Question Statement

The diagram shows the velocity-time graph of the motion of an athlete running along a straight track. For the first $4\text{s}$, he accelerates uniformly from rest to a velocity of $9\text{m/s}$.

The velocity is then maintained for a further $8\text{s}$. Find:

(i) the rate at which the athlete accelerates

(ii) the displacement from the starting point of the athlete after $12\text{s}$.

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Background and Explanation

This problem involves analyzing motion with two distinct phases: uniform acceleration followed by constant velocity. The key concepts are calculating acceleration from the rate of change of velocity, and determining total displacement by summing the displacements from each phase of motion.

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Solution

First, let us identify the given quantities from the problem:

$\begin{array}{rl}{v}_i& =0\text{(starts from rest)}\\ v_f& =9\text{m/s}\\ t& =4\text{s}\\ a& =?\end{array}$

Part (i): Finding the acceleration

Since the athlete accelerates uniformly, we use the definition of uniform acceleration:

$\begin{array}{rl}a& =\frac{v_f-v_i}{t}\\ & =\frac{9-0}{4}\\ & =2.25{\text{m/s}}^2\end{array}$

Thus, the athlete accelerates at a rate of $2.25{\text{m/s}}^2$.

Part (ii): Finding the displacement after $12\text{s}$

The total time of $12\text{s}$ consists of two distinct intervals:

• First interval: $0$ to $4\text{s}$ (uniform acceleration)
• Second interval: $4\text{s}$ to $12\text{s}$ (constant velocity for $8\text{s}$)

Displacement during the first $4$ seconds ($S_1$):

Using the kinematic equation for displacement under uniform acceleration:

$\begin{array}{rl}{S}_1& =v_{i}t+\frac{1}{2}a{t}^2\\ & =0\times 4+\frac{1}{2}\times 2.25\times (4{)}^2\\ & =0+\frac{1}{2}\times 2.25\times 16\\ & =18\text{m}\end{array}$

Displacement during the last $8$ seconds ($S_2$):

During this interval, the velocity remains constant at $9\text{m/s}$. For constant velocity, displacement is simply velocity multiplied by time:

$\begin{array}{rl}{S}_2& =vt\\ & =9\times 8\\ & =72\text{m}\end{array}$

Total displacement:

$\text{Total displacement}=S_1+S_2=18+72=90\text{m}$

Therefore, the athlete is $90\text{m}$ from the starting point after $12\text{s}$.

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Key Formulas or Methods Used

• Uniform acceleration formula: $a=\frac{v_f-v_i}{t}$
• Displacement with uniform acceleration: $s=v_{i}t+\frac{1}{2}a{t}^2$
• Displacement with constant velocity: $s=vt$ (or equivalently, area of rectangle under velocity-time graph)
• Total displacement: Sum of displacements from individual intervals ($s_{\text{total}}=s_1+s_2$)

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Summary of Steps

1. Identify given values: Note that $v_i=0$, $v_f=9\text{m/s}$, and $t=4\text{s}$ for the acceleration phase.
2. Calculate acceleration: Apply $a=\frac{v_f-v_i}{t}$ to find $a=2.25{\text{m/s}}^2$.
3. Calculate first displacement: Use $s=v_{i}t+\frac{1}{2}a{t}^2$ for the first $4\text{s}$ to get $18\text{m}$.
4. Calculate second displacement: Use $s=vt$ for the remaining $8\text{s}$ at constant velocity to get $72\text{m}$.
5. Sum the results: Add the two displacements ($18\text{m}+72\text{m}$) to obtain the total displacement of $90\text{m}$.