Question

An Air India flight from Mumbai to Colombo was delayed by 60 minutes due to the emergency landing of the plane in Vizag as a passenger in the flight suddenly had a cardiac arrest. Now to reach the destination of 1800 km from Vizag to Colombo in time, so that passengers could catch their connecting flights, the speed of the plane was increased by 300 km/h than the usual speed.

Based on the above information, answer the following questions:

1. Taking the usual speed of plane as x km/h, form a quadratic equation for the situation described above. 
2. Find the nature of the roots of the quadratic equation. 
3. Find the usual speed of the plane.

Answer 1

(i) Let the usual speed of the plane be x km/hr.

Distance from Vizag to Colombo = 1800 km

Time taken at usual speed = `"Distance"/"Speed"`= 1800hr.

Increased speed = x + 300

Time taken at increased speed = `"Distance"/"Speed"=1800/(x+300)`

Since flight delayed by 60 minutes = 1 hour,

`1800/x - 1800/(x+300) = 1`

⇒ `(1800 (x + 300) - 1800x)/(x(x + 300)) = 1`

⇒ `(1800x + 540000 - 1800x)/(x(x + 300)) = 1`

⇒ 1800x + 540000 −1800x = x(x + 300)

⇒ 540000 = x² + 300x

⇒ x² + 300x − 540000 = 0

Hence, the required equation is x² + 300x − 540000 = 0.

(ii) Comparing x² + 300x − 540000 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 300 and c = −540000.

We know that,

D = b² − 4ac

= (300)² − 4(1)(−540000)

= 90000 + 2160000

= 2250000

∴ D > 0

Hence, roots are real and unequal.

(iii) Comparing x² + 300x − 540000 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 300 and c = −540000.

We know that,

`x = (-b +- sqrtD)/(2a)`

Substituting values of a, b and c in above equation we get,

`x = ((-300) +- sqrt2250000)/(2(1))`

= `(-300 +- 1500)/2`

= `(-300 + 1500)/2 or (-300 - 1500)/2`

= `1200/2 or (-1800)/2`

= 600 or −900.

Speed cannot be negative,

∴ x = 600 km/h

Hence, usual speed of the plane = 600 km/h.