Question Statement

A cyclist is traveling along a straight path, and the distance traveled $s(t)$ (in meters) is given by:

$s(t)=5t^2+3t$

where $t$ is the time in seconds.

a. Find the speed (slope of the distance-time graph) at any time $t$.

b. Determine the speed at $t=4$ seconds.

c. Interpret the significance of the slope in this context.

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Background and Explanation

This problem applies differential calculus to kinematics. The derivative of a distance-time function gives the instantaneous rate of change of position, which is the speed of the object. The power rule for differentiation allows us to find this rate of change at any moment.

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Solution

Part (a)

To find the speed at any time $t$, we differentiate the distance function $s(t)$ with respect to time $t$. The derivative $\frac{ds}{dt}$ represents the slope of the distance-time graph and gives the instantaneous speed.

Given: $s(t)=5t^2+3t$

Differentiating with respect to $t$ using the power rule:

$\begin{array}{rl}\frac{ds}{dt}& =5\cdot \frac{d}{dt}(t^2)+3\cdot \frac{d}{dt}(t)\\ & =5(2t)+3(1)\\ & =10t+3\end{array}$

Therefore, the speed at any time $t$ is: $\text{Speed}=10t+3\text{ m/s}$

Part (b)

To find the speed at $t=4$ seconds, we substitute $t=4$ into the velocity function derived in part (a):

$\begin{array}{rl}\text{Speed}& =\frac{ds}{dt}{|}_{t=4}\\ & =10(4)+3\\ & =40+3\\ & =43\text{ m/s}\end{array}$

The cyclist is traveling at 43 m/s at $t=4$ seconds.

Part (c)

The slope of the distance-time graph is given by $\frac{ds}{dt}=10t+3$.

Observe that as $t$ increases, the value of $10t+3$ also increases (since the coefficient of $t$ is positive). This means the speed of the cyclist is not constant but increases linearly with time.

In physical terms, this indicates that the cyclist is accelerating—specifically, the cyclist has a constant acceleration of $10{\text{ m/s}}^2$ (the rate of change of speed). The slope represents the instantaneous velocity, and the fact that it changes with time tells us the motion is non-uniform.

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Key Formulas or Methods Used

• Power Rule for Differentiation: $\frac{d}{dt}(t^n)=n{t}^{n-1}$
• Constant Multiple Rule: $\frac{d}{dt}[c\cdot f(t)]=c\cdot \frac{d}{dt}[f(t)]$
• Sum Rule: $\frac{d}{dt}[f(t)+g(t)]=\frac{d}{dt}[f(t)]+\frac{d}{dt}[g(t)]$
• Interpretation of Derivative: $\frac{ds}{dt}$ represents instantaneous speed (slope of distance-time graph)

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Summary of Steps

1. Differentiate the distance function $s(t)=5t^2+3t$ using the power rule to obtain the speed function $v(t)=10t+3$.
2. Evaluate the speed function at $t=4$ by substitution: $10(4)+3=43$ m/s.
3. Analyze the slope function $10t+3$ to determine that speed increases linearly with time, indicating constant positive acceleration.