Question

A passenger , while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for the treatment. Due to this, the plane got delayed by half an hour. To reach the destination $1500km$ away in time. so that the passengers could catch the connecting flight, the speed of the plane was increased by $250km/hr$ than the usual speed. Find the usual speed of the plane What value is depicted in this question?

Answer 1

Let the usual speed of the plane is $=s$ km/hr

and the usual time it takes to reach the destination $=t$ hr

Distance , $d=1500$ km

Now, $speed=\frac{distance}{time}$

$s=\frac{1500}{t}$

$t=\frac{1500}{s}$ .......... (1)

The plane got delayed by half an hour and speed was increased by $250$ km/hr to reach the destination on time.

Now, speed $=s+250$ km/hr

and time $=t-\frac{1}{2}$ hr

Now, speed $\times$ time $=$ distance

$(s+250)(t-\frac{1}{2})=1500$

$(s+250)(\frac{1500}{s}-\frac{1}{2})=1500$ (from (1) $t=\frac{1500}{s}$ )

$(s+250)\left(\frac{3000-s}{2s}\right)=1500$

$3000s-s^2+250\times 3000-250s=3000s$

$-s^2+250\times 3000-250s=0$

$s^2+250s-250\times 3000=0$

$s^2+1000s-750s-250\times 3000=0$

$(s+1000)(s-750)=0$

$s=-1000$ or $s=750$

Speed cannot be negative.

So, usual speed $s=750$ km/hr

usual time $t=\frac{1500}{750}=2$ hr