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A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time, it has to increase its speed by 250 km/hr from its usual speed. Find the usual speed of the plane.

 

Answers (1)

Given  \rightarrow speed of plane normally, \vartheta = x \, km/hr
distance to be covered , d = 1500 km
time , t= \frac{d}{\vartheta }
t= \frac{1500}{x}\, hr
According to question
\frac{1500}{x}-\frac{1500}{x+250}= \frac{30}{60}\, hr
\Rightarrow 1500\left ( \frac{1}{x}-\frac{1}{x+250} \right )= \frac{1}{2}
\Rightarrow 1500\left ( \frac{x+250-x}{x\left ( x+250 \right )} \right )= \frac{1}{2}
\Rightarrow \left ( 1500\times 250 \right )\times 2= x\left ( x+250 \right )
\Rightarrow x^{2}+250x-75000= 0
\Rightarrow x^{2}+\left ( 1000x-750x \right )-75000= 0 (by factor method)
\Rightarrow x^{2}+ 1000x-750x-75000= 0
\Rightarrow x\left ( x+1000 \right )-750\left ( x+1000 \right )= 0
\Rightarrow \left ( x-750 \right )\left ( x+1000 \right )= 0
\Rightarrow x= +750 ; x= -1000
Since speed can not be negative
Hence speed of the plane  = 750 km/hr

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Safeer PP

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