Question Statement

The diagram shows the velocity-time graph of the motion of a cyclist riding along a straight road. He accelerates uniformly from rest to $8m/s$ in $20\mathrm{s}$. He then travels at a constant velocity of $8m/s$ for $40\mathrm{s}$. Then he decelerates uniformly to rest in $15\mathrm{s}$. Find:

(i) The acceleration of the cyclist in the first $20\mathrm{s}$ of motion.

(ii) The deceleration of the cyclist in the last $15\mathrm{s}$ of motion.

(iii) The displacement from the starting point of the cyclist after $75\mathrm{s}$.

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Background and Explanation

This problem involves analyzing motion using a velocity-time graph for three distinct phases: uniform acceleration, constant velocity, and uniform deceleration. Remember that acceleration is the slope of the velocity-time graph, while displacement equals the total area under the curve (or can be calculated using kinematic equations for each phase).

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Solution

Part (i): Acceleration in the first $20\mathrm{s}$

For the first phase, the cyclist starts from rest and reaches $8m/s$ in $20\mathrm{s}$. Using the definition of uniform acceleration:

$a=\frac{v_f-v_i}{t}$

Where $v_i=0m/s$, $v_f=8m/s$, and $t=20\mathrm{s}$.

$\begin{array}{rl}a& =\frac{8-0}{20}\\ & =\frac{8}{20}\\ & =0.4{m/s}^2\end{array}$

Part (ii): Deceleration in the last $15\mathrm{s}$

For the deceleration phase, the velocity drops from $8m/s$ to $0m/s$ in $15\mathrm{s}$. Using the same acceleration formula:

$a=\frac{v_f-v_i}{t}$

Where $v_i=8m/s$, $v_f=0m/s$, and $t=15\mathrm{s}$.

$\begin{array}{rl}a& =\frac{0-8}{15}\\ & =-\frac{8}{15}{m/s}^2\\ & \approx -0.533{m/s}^2\end{array}$

The negative sign indicates deceleration. Therefore, the magnitude of deceleration is $\frac{8}{15}{m/s}^2$ (or approximately $0.533{m/s}^2$).

Part (iii): Displacement after $75\mathrm{s}$

To find the total displacement, we calculate the distance traveled during each of the three phases and sum them.

Phase 1: Acceleration ($0$ to $20\mathrm{s}$)

Using the kinematic equation $s=v_{i}t+\frac{1}{2}a{t}^2$:

$\begin{array}{rl}{s}_1& =0\times 20+\frac{1}{2}\times 0.4\times 20^2\\ & =0+\frac{1}{2}\times 0.4\times 400\\ & =80\mathrm{m}\end{array}$

Phase 2: Constant Velocity ($20\mathrm{s}$ to $60\mathrm{s}$)

During this $40\mathrm{s}$ interval, the velocity remains constant at $8m/s$:

$\begin{array}{rl}{s}_2& =v\times t\\ & =8\times 40\\ & =320\mathrm{m}\end{array}$

Phase 3: Deceleration ($60\mathrm{s}$ to $75\mathrm{s}$)

Using $s=v_{i}t+\frac{1}{2}a{t}^2$ with $a=-\frac{8}{15}{m/s}^2$:

$\begin{array}{rl}{s}_3& =8\times 15+\frac{1}{2}\times \left(-\frac{8}{15}\right)\times 15^2\\ & =120+\frac{1}{2}\times \left(-\frac{8}{15}\right)\times 225\\ & =120-60\\ & =60\mathrm{m}\end{array}$

Total Displacement:

$\begin{array}{rl}\text{Total displacement}& =s_1+s_2+s_3\\ & =80+320+60\\ & =460\mathrm{m}\end{array}$

Alternative verification: The displacement equals the area under the velocity-time graph: area of triangle (phase 1) + area of rectangle (phase 2) + area of triangle (phase 3) = $\frac{1}{2}(20)(8)+(40)(8)+\frac{1}{2}(15)(8)=80+320+60=460\mathrm{m}$.

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Key Formulas or Methods Used

• Acceleration: $a=\frac{v_f-v_i}{t}=\frac{\Delta v}{\Delta t}$
• Displacement with constant acceleration: $s=v_{i}t+\frac{1}{2}a{t}^2$
• Displacement with constant velocity: $s=v\times t$
• Geometric interpretation: Displacement equals the area under the velocity-time graph

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Summary of Steps

1. Find acceleration (part i): Apply $a=\frac{v_f-v_i}{t}$ to the first $20\mathrm{s}$ to obtain $0.4{m/s}^2$.
2. Find deceleration (part ii): Apply the same formula to the last $15\mathrm{s}$ to get $-\frac{8}{15}{m/s}^2$ (negative indicates deceleration).
3. Calculate phase 1 displacement: Use $s=v_{i}t+\frac{1}{2}a{t}^2$ to find $s_1=80\mathrm{m}$.
4. Calculate phase 2 displacement: Multiply constant velocity by time ($8\times 40$) to get $s_2=320\mathrm{m}$.
5. Calculate phase 3 displacement: Use the kinematic equation with negative acceleration to find $s_3=60\mathrm{m}$.
6. Sum displacements: Add $80+320+60$ to get the total displacement of $460\mathrm{m}$.