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A-team-that-is-100-meters-long-is-moving-forward-in-a-straight-line-at-a-constant-speed-A-messenger-runs-at-a-constant-speed-from-the-rear-of-the-team-to-the-front-to-deliver-a-message-Then-with

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Question Number 227824 by mr W last updated on 20/Feb/26
A team that is 100 meters long is moving forward in a straight line at a constant speed. A messenger runs at a constant speed from the rear of the team to the front to deliver a message. Then without changing the speed he runs back to the rear of the team. By the time he returns to the rear the team has advanced 240 meters. How far has the messenger traveled?
$$\mathrm{A}\:\mathrm{team}\:\mathrm{that}\:\mathrm{is}\:\mathrm{100}\:\mathrm{meters}\:\mathrm{long}\:\mathrm{is} \\ $$$$\mathrm{moving}\:\mathrm{forward}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\: \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}.\:\mathrm{A}\:\mathrm{messenger}\: \\ $$$$\mathrm{runs}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}\:\mathrm{to}\:\mathrm{the}\:\mathrm{front}\:\mathrm{to}\: \\ $$$$\mathrm{deliver}\:\mathrm{a}\:\mathrm{message}.\:\mathrm{Then}\:\mathrm{without}\: \\ $$$$\mathrm{changing}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{he}\:\mathrm{runs}\:\mathrm{back}\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}.\:\mathrm{By}\:\mathrm{the}\:\mathrm{time}\:\mathrm{he}\: \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rear}\:\mathrm{the}\:\mathrm{team}\:\mathrm{has} \\ $$$$\mathrm{advanced}\:\mathrm{240}\:\mathrm{meters}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{has}\: \\ $$$$\mathrm{the}\:\mathrm{messenger}\:\mathrm{traveled}? \\ $$
Answered by Ghisom_ last updated on 21/Feb/26
t_1 =((100)/(v_m −v_t )) t_2 =((100)/(v_m +v_t )) t_1 +t_2 =((240)/v_t ) v_m ^2 −(5/6)v_m v_t −v_t ^2 =0 v_m =(3/2)v_t s_m =(t_1 +t_2 )v_m =360 meters
$${t}_{\mathrm{1}} =\frac{\mathrm{100}}{{v}_{{m}} −{v}_{{t}} } \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{100}}{{v}_{{m}} +{v}_{{t}} } \\ $$$${t}_{\mathrm{1}} +{t}_{\mathrm{2}} =\frac{\mathrm{240}}{{v}_{{t}} } \\ $$$${v}_{{m}} ^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{6}}{v}_{{m}} {v}_{{t}} −{v}_{{t}} ^{\mathrm{2}} =\mathrm{0} \\ $$$${v}_{{m}} =\frac{\mathrm{3}}{\mathrm{2}}{v}_{{t}} \\ $$$${s}_{{m}} =\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right){v}_{{m}} =\mathrm{360}\:\mathrm{meters} \\ $$
Commented by mr W last updated on 21/Feb/26
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Answered by mr W last updated on 21/Feb/26
Commented by mr W last updated on 21/Feb/26
((100×240)/(100+x))+100=240+((100+x)/2) x^2 +480x−10000=0 (x−20)(x+500)=0 x=20 l=240+100+20=360 m ✓
$$\frac{\mathrm{100}×\mathrm{240}}{\mathrm{100}+{x}}+\mathrm{100}=\mathrm{240}+\frac{\mathrm{100}+{x}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{480}{x}−\mathrm{10000}=\mathrm{0} \\ $$$$\left({x}−\mathrm{20}\right)\left({x}+\mathrm{500}\right)=\mathrm{0} \\ $$$${x}=\mathrm{20} \\ $$$${l}=\mathrm{240}+\mathrm{100}+\mathrm{20}=\mathrm{360}\:{m}\:\checkmark \\ $$

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