Question

The speed of a train pulling out of a station is given by the equation \(s=t^{2}+t,\) where \(s\) is the speed in kilometers per hour and \(t\) is the time in seconds from when the train starts moving. The equation holds for all situations where \(0 \leq t \leq 4\). In kilometers per hour, what is the positive difference in the speed of the train 4 seconds after it starts moving compared to the speed 2 seconds after it starts moving?

Short answer

14 km/h

Step-by-step solution

Identify Variables and Equation

The given equation is \(s = t^2 + t\), where \(s\) is the speed in km/h and \(t\) is the time in seconds. The task is to find the positive difference in speed at \(t = 4\) seconds and \(t = 2\) seconds.

Calculate Speed at t = 4 seconds

Substitute \(t = 4\) into the equation: \[s(4) = 4^2 + 4 = 16 + 4 = 20 \text{ km/h}\]

Calculate Speed at t = 2 seconds

Substitute \(t = 2\) into the equation: \[s(2) = 2^2 + 2 = 4 + 2 = 6 \text{ km/h}\]

Find the Positive Difference

Compute the positive difference between the speeds at \(t = 4\) seconds and \(t = 2\) seconds: \[|20 - 6| = 14 \text{ km/h}\]

Key concepts

Train Speed Calculation

Train speed calculation often involves understanding how speed changes over time. In our exercise, the speed of the train is described by the equation: \( s = t^2 + t \) where \( s \) stands for speed in kilometers per hour and \( t \) represents time in seconds. You are tasked with finding the train's speed at different moments and the positive difference between those speeds.

For instance, to find the speed at \( t = 4 \) seconds, substitute 4 into the equation: \[ s(4) = 4^2 + 4 = 16 + 4 = 20 \text{ km/h} \] Similarly, to determine the speed at \( t = 2 \) seconds, replace \( t \) with 2 in the equation: \[ s(2) = 2^2 + 2 = 4 + 2 = 6 \text{ km/h} \] Finally, calculate the positive difference in speed using the absolute value of the difference between the speeds at \( t = 4 \) and \( t = 2 \).

\[ |20 - 6| = 14 \text{ km/h} \]

Equation Solving

Equation solving is critical in determining specific values wherever needed. This typically means substituting a given variable with its value and simplifying the expression.

Begin by identifying the variable and its value. In our scenario, the key equation is: \( s = t^2 + t \) Let's solve for \( t = 4 \) first: Substitute 4 into the variable \( t \): \[ s(4) = 4^2 + 4 = 16 + 4 = 20 \text{ km/h} \] Next, solve for \( t = 2 \): Substitute 2 into \( t \): \[ s(2) = 2^2 + 2 = 4 + 2 = 6 \text{ km/h} \] Breaking down each step helps decipher how substituting values simplifies the equation into a precise numerical answer. This is fundamental in solving any relevant mathematical problems.

Time and Speed Relationship

Understanding the relationship between time and speed is crucial in this exercise. Here, the speed of the train is not constant but changes as time progresses. The given equation: \( s = t^2 + t \) shows how speed (\( s \)) depends on time (\( t \)). As time increases from 0 to 4 seconds, the speed calculated from the equation shows the train accelerates over time.

For example: * At \( t = 2 \) seconds: \[ s = 2^2 + 2 = 4 + 2 = 6 \text{ km/h} \]

* At \( t = 4 \) seconds: \[ s = 4^2 + 4 = 16 + 4 = 20 \text{ km/h} \]

The relationship is clearly quadratic, indicating non-linear acceleration of the train. Subsequently, recognizing this helps in grasping how different time intervals affect speed and ultimately enables tackling similar problems with confidence.