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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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 .
❌Don't Spam ❌
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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
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
⇒ 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)
Verified answer
Answer:
[tex] \boxed{\mathfrak{Average \ speed \ (v_{avg} )= \dfrac{2v_{1}(v_{2} + v_{3}) }{2v_{1} + v_{2} + v_{3} }}} [/tex]
Explanation:
Let the total distance covered by man be 'd'
Velocity to cover first half distance is given as [tex] \rm v_1 [/tex]
Let time taken to cover first half distance be [tex] \rm t_1 [/tex]
[tex] \rm \implies t_1 = \dfrac{d}{2v_1} [/tex]
Velocity to cover second half distance is given as [tex] \rm v_2 [/tex] & [tex] \rm v_3 [/tex].
Time interval for [tex] \rm v_2 [/tex] & [tex] \rm v_3 [/tex] are equal let it be 't'
Let the distance travelled by [tex] \rm v_2 [/tex] be [tex] \rm d_2 [/tex] & [tex] \rm v_3 [/tex] be [tex] \rm d_3 [/tex]
So,
[tex] \rm \implies t = \dfrac{ d_2}{v_2} \: \: \: \: \: ...eq_1 \\ \rm \implies t = \dfrac{d_3}{v_ 3} \: \: \: \: \: ...eq_2[/tex]
[tex] \rm d_2 [/tex] & [tex] \rm d_3 [/tex] is equal to second half distance i.e.
[tex] \rm d_2 + d_3 = \dfrac{d}{2} \: \: \: \: \: ...eq_3 [/tex]
From [tex] \rm eq_1 [/tex] & [tex] \rm eq_2 [/tex] we can find [tex] \rm d_2 [/tex] & [tex] \rm d_3 [/tex] and substitute it in [tex] \rm eq_3 [/tex]
[tex] \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) } [/tex]
Total time = [tex] \rm t_1 + t + t [/tex]
[tex] \rm Average \: speed \: ( v_{avg})= \dfrac{Total \: distance \: travelled}{Total \: time} [/tex]
[tex] \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} }
[/tex]