Question Statement

A particle moves $100\mathrm{m}$ in a straight line. The diagram shows a sketch of the velocity-time graph of the motion. The particle starts with velocity $u\mathrm{m}/\mathrm{s}$ and accelerates to a velocity of $10\mathrm{m}/\mathrm{s}$. This velocity is maintained for $7\mathrm{s}$, and then the particle decelerates to rest in $2\mathrm{s}$.

(i) Find the value of $u$.

(ii) Find the acceleration of the particle in the first part of the motion.

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Background and Explanation

This problem involves analyzing motion with constant acceleration using a velocity-time graph. The total distance traveled equals the area under the velocity-time graph. For each segment of the motion, we can apply kinematic equations or use the average velocity formula to find unknown quantities.

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Solution

The motion consists of three distinct stages. We analyze stages 2 and 3 first to determine their distances, then use the total distance to find the unknowns for stage 1.

Stage 1: Initial Acceleration

The particle accelerates from initial velocity $u$ to $10\mathrm{m}/\mathrm{s}$ in $3\mathrm{s}$.

$\begin{array}{rl}{v}_i& =u\mathrm{m}/\mathrm{s}\\ t& =3\mathrm{s}\\ v_f& =10\mathrm{m}/\mathrm{s}\\ S_1& =?\\ a& =?\end{array}$

Stage 2: Constant Velocity

The particle travels at constant velocity.

$\begin{array}{rl}v& =10\mathrm{m}/\mathrm{s}\\ t& =7\mathrm{s}\\ S_2& =vt=10\times 7=70\mathrm{m}\end{array}$

Stage 3: Deceleration to Rest

The particle decelerates from $10\mathrm{m}/\mathrm{s}$ to rest.

$\begin{array}{rl}{v}_i& =10\mathrm{m}/\mathrm{s}\\ v_f& =0\\ t& =2\mathrm{s}\\ a& =\frac{v_f-v_i}{t}\\ & =\frac{0-10}{2}=-5\mathrm{m}/{\mathrm{s}}^2\\ S_3& =v_{i}t+\frac{1}{2}a{t}^2\\ & =10\times 2+\frac{1}{2}(-5)2^2\\ & =20-10=10\mathrm{m}\end{array}$

Finding $S_1$ and the Initial Velocity $u$

The total distance is given as $100\mathrm{m}$:

$S_1+S_2+S_3=100\mathrm{m}$

Substituting the known values:

$S_1+70+10=100$

Solving for $S_1$:

$S_1=100-80=20\mathrm{m}$

Now using the average velocity formula for Stage 1 $\left(S=\frac{v_f+v_i}{2}\times t\right)$:

$\begin{array}{rl}{S}_1& =\frac{u+10}{2}\times 3\\ 20& =\frac{u+10}{2}\times 3\\ \frac{40}{3}& =u+10\\ u& =\frac{40}{3}-10\\ & =\frac{10}{3}\mathrm{m}/\mathrm{s}\end{array}$

Finding the Acceleration in Stage 1

Using the definition of acceleration:

$\begin{array}{rl}a& =\frac{v_f-v_i}{t}\\ & =\frac{10-\frac{10}{3}}{3}\\ & =\frac{20/3}{3}=\frac{20}{9}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore:

• (i) $u=\frac{10}{3}\mathrm{m}/\mathrm{s}$ (or approximately $3.33\mathrm{m}/\mathrm{s}$)
• (ii) The acceleration in the first part is $\frac{20}{9}\mathrm{m}/{\mathrm{s}}^2$ (or approximately $2.22\mathrm{m}/{\mathrm{s}}^2$)

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Key Formulas or Methods Used

• Distance as area under velocity-time graph: For constant velocity, $s=vt$; for constant acceleration, $s=\frac{u+v}{2}\times t$
• Kinematic equation for distance: $s=ut+\frac{1}{2}a{t}^2$
• Definition of acceleration: $a=\frac{v-u}{t}$ (or $a=\frac{\Delta v}{\Delta t}$)
• Total distance constraint: Sum of distances from all stages equals the total displacement

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Summary of Steps

1. Analyze Stage 2 (constant velocity): Calculate $S_2=10\times 7=70\mathrm{m}$
2. Analyze Stage 3 (deceleration): Calculate acceleration $a=\frac{0-10}{2}=-5\mathrm{m}/{\mathrm{s}}^2$, then distance $S_3=10(2)+\frac{1}{2}(-5)(2{)}^2=10\mathrm{m}$
3. Find Stage 1 distance: Use total distance $100\mathrm{m}$ to get $S_1=100-70-10=20\mathrm{m}$
4. Calculate initial velocity $u$: Use average velocity formula $20=\frac{u+10}{2}\times 3$ to find $u=\frac{10}{3}\mathrm{m}/\mathrm{s}$
5. Calculate Stage 1 acceleration: Use $a=\frac{10-\frac{10}{3}}{3}=\frac{20}{9}\mathrm{m}/{\mathrm{s}}^2$