Question Statement

A particle moves along a straight line. The particle accelerates from rest to a velocity of $20m/s$ in $15\mathrm{s}$. The particle then moves at a constant velocity of $20m/s$ for a period of time. The particle then decelerates uniformly to rest. The period of time for which the particle is travelling at a constant velocity is $4$ times the period of time for which it is decelerating.

(i) Sketch a velocity-time graph to illustrate the motion of the particle.

Given that the displacement from the starting point of the particle after it comes to rest is $480\mathrm{m}$:

(ii) Find the time for which the particle is moving.

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Background and Explanation

This problem involves motion with constant acceleration and constant velocity. The key principle is that the area under a velocity-time graph represents displacement. For motion with uniform acceleration, displacement can be calculated using the average velocity formula $s=\frac{u+v}{2}\times t$, while for constant velocity, displacement is simply $s=v\times t$.

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Solution

Part (i): Velocity-Time Graph

The velocity-time graph consists of three distinct sections:

1. Stage 1 (Acceleration): A straight line rising from $(0,0)$ to $(15,20)$, representing uniform acceleration from rest to $20m/s$ in $15\mathrm{s}$.
2. Stage 2 (Constant Velocity): A horizontal line at $v=20m/s$. If we let the deceleration time be $t$, this stage lasts for $4t$ seconds.
3. Stage 3 (Deceleration): A straight line falling from $(15+4t,20)$ to $(15+5t,0)$, representing uniform deceleration from $20m/s$ to rest in time $t$.

The graph forms a trapezoid shape: a triangle (acceleration), followed by a rectangle (constant velocity), followed by another triangle (deceleration).

Part (ii): Finding the Total Time

Step 1: Define the time variables

Let $t$ be the time taken for the deceleration phase (Stage 3). Then, the time for constant velocity (Stage 2) is $4t$. The time for acceleration (Stage 1) is given as $15\mathrm{s}$.

Step 2: Calculate displacement for each stage

For Stage 1 (acceleration from $0$ to $20m/s$): Using the formula for displacement with uniform acceleration $s=\frac{u+v}{2}\times t$:

$S_1=\frac{0+20}{2}\times 15=10\times 15=150\mathrm{m}$

For Stage 2 (constant velocity): $S_2=v\times \text{time}=20\times 4t=80t\mathrm{m}$

For Stage 3 (deceleration from $20m/s$ to $0$): Using the average velocity formula again: $S_3=\frac{20+0}{2}\times t=10t\mathrm{m}$

Step 3: Set up the total displacement equation

The total displacement is given as $480\mathrm{m}$:

$S_{\text{total}}=S_1+S_2+S_3=480$

Substituting the expressions: $150+80t+10t=480$

Step 4: Solve for $t$

Combine like terms: $150+90t=480$

Subtract $150$ from both sides: $90t=330$

Divide by $90$: $t=\frac{330}{90}=\frac{11}{3}\mathrm{s}$

Step 5: Calculate the total time

Now we find the duration of each stage:

• Stage 1: $15\mathrm{s}$
• Stage 2: $4t=4\times \frac{11}{3}=\frac{44}{3}\mathrm{s}$
• Stage 3: $t=\frac{11}{3}\mathrm{s}$

Total time: $\text{Total Time}=15+\frac{44}{3}+\frac{11}{3}=15+\frac{55}{3}$

Convert to a common denominator: $=\frac{45}{3}+\frac{55}{3}=\frac{100}{3}\mathrm{s}$

Or as a decimal: $\approx 33.333\mathrm{s}$

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Key Formulas or Methods Used

• Displacement with uniform acceleration: $s=\frac{u+v}{2}\times t$ (where $u$ is initial velocity, $v$ is final velocity)
• Displacement with constant velocity: $s=v\times t$
• Area under velocity-time graph: Represents total displacement (geometric approach using trapezoid/triangle areas)
• Algebraic ratio handling: Setting up variables based on the given $4:1$ time ratio

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Summary of Steps

1. Define variables: Let $t$ = deceleration time, so constant velocity time = $4t$, and acceleration time = $15\mathrm{s}$.
2. Calculate Stage 1 displacement: $S_1=\frac{0+20}{2}\times 15=150\mathrm{m}$.
3. Calculate Stage 2 displacement: $S_2=20\times 4t=80t\mathrm{m}$.
4. Calculate Stage 3 displacement: $S_3=\frac{20+0}{2}\times t=10t\mathrm{m}$.
5. Form equation: $150+80t+10t=480$, which simplifies to $90t=330$.
6. Solve for $t$: $t=\frac{11}{3}\mathrm{s}$ (or $3.67\mathrm{s}$).
7. Compute total time: $15+\frac{44}{3}+\frac{11}{3}=\frac{100}{3}\approx 33.33\mathrm{s}$.