Question

A plane took off 1 hour late from Vadodara. To reach in time at a distance of 1200 km, its speed is increased by 100 km/hr. Find its usual speed.

Answer 1

Hint: First we will formulate the given statement in the form of an algebraic expression. Then we will solve this expression using the quadratic formula, $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We will also use the distance-speed relationship to solve this problem. The distance-speed relationship is given as,
$\Rightarrow v=\dfrac{s}{t}$.
Here, $s$ is the distance travelled by the plane, $v$ is the speed of the plane and $t$ is the time taken to cover the distance $s$.

Complete step-by-step answer:

In this problem it is given that the distance to be travelled is $s$= 1200 km.
If $v$ is the usual speed, then we can represent it using the distance-speed relationship as,
$\begin{align}
  & \Rightarrow v=\dfrac{s}{t} \\
 & \Rightarrow v=\dfrac{1200}{t}......(i) \\
\end{align}$
With the usual speed, the time taken by the plane to reach Vadodara is given as,
$\Rightarrow t=\dfrac{1200}{v}.........(ii)$
Now, in the question it is given that the speed of the plane is increased by 100 km/hr. This can be represented as,
\[\Rightarrow {{t}_{1}}=\dfrac{1200}{v+100}........(iii)\]
In equation (iii), ${{t}_{1}}$ represents the new time when the speed increases by 100 km/hr.
It is also given that the plane took off 1 hour late from Vadodara. Now, we can represent this as,
$\Rightarrow t-1={{t}_{1}}......(iv)$
On substituting the values of $t$ and ${{t}_{1}}$ from equations (ii) and (iii) in equation (iv) we get,
$\Rightarrow \dfrac{1200}{v}-1=\dfrac{1200}{v+100}......(v)$
On simplifying equation (v) we get,
$\begin{align}
  & \Rightarrow \dfrac{1200}{v}-\dfrac{1200}{v+100}=1 \\
 & \Rightarrow \dfrac{1200\left( v+100 \right)-1200v}{v(v+100)}=1......(vi) \\
\end{align}$
On further simplifying equation (vi) by cross multiplication we get,
$\begin{align}
  & \Rightarrow 1200(v+100)-1200v=v(v+100) \\
 & \Rightarrow {1200v}+120000-{1200v}={{v}^{2}}+100v \\
 & \Rightarrow {{v}^{2}}+100v-120000=0........(vii) \\
\end{align}$
Equation (vii) can be solved using the quadratic formula,
$v=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.........(viii)$
Comparing equation (viii) with equation (vii) we know that,
$\begin{align}
  & \Rightarrow a=1 \\
 & \Rightarrow b=100 \\
 & \Rightarrow c=-120000 \\
\end{align}$
Substituting these values in equation (viii) we get,
$\begin{align}
  & \Rightarrow v=\dfrac{-100\pm \sqrt{{{100}^{2}}-\left( 4\times 1\times -120000 \right)}}{2\times 1} \\
 & \Rightarrow v=\dfrac{-100\pm \sqrt{10000+480000)}}{2\times 1} \\
 & \Rightarrow v=\dfrac{-100\pm \sqrt{490000}}{2} \\
 & \Rightarrow v=\dfrac{-100\pm 700}{2} \\
 & \Rightarrow v=\dfrac{-100+700}{2}=\dfrac{600}{2}=300.......(ix) \\
 & \Rightarrow v=\dfrac{-100-700}{2}=\dfrac{-800}{2}=-400.......(x) \\
\end{align}$
From equation (ix) and (x) we got the values of $v$ as, 300 and -400.
We cannot take the value -400 because speed cannot be a negative value.
Therefore, the usual speed of the plane is $v$= 300 km/hr.

Note: In this problem the main idea is representing the time in terms of speed of the plane. This is derived from the distance-speed relationship,
$\begin{align}
  & \Rightarrow v=\dfrac{s}{t} \\
 & \Rightarrow t=\dfrac{s}{v} \\
\end{align}$
Also make sure that all the units of time, speed and distance are uniform throughout solving the problem.