v_{avg} = \dfrac{1}{ \dfrac{1}{2v_{1}} + \dfrac{1}{(v_{2} + v_{3}) } } \\ \\ \rm \implies v_{avg} = \dfrac{1}{ \dfrac{(v_{2} + v_{3}) + 2v_{1}}{2v_{1}(v_{2} + v_{3})} } \\ \\ \rm \implies v_{avg} = \dfrac{2v_{1}(v_{2} + v_{3}) }{2v_{1} + v_{2} + v_{3} }</p><p>$