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(C) Let the speed of the engine be $v \ m/s$.The initial distance between the engine and the hill is $d = 0.9 \ km = 900 \ m$.In $t = 5 \ s$,the engin

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) Let the speed of the engine be $v \ m/s$.The initial distance between the engine and the hill is $d = 0.9 \ km = 900 \ m$.In $t = 5 \ s$,the engine travels a distance of $d_{engine} = v 	imes t = 5v \ m$.When the echo is heard,the engine has moved closer to the hill,so its new distance from the hill is $(900 - 5v) \ m$.The total distance traveled by the sound is the distance to the hill plus the distance from the hill to the engine's new position:$D_{sound} = 900 + (900 - 5v) = 1800 - 5v$.Given the speed of sound $V_s = 330 \ m/s$ and the time taken $t = 5 \ s$,the distance traveled by sound is also $D_{sound} = V_s 	imes t = 330 	imes 5 = 1650 \ m$.Equating the two expressions for distance:$1800 - 5v = 1650$$5v = 1800 - 1650$$5v = 150$$v = 30 \ m/s$.Thus,the speed of the engine is $30 \ m/s$.

An engine approaches a hill with a constant speed. When it is at a distance of $0.9 \ km$,it blows a whistle,and the echo is heard by the driver after $5 \ s$. If the speed of sound in air is $330 \ m/s$,the speed of the engine is .... $m/s$.

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