Question

A bike travels a distance of 200 km at a constant speed. If the speed of the bike is increased by 5 km. an hour, the journey would have taken 2 hours less. What is the speed of the bike?
(a) 30 km/hour
(b) 25 km/hour
(c) 20 km/hour
(d) 15 km/hour

Answer 1

Hint: To solve this question, we have to recall one of the kinematic equations of motion from physics. This equation is $v=\dfrac{s}{t}$, where v is the constant speed, s is the distance and t is the time required to cover this distance. We will use this equation to form a quadratic equation and then use one of the three available methods to solve the quadratic equation. This will give us the speed of the bike.

Complete step by step answer:
Now, according to the given information, s = 200 km. We assume that it is travelling at a constant speed v and time taken to cover 200 km at this speed is t.
\[\begin{align}
  & \Rightarrow v=\dfrac{200}{t} \\
 & \Rightarrow t=\dfrac{200}{v}......\left( 1 \right) \\
\end{align}\]
Now, it is given that if speed is increased by 5 km per hour, then time is reduced by 2 hours.
$\Rightarrow v+5=\dfrac{200}{t-2}$
We will cross multiply (t – 2) from RHS to LHS.
 $\Rightarrow \left( v+5 \right)\left( t-2 \right)=200......\left( 2 \right)$
We will substitute the value of t from (1) into (2).
$\Rightarrow \left( v+5 \right)\left( \dfrac{200}{v}-2 \right)=200$
Now, we will solve the parenthesis and form an equation.
$\begin{align}
  & \Rightarrow \left( v+5 \right)\left( \dfrac{200-2v}{v} \right)=200 \\
 & \Rightarrow \left( v+5 \right)\left( 200-2v \right)=200v \\
 & \Rightarrow 200v-2{{v}^{2}}+1000-10v=200v \\
 & \Rightarrow 2{{v}^{2}}+10v-1000=0 \\
 & \Rightarrow {{v}^{2}}+5v-500=0 \\
\end{align}$
Now, we will solve this quadratic equation by factorisation method.
We have to find factors such that their sum is 5 and product is -500.
The two factors will be 25 and -20.
$\begin{align}
  & \Rightarrow {{v}^{2}}+25v-20v-500=0 \\
 & \Rightarrow v\left( v+25 \right)-20\left( v+25 \right)=0 \\
 & \Rightarrow \left( v-20 \right)\left( v+25 \right)=0 \\
\end{align}$
So, either v – 20 = 0 or v + 25 = 0, this means =, either v = 20 or v = ─25. But speed cannot be negative, thus, v = 20.
Therefore, the speed of the bike is 20 km/hour.

Note: Students should be careful while forming the equation as that can be tricky. Quadratic equations can also be solved using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where a, b and c are obtained from the quadratic equation in standard form $a{{x}^{2}}+bx+c=0$.