Question Statement

A particle $P$ moves along the positive $x$-axis. At time $t$ seconds the velocity $v$ of $P$ is given by

$v=\{\begin{array}{ll}8t-\frac{3}{2}{t}^2,& 0\le t\le 4\\ 16-2t,& t>4\end{array}$

When $t=0$, $P$ is at the origin $O$. Find:

(i) the greatest speed of $P$ in the interval $0\le t\le 4$,

(ii) the distance of $P$ from $O$ when $t=4$,

(iii) the time at which $P$ is instantaneously at rest for $t>4$,

(iv) the total distance travelled by $P$ in the first $10\text{ s}$ of its motion.

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Background and Explanation

This problem involves kinematics with a piecewise velocity function. You will need to use differentiation to find maximum values (extrema) and integration to determine position from velocity. Remember that total distance travelled accounts for any changes in direction (backward motion), while displacement only measures net change in position.

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Solution

Part (i)

To find the greatest speed in the interval $[0,4]$, we analyze the velocity function $v=8t-\frac{3}{2}{t}^2$.

First, differentiate $v$ with respect to $t$:

$\begin{array}{rl}\frac{dv}{dt}& =\frac{d}{dt}\left(8t-\frac{3}{2}{t}^2\right)\\ & =8-\frac{3}{2}\cdot 2t\\ & =8-3t\end{array}$

For extreme values, set $\frac{dv}{dt}=0$:

$\begin{array}{rl}8-3t& =0\\ \Rightarrow t& =\frac{8}{3}\end{array}$

Substitute $t=\frac{8}{3}$ back into the velocity equation:

$\begin{array}{rl}v& =8\left(\frac{8}{3}\right)-\frac{3}{2}{\left(\frac{8}{3}\right)}^2\\ & =\frac{64}{3}-\frac{3}{2}\times \frac{64}{9}\\ & =\frac{64}{3}-\frac{32}{3}\\ & =\frac{32}{3}\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the greatest speed is $\boxed{\frac{32}{3}\text{ m/s}}$ (or approximately $10.67\text{ m/s}$).

Part (ii)

To find the distance from $O$ at $t=4$, we integrate the velocity function over $[0,4]$:

$\begin{array}{rl}S& =\int vdt\\ & =\int \left(8t-\frac{3}{2}{t}^2\right)dt\\ & =\frac{8t^2}{2}-\frac{3t^3}{2\cdot 3}+C\\ S& =4t^2-\frac{1}{2}{t}^3+C\end{array}$

Apply the initial condition: at $t=0$, $S=0$:

$\begin{array}{rl}0& =0-0+C\\ \Rightarrow C& =0\end{array}$

So the position function is $S=4t^2-\frac{1}{2}{t}^3$.

At $t=4$:

$\begin{array}{rl}S& =4(4{)}^2-\frac{1}{2}(4{)}^3\\ & =64-\frac{64}{2}\\ & =32\mathrm{m}\end{array}$

Therefore, the distance from $O$ when $t=4$ is $\boxed{32\text{ m}}$.

Part (iii)

For $t>4$, the velocity is given by $v=16-2t$. The particle is instantaneously at rest when $v=0$:

$\begin{array}{rl}16-2t& =0\\ -2t& =-16\\ t& =8\mathrm{s}\end{array}$

Therefore, the particle is at rest at $\boxed{t=8\text{ s}}$.

Part (iv)

To find the total distance travelled in the first $10\text{ s}$, we analyze the motion in separate intervals. First, integrate the velocity function for $t>4$:

$\begin{array}{rl}S& =\int vdt\\ & =\int (16-2t)dt\\ S& =16t-\frac{2t^2}{2}+C=16t-t^2+C\end{array}$

Using the condition $S=0$ at $t=0$ to determine $C$:

$0=0-0+C\Rightarrow C=0$

Thus $S=16t-t^2$.

Calculate the position at key times:

• At $t=8$ s: $S=16\times 8-8^2=128-64=64\mathrm{m}$

• At $t=10$ s: $S=16\times 10-10^2=160-100=60\mathrm{m}$

The distance covered in the last $2\text{ s}$ (from $t=8$ to $t=10$) is: $|60-64|=4\mathrm{m}$

Now calculate the total distance travelled:

• Distance from $t=0$ to $t=4$: $32\text{ m}$ (from part ii)
• Distance from $t=4$ to $t=8$: $|64-32|=32\text{ m}$
• Distance from $t=8$ to $t=10$: $|-4|=4\text{ m}$

$\begin{array}{rl}\text{Total distance in }10\text{ s}& =32+|64-32|+|-4|\\ & =32+32+4\\ & =68\mathrm{m}\end{array}$

Therefore, the total distance travelled in the first $10\text{ s}$ is $\boxed{68\text{ m}}$.

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Key Formulas or Methods Used

• Differentiation for extrema: $\frac{dv}{dt}=0$ to find critical points where speed is maximized
• Power rule for integration: $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$
• Initial value problems: Using $S=0$ at $t=0$ to determine constants of integration
• Rest condition: Setting $v=0$ to find when the particle stops
• Total distance calculation: Summing absolute displacements over intervals: $\sum |S(t_i)-S(t_{i-1})|$

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Summary of Steps

1. Greatest speed (i): Differentiate $v=8t-\frac{3}{2}{t}^2$ to get $\frac{dv}{dt}=8-3t$, set to zero yielding $t=\frac{8}{3}$, substitute to find $v=\frac{32}{3}$ m/s.
2. Position at $t=4$ (ii): Integrate to get $S=4t^2-\frac{1}{2}{t}^3$, use $S(0)=0$ to find $C=0$, evaluate at $t=4$ to get $32$ m.
3. Time at rest (iii): Solve $16-2t=0$ for $t>4$ to obtain $t=8$ s.
4. Total distance (iv):
   • Integrate $v=16-2t$ to get $S=16t-t^2$ (with $C=0$)
   • Calculate position at $t=8$ (64 m) and $t=10$ (60 m)
   • Find distance in last 2 s: $|60-64|=4$ m
   • Sum distances: $32+32+4=68$ m