Solving a Real-World Application of Speed and Current with Algebra

Solving a Real-World Application of Speed and Current with Algebra

Ever faced a problem where you had to calculate the speed of a boat in still water when you know the distance it covers downstream and upstream, along with the time and speed of the current? This article guides you through a real-world application of speed and current using algebra, helping you understand how to solve such problems step by step.

Problem Statement

A boat goes 28 km downstream and while returning covers only 75% of the distance covered downstream. If the boat takes 3 hours more to cover the upstream distance than the downstream distance, what is the speed of the boat in still water if the speed of the current is 5/9 meters per second?

Step-by-Step Solution

Step 1: Define Variables

Let the speed of the boat in still water be b km/h. The speed of the current is given as c 5/9 m/s. The distance covered downstream is d 28 km. The distance covered upstream is 0.75 * 28 21 km (75% of 28 km).

Step 2: Convert Current Speed to Consistent Units

To ensure our calculations are consistent, we convert the speed of the current from meters per second (m/s) to kilometers per hour (km/h).

c (5/9) * (3600/1000) 20 km/h

Step 3: Determine Effective Speeds

Downstream speed: b c b 20 km/h Upstream speed: b - c b - 20 km/h

Step 4: Calculate Time Taken for Journeys

Time taken downstream: t_d 28 / (b 20) Time taken upstream: t_u 21 / (b - 20)

Step 5: Set Up the Time Difference Equation

Given that the boat takes 3 hours more to cover the upstream distance than the downstream distance, we can write:

t_u t_d 3

Substituting the expressions for t_u and t_d:

21 / (b - 20) 28 / (b 20) 3

Step 6: Solve the Equation

First, eliminate the denominators by multiplying both sides by (b - 20)(b 20):

21(b 20) 28(b - 20) 3(b - 20)(b 20)

Expanding and rearranging terms:

21b 420 28b - 560 3b^2 - 400

0 3b^2 - 7b - 2180

Step 7: Quadratic Equation and Solution

We now have a standard quadratic equation in the form:

3b^2 - 7b - 2180 0

To solve this, we use the quadratic formula:

b (-B ± sqrt(B^2 - 4AC)) / (2A)

Where A 3, B -7, C -2180.

First, calculate the discriminant:

D B^2 - 4AC (-7)^2 - 4 * 3 * (-2180) 49 26160 26109

Now, substitute the values into the quadratic formula:

b (-(-7) ± sqrt(26109)) / (2 * 3) (7 ± 161.5) / 6

Calculating the two possible values for b:

b (154.5) / 6 25.75 km/h (valid speed since it is positive)

b (-168.5) / 6 (not valid as speed cannot be negative)

Thus, the speed of the boat in still water is approximately: boxed{25.75 km/h}

Keywords

speed of the boat in still water algebraic problem solving speed of current real-world application