Question

On a track of lenght 100km, the train covers first 30km with a constant speed of 30kmh-1 .What should be the speed of the train for the next 70km so that the average speed of the train remains 50kmh-1 for the entire trip?​

Answer 1

Answer :-

Speed of the train for the next 70 km should be 70 km h⁻¹ .

Explanation :-

We have :-

→ Length of the track = 100 km

→ Speed during first 30 km = 30 km h⁻¹

→ Average speed = 50 km h⁻¹

______________________________

Let the required speed of the train be ' v₂ km h⁻¹ ' .

Time taken to cover 1st 30 km :-

= Distance/Speed

= 30/30

= 1 hour

Time taken to cover next 70 km :-

= Distance/Speed

= 70/v₂ hour

______________________________

Average speed = Total distance/Total time

Substituting values, we get :-

⇒ 50 = 100/[1 + 70/v₂]

⇒ 50 = 100/[(v₂ + 70)/v₂]

⇒ 50 = 100v₂/(v₂ + 70)

⇒ 50(v₂ + 70) = 100v₂

⇒ 50v₂ + 3500 = 100v₂

⇒ 3500 = 100v₂ - 50v₂

⇒ 3500 = 50v₂

⇒ v₂ = 3500/50

⇒ v₂ = 70 km h⁻¹

Answer 2

Given : The Length of track is 100 km , The trains covers first 30 km with a constant speed of 30 km/h & the average speed of the train remains 50km/h for entire trip .

Need To Find : The speed of the train for the next 70km .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━

❍ Let's Consider the speed for next 70 km be a km/hr⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding the Time taken ( t ) for both :

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\\qquad\maltese\: \bf Formula \:for \:time \::$

$\qquad \dag\:\:\bigg\lgroup \pmb{\frak{ \qquad Time \:=\:\dfrac{Distance}{Speed}\qquad}}\bigg\rgroup$

⠀⠀⠀⠀⠀➠ Time taken when the train covers first 30 km with a constant speed of 30 km/h .

$\qquad:\implies \sf Time \:=\:\dfrac{Distance}{Speed}$

$\qquad:\implies \sf Time \:=\:\dfrac{30}{30}$

$\qquad:\implies \sf Time \:=\:\cancel{\dfrac{30}{30}}$

$\qquad:\implies \sf Time \:=\:1 \:$

$\qquad \therefore \:\:\pmb{\underline{\purple{\frak{ \:Time_{(\:First\:30\:km\:)}\: \:=\:1 \:hr }}} }\:\:\bigstar$

⠀⠀⠀⠀⠀➠ Time taken when the train covers first 70 km with a constant speed of a km/h .

$\qquad:\implies \sf Time \:=\:\dfrac{Distance}{Speed}$

$\qquad:\implies \sf Time \:=\:\dfrac{70}{a}$

$\qquad \therefore \:\:\pmb{\underline{\purple{\frak{ \:Time_{(\:Next\:70\:km\:)}\: \:=\:\dfrac{70}{a} \:hr }}} }\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\\qquad\maltese\: \bf Formula \:for \:Average \:speed\::$

$\qquad \dag\:\:\bigg\lgroup \pmb{\frak{ \qquad Average \:Speed\:=\:\dfrac{Total \: Distance \:Travelled}{Total \:Time \:taken }\: \qquad}}\bigg\rgroup$

⠀⠀⠀⠀⠀Here , Total Distance Travelled is 100 km & Average speed is 50 km / h .

$\qquad \dashrightarrow \sf Average \:Speed\:=\:\dfrac{Total \: Distance \:Travelled}{Total \:Time \:taken }\: \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf Average \:Speed\:=\:\dfrac{Total \: Distance \:Travelled}{Total \:Time \:taken }\: \:$

$\qquad \dashrightarrow \sf 50\:=\:\dfrac{100}{1 + \bigg( \dfrac{70}{a}\:\bigg) }\: \:$

$\qquad \dashrightarrow \sf 50\:=\:\dfrac{100}{1 + \dfrac{70}{a}\: }\: \:$

$\qquad \dashrightarrow \sf 50\:=\:\dfrac{100}{ \dfrac{a + 70}{a}\: }\: \:$

$\qquad \dashrightarrow \sf 50\:=\:\dfrac{100a }{ a + 70\: }\: \:$

$\qquad \dashrightarrow \sf 50 ( a + 70 ) \:=\:100a \: \:$

$\qquad \dashrightarrow \sf 50a + 3500 \:=\:100a \: \:$

$\qquad \dashrightarrow \sf 3500 \:=\:100a - 50a \: \:$

$\qquad \dashrightarrow \sf 3500 \:=\:50a \: \:$

$\qquad \dashrightarrow \sf a\:=\:\dfrac{3500}{50} \: \:$

$\qquad \dashrightarrow \sf a\:=\:\cancel {\dfrac{3500}{50}} \: \:$

$\qquad \dashrightarrow \sf a\:=\:70 \: \:$

$\qquad \therefore \:\:\pmb{\underline{\purple{\frak{ \:a \:( \:or \: Speed \:for\:Next\:70\:km\:)\: \:=\:70\:km/hr }}} }\:\:\bigstar$

• Here a denotes speed of next 70 km which is 70 km/hr

$\qquad \therefore \:\:\underline {\sf Hence , \:The \: speed \: for \: next \: 70\:km \: is \:\bf 70 \: km/h \:}$