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MrRa [10]
2 years ago
3

An airplane travels 2100 km at a speed of 500 km/hr encounters a head wind that decreases its speed to 400km/h for the next thre

e hours and then travels the last 400km to complete the trip at an average speed for the entire trip of 440 km/h what was the speed of the plane for the last part of the trip
Mathematics
2 answers:
Ket [755]2 years ago
7 0
Need to work out the time and distance
it flew each segment.
2100/500=4.2 hrs
400 km for 3 hrs = 1200km
total distance = 2100 + 1200 + 400 = 3700km
av speed is 440km/h so total time travelled is 3700/440= 8.40909hrs so flew last distance in 8.4909-4.2-3=1.20909hrs
400km in 1.20909hrs
=330.827km/hr
which means I've read question wrong


zubka84 [21]2 years ago
4 0

Answer with explanation:

⇒Total Distance which needs to be Traveled =2100 Km

 V_{1}=500 \frac{Km}{hr},t_{1}=x hr\\\\V_{2}=400 \frac{Km}{hr},t_{2}=3 hr\\\\V=440 \frac{Km}{hr}\\\\S_{3}=400 Km\\\\2100=500 \times x+400 \times 3+400\\\\2100-1600=500 x\\\\500=500 x\\\\x=1 \text{hr}

Plane traveled ,at a speed of 500 km/hr , for 1 hour.

⇒Average speed for the entire trip=440 Km/hr

440=\frac{500+400 +y}{3}\\\\440 \times 3=900+y\\\\1320-900=y\\\\y=420 \frac{Km}{hr}

Speed of the plane for the last part of the trip=420 km/hr

     

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