Question

In a flight of $6000km$ an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by $400$$km/hour$ and time increased by $30$ minutes. Find the original duration of flight.

Answer 1

Total distance given$=$$6000$km

Let${}^{\prime }{t}^{\prime }$ be the original duration of flight

Thus, the original speed $=$$\frac{6000}{t}$

The speed for the trip was reduced by $400$km/hr and time increased by 30 minutes

Thus,new speed$=$$\frac{6000}{t+\frac{1}{2}}$

Difference in speed$=$$400$km/hr

$\frac{6000}{t}$$-$$\frac{6000}{t+\frac{1}{2}}$$=$$400$

$\frac{6000(t+\frac{1}{2})}{t(t+\frac{1}{2})}$$-$$\frac{6000t}{t(t+\frac{1}{2})}$=$400$

$3000$$=$$400$$t$$[$$t$$+$$\frac{1}{2}$$]$

$400$$t^2$$+$$200$$t$$=$$3000$

$400$$t^2$$+$$200$$t$$-$$3000$$=$$0$

$2$$t^2$$+$$t$$-$$15$$=$$0$

$2$$t^2$$+$$6$$t$$-$$5$$t$$-$$15$$=$$0$

$2$$t$$($$t$$+$$3$$)$$-$$5$$($$t$$+$$3$$)$$=$$0$

$($$2$$t$$-$$5$$)$$($$t+$$3$$)$$=$$0$

$t$$=$$\frac{5}{2}$ or $t$$=$$-3$

Time cannot be negative

$t$$=$$\frac{5}{2}$ hour$=$$60\times \frac{5}{2}$$=$$150$min

original duration of flight $=$ $150$min