Question

In a flight for 3000 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 100 km/h and consequently the time of flight increased by 1 h. Find the original duration of the flight.

Answer 1

Hint: Assume that the fixed distance travelled by the flight is d km. Assume that the original speed of the flight is v km/h and time taken to cover the distance at original speed is t hours. Now, apply the speed – time formula given as $d=v\times t$ to form two expressions in v and t. Substitute the value of v in terms of t and solve the quadratic equation in t to get the answer.

Complete step by step answer:
Let us assume that the distance travelled by the flight is d km and the original speed of the flight is v km/h and time taken to cover the distance at original speed is t hours. The provided distance of the trip is 3000 km, so we have d = 3000 km.
(1) Assuming that the flight runs by its original speed v km/h and time t hours, so applying the speed – time formula given as $d=v\times t$ we get,
$\Rightarrow v\times t=3000.........\left( i \right)$
(2) In this case we are considering that the speed of the flight gets reduced by 100 km/h and the time of flight increased by 1 h, so we have,
New speed of the flight = (v – 100) km/h
New time taken = (t + 1) hour
Again applying the speed – time formula for the new speed and time of flight we get,
$\begin{align}
  & \Rightarrow \left( v-100 \right)\times \left( t+1 \right)=3000 \\
 & \Rightarrow vt+v-100t-100=3000........\left( ii \right) \\
\end{align}$
Substituting the value of v from equation (i) in equation (ii) we get,
$\begin{align}
  & \Rightarrow \dfrac{3000t}{t}+\dfrac{3000}{t}-100t-100=3000 \\
 & \Rightarrow 3000+\dfrac{3000}{t}-100t-100=3000 \\
 & \Rightarrow \dfrac{30}{t}-t-1=0 \\
\end{align}$
Multiplying both the sides with t and forming a quadratic equation in t we get,
$\begin{align}
  & \Rightarrow 30-{{t}^{2}}-t=0 \\
 & \Rightarrow {{t}^{2}}+t-30=0 \\
\end{align}$
Using the middle term split method we get,
$\begin{align}
  & \Rightarrow {{t}^{2}}+6t-5t-30=0 \\
 & \Rightarrow \left( t+6 \right)\left( t-5 \right)=0 \\
\end{align}$
Substituting each term equal to 0 we get,
$\Rightarrow t=-6$ or $t=5$
Since time of the flight cannot be negative, therefore neglecting $t=-6$ we get $t=5$.
Hence, the original duration of the flight is 5 hours.

Note: Note that it is necessary to convert the obtained equations into a quadratic equation because the two equations involve multiplication of variables which can only be solved by the substitution of one variable in terms of the other. You can use the discriminant formula to get the answer but do not forget to reject the negative values because distance, speed and time are never negative.