A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger to the emergency clinic at the airport for 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 $250 km/hr$ than the usual speed. Find the usual speed of the plane. What value is depicted in this question?

Hint: The speed is given by the formula, $s = \dfrac{d}{t}$, s is the speed, d is the distance and t is the time traveled. In the above question, distance is constant. i.e., $1500km.$ The time varies by half an hour.

Complete step-by-step answer:
Let the usual speed of the plane by $x km/hr$
and usual time it takes to reach the destination = $t$ $ hr$
Distance(d)= $1500 km$
Then, according to the question, time taken by the plane to cover $1500km$ by the usual speed , $t = \dfrac{{1500}}{x}$..........(1)
Again, time taken by the plane to cover $1500km$ when speed is increased by $250 km/hr$,
$t - \dfrac{{1}}{{2}} = \dfrac{{1500}}{{x + 250}}$.............(2)

According to the question, the difference in the above two cases is half an hour.
Substitute the $t$ value from equation (1) to equation (2), we get
$
   \Rightarrow \dfrac{{1500}}{x} - \dfrac{{1500}}{{x + 250}} = \dfrac{1}{2} \\
   \Rightarrow \dfrac{{1500(x + 250) - 1500(x)}}{{(x)(x + 250)}} = \dfrac{1}{2} \\
   \Rightarrow \dfrac{{1500 \times 250}}{{(x)(x + 250)}} = \dfrac{1}{2} \\
   \Rightarrow 75000 = {x^2} + 250x \\
   \Rightarrow {x^2} + 250x - 75000 = 0 \\
   \Rightarrow x = 750, - 1000 \\
$
Since speed cannot be negative.
 Hence, the usual speed of the plane comes out to be $750 km/hr$

Note: Important aspect to notice is that we neglected the value -1000 as speed because speed cannot be negative as it is a scalar quantity that only has magnitude. On the other hand, Velocity is a vector quantity that has both magnitude and direction.