Question

A minibus takes 6 hours less to cover \(1680 \mathrm{~km}\) distance, if its speed is increased by \(14 \mathrm{~km} / \mathrm{h} ?\) What is the usual time taken by minibus? (a) \(15 \mathrm{~h}\) (b) \(24 \mathrm{~h}\) (c) \(25 \mathrm{~h}\) (d) \(30 \mathrm{~h}\)

Short answer

Answer: The usual time taken by the minibus to cover 1680 km is 24 hours.

Step-by-step solution

Define the variables needed

Let's denote the usual speed of the minibus as \(s\) km/h, and the time it takes to cover the distance at its usual speed as \(t\) hours.

Write the equation for the initial distance

Using the distance, speed, and time relationship, we have: Distance = Speed × Time \(1680 = s \times t\)

Write the equation for increased speed and decreased time

When the speed is increased by 14 km/h, it takes 6 hours less. So, the new speed is \((s + 14)\) km/h, and the new time is \((t - 6)\) hours: \(1680 = (s + 14) \times (t - 6)\)

Solve the system of equations

Now, we have a system of two equations: 1) \(1680 = s \times t\) 2) \(1680 = (s + 14) \times (t - 6)\) We can solve this system of equations using substitution or elimination. From equation (1), we can get: \(t = \frac{1680}{s}\) Substitute this value of \(t\) into equation (2): \(1680 = (s + 14) \times (\frac{1680}{s} - 6)\) Now, solve for \(s\): \(1680s = (s + 14)(1680 - 6s)\) Expand the brackets: \(1680s = 1680s - 6s^2 + 23520 - 84s\) Now, simplify and move everything to the left side: \(6s^2 + 84s - 23520 = 0\)

Solve the quadratic equation

To solve the quadratic equation, \(6s^2 + 84s - 23520 = 0\) Divide the equation by 6: \(s^2 + 14s - 3920=0\) Now, factorize the equation or use the quadratic formula to solve for \(s\): \((s+56)(s-70)=0\) s = -56 or s = 70 The speed cannot be negative, so the usual speed of the minibus is 70 km/h.

Find the value of t

Substitute the value of \(s\) into the equation for \(t\): \(t = \frac{1680}{s} = \frac{1680}{70}\) \(t = 24\) So the usual time taken by the minibus is 24 hours, which corresponds to option (b).

Key concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that involve terms up to the second degree, meaning they include a variable raised to the power of two. They take the general form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients, and \(a \eq 0\). Solving these equations involves finding the values of the variable \(x\) that satisfy the equation.

There are several methods to solve these equations, including factoring, completing the square, and using the quadratic formula.

• Factoring: This method involves writing the quadratic as a product of two binomials, if possible.
• Completing the Square: This technique converts the equation into a perfect square trinomial, making it easier to solve.
• Quadratic Formula: This is a universal method that uses the formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\) to find the solutions.

In our exercise, since we have a quadratic equation, we can solve it by factoring or using the quadratic formula. Factoring leads us to find that \(s + 56)(s - 70) = 0\), guiding us to the conclusion that the speed of the minibus cannot be negative, and hence, \(s = 70 \text{{km/h}}\) is the practical solution.

System of Equations

A system of equations is a set of two or more equations that have the same variables and are solved simultaneously. The goal is to find a common solution that satisfies all equations within the system.

• Substitution Method: This involves solving one equation for one variable and then substituting this expression into the other equation.
• Elimination Method: This method involves adding or subtracting the equations to eliminate one variable, making it possible to solve for the remaining variable.
• Graphical Method: This involves plotting the equations on a graph to find the intersection point, which is the solution to the system.

In the context of our minibus problem, we dealt with two equations representing the distance-speed-time relationship. After expressing the time \(t\) in terms of speed \(s\) from the first equation and substituting into the second, we ended up with a quadratic equation that we then solved for \(s\), illustrating the substitution method.

Distance-Speed-Time Relationship

The relationship between distance, speed, and time is a fundamental concept in physics and various practical scenarios. The basic formula is:\(\text{{Distance}} = \text{{Speed}} \times \text{{Time}}\)This formula shows that distance traveled is the product of the speed at which an object is moving and the time it spends moving. Each of the three variables is interdependent:

• To find distance, multiply speed by time.
• To find speed, divide distance by time.
• To find time, divide distance by speed.

In our exercise, we had to find the time taken by a minibus to cover a certain distance at two different speeds. By setting up an initial equation with the minibus's usual speed and time, and a second equation representing the scenario when the speed was increased, we were able to use the system of equations to resolve the problem. The successful application of these concepts allowed us to determine that the usual time taken by the minibus was 24 hours.