Question Statement

A particle $P$ moves along the $x$-axis. The acceleration of $P$ at time $t$ seconds is $a=(4t-8)\mathrm{m}/{\mathrm{s}}^2$, measured in the increasing direction of $x$. The velocity of $P$ at time $t$ seconds is $v\mathrm{m}/\mathrm{s}$. Given that $v=6$ when $t=0$, find:

(i) $v$ in terms of $t$.

(ii) the distance between the two points where $P$ is instantaneously at rest.

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Background and Explanation

This problem involves kinematics with variable acceleration. Since acceleration is given as a function of time, we use integration to find velocity from acceleration, and displacement from velocity. The constant of integration is determined using the given initial conditions.

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Solution

Part (i): Finding $v$ in terms of $t$

Given the acceleration: $a=4t-8$

To find the velocity, we integrate the acceleration with respect to time: $\begin{array}{rl}v& =\int adt\\ & =\int (4t-8)dt\\ v& =\frac{4t^2}{2}-8t+C\\ v& =2t^2-8t+C\end{array}$

Using the initial condition when $t=0$, then $v=0$: $\begin{array}{rl}0& =0-0+C\\ \Rightarrow C& =0\end{array}$

Therefore, the velocity in terms of $t$ is: $v=2t^2-8t$

Part (ii): Finding the distance between rest points

First, we determine when the particle is instantaneously at rest by setting $v=0$: $\begin{array}{rl}v& =0\\ 2t^2-8t& =0\\ 2t(t-4)& =0\end{array}$

This gives $t=0$ or $t=4$ seconds.

To find the displacement, we integrate the velocity function: $\begin{array}{rl}S& =\int vdt\\ & =\int \left(2t^2-8t\right)dt\\ & =\frac{2t^3}{3}-\frac{8t^2}{2}+C\\ & =\frac{2t^3}{3}-4t^2+C\end{array}$

When $t=0$, $v=0$, so $S=0\Rightarrow C=0$: $S=\frac{2}{3}{t}^3-4t^2$

At $t=4$: $\begin{array}{rl}S& =\frac{2}{3}(4{)}^3-4(4{)}^2\\ & =\frac{2}{3}\times 64-64\\ & =\frac{128}{3}-64\\ & =\frac{128}{3}-\frac{192}{3}\\ & =-\frac{64}{3}\mathrm{m}\end{array}$

The distance between the two points where $P$ is instantaneously at rest (at $t=0$ and $t=4$) is the magnitude of this displacement:

$\text{Distance}=\frac{64}{3}\mathrm{m}$

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Key Formulas or Methods Used

• $v=\int adt$ (Velocity is the integral of acceleration with respect to time)
• $s=\int vdt$ (Displacement is the integral of velocity with respect to time)
• Constant of integration $C$ determined from initial boundary conditions
• Particle is at rest when $v=0$
• Distance is the absolute value of displacement between two points

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Summary of Steps

1. Integrate acceleration $a=4t-8$ to obtain $v=2t^2-8t+C$
2. Apply initial condition $v=0$ at $t=0$ to find $C=0$, giving $v=2t^2-8t$
3. Find rest times by solving $v=0$: factor $2t(t-4)=0$ to get $t=0$ and $t=4$
4. Integrate velocity to get displacement $S=\frac{2}{3}{t}^3-4t^2+C$
5. Apply initial position $S=0$ at $t=0$ to find $C=0$
6. Calculate displacement at $t=4$: $S=-\frac{64}{3}$ m
7. Take absolute value to find distance: $\frac{64}{3}$ m (approximately $21.33$ m)