Understanding the Speed, Distance, and Time Relationship

This problem involves the fundamental relationship between speed, distance, and time. The formula connecting these three quantities is:

\(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\)

We are given two scenarios and asked to find the ratio of the speeds in these two scenarios.

Scenario 1: Initial Conditions

Let's define the variables for the first scenario:

• Distance = \(D\)
• Speed = \(S\)
• Time = \(T\)

Using the speed formula, we can write the equation for the first scenario:

\(S = \frac{D}{T}\)   (Equation 1)

Scenario 2: Modified Conditions

In the second scenario, the conditions are changed:

• Distance = Half the original distance = \(\frac{D}{2}\)
• Time = Double the original time = \(2T\)
• Let the new speed be \(S'\)

Applying the speed formula to the second scenario:

\(S' = \frac{\text{New Distance}}{\text{New Time}}\)

\(S' = \frac{\frac{D}{2}}{2T}\)

Simplifying the Second Speed Equation

Let's simplify the expression for \(S'\):

\(S' = \frac{D}{2} \times \frac{1}{2T}\)

\(S' = \frac{D}{4T}\)    (Equation 2)

Calculating the Ratio of the Two Speeds

We need to find the ratio of the two speeds, \(S : S'\), which can be written as \(\frac{S}{S'}\). Let's divide Equation 1 by Equation 2:

\(\frac{S}{S'} = \frac{\frac{D}{T}}{\frac{D}{4T}}\)

To simplify this fraction, we can multiply the numerator by the reciprocal of the denominator:

\(\frac{S}{S'} = \frac{D}{T} \times \frac{4T}{D}\)

We can cancel out the common terms \(D\) and \(T\):

\(\frac{S}{S'} = \frac{\cancel{D}}{\cancel{T}} \times \frac{4\cancel{T}}{\cancel{D}}\)

\(\frac{S}{S'} = \frac{4}{1}\)

So, the ratio of the two speeds \(S : S'\) is \(4 : 1\).

The ratio of the initial speed to the second speed is \(4 : 1\).

Revision Table: Speed, Distance, Time Concepts

Additional Information on Ratios and Proportionality

In this problem, we saw how changes in distance and time affect speed. Speed is directly proportional to distance (if time is constant) and inversely proportional to time (if distance is constant).

• If distance increases and time is constant, speed increases.
• If time increases and distance is constant, speed decreases.

In our case, distance was halved (\(\times \frac{1}{2}\)) and time was doubled (\(\times 2\)). Let's see how these changes affect speed:

\(S' = \frac{\text{New Distance}}{\text{New Time}} = \frac{\frac{1}{2} \times \text{Original Distance}}{2 \times \text{Original Time}} = \frac{1/2}{2} \times \frac{\text{Original Distance}}{\text{Original Time}} = \frac{1}{4} \times \text{Original Speed}\)

So, the new speed \(S'\) is one-fourth of the original speed \(S\). This means \(S = 4S'\), which gives the ratio \(S : S' = 4 : 1\).

Understanding these proportional relationships can help solve speed, distance, and time problems quickly.