4. While boarding an aeroplane, a passenger got hurt. The pilot showing
promptness and concern, made arrangements to hospitalise the injured
and so the plane started late by 30 minutes. To reach the destination,
1500 km away, in time, the pilot increased the speed by 100 km/hour.
Find the original speed of the plane.
Do you appreciate the values shown by the pilot, namely promptness
in providing help to the injured and his efforts to reach in time?
PLEASE ANSWER PROPERLY WITH STEPS
I WILL MARK YOU AS BRAINLIEST!
Answers
Explanation:
The distance travelled =1500 km
The initial speed of aeroplane=vkm/hr
New speed =v+100km/hr
The initial time taken=
v
1500
hr
The final time taken=
v+100
1500
hr
As the plane started
2
1
hr late.
v
1500
=
v+100
1500
+
2
1
Solving the above quadratic equation.
v=−600,v=500
Hence the speed of the plane = 500km/hr
Answer:
500 kmph
Explanation:
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Answered
while boarding an aeroplane a passenger got hurt .the pilot showing promptness and concern made arrangements to hospitalise the injured and so the plane started late by 30 minutes.to reach the destination,1500 km away the pilot increased the speed by 100 km/h.find the original speed
Asked by rosetta | 9th Mar, 2013, 09:05: PM
Expert Answer:
1.
We are given that the distance = 1500 km
Let the time taken by the aeroplane at its usual speed be T hours and its usual speed be S kmph
Case 1(usual speed):
T = D/S = 1500/S ....(1)
Case 2(speed was increased):
Speed = (S+100) kmph
As the aeroplane left half an hour late and still reached the destination in time, time taken = (T - 1/2) hours
Therefore we have:
T - 1/2 = D/(S+100) = 1500/(S+100)
From (1):
1500/S - 1/2 = 1500/(S+100)
⇒1500/S - 1500/(S+100) = 1/2
⇒1/S - 1/(S+100) = 1/3000
⇒(S+100-S)3000 = S(S+100)
⇒300000 = S^2 + 100S
⇒S^2 + 100S - 300000 = 0
⇒S^2+600S-500S-300000 = 0
⇒S(S+600)-500(S-600) = 0
⇒(S-500)(S+600) = 0
Therefore, either S = 500 or S = -600. Since speed of the aeroplane cannot be negative, we reject S = -600. Therefore, the usual speed of the aeroplane is 500 kmph.