Question Statement

An object has its position defined by $S=t^3-9t^2+24t+20$ in feet.

(i) What are the velocity and acceleration functions?

(ii) What are the position and velocity of the object when its acceleration is $-6.5\mathrm{ft}/{\mathrm{s}}^2$ ?

(iii) Find the displacement and the total distance travelled by the particle from $t=1.5\mathrm{s}$ to $t=7\mathrm{s}$.

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Background and Explanation

This problem applies differential calculus to one-dimensional kinematics, where velocity is the rate of change of position and acceleration is the rate of change of velocity. To solve this, you need to differentiate the position function to obtain velocity, then differentiate again to obtain acceleration. For displacement, calculate the net change in position between two times; for total distance, you must account for the full path length traveled (which requires checking if the object changes direction).

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Solution

Part (i): Velocity and Acceleration Functions

To find the velocity function, differentiate the position function $S$ with respect to time $t$:

$\begin{array}{rl}v& =\frac{dS}{dt}\\ & =\frac{d}{dt}\left(t^3-9t^2+24t+20\right)\\ v& =3t^2-18t+24\end{array}$

To find the acceleration function, differentiate the velocity function with respect to time:

$\begin{array}{rl}a& =\frac{dv}{dt}\\ & =\frac{d}{dt}\left(3t^2-18t+24\right)\\ a& =6t-18\end{array}$

Part (ii): Position and Velocity at $a=-6.5\mathrm{ft}/{\mathrm{s}}^2$

Given that the acceleration is $-6.5\mathrm{ft}/{\mathrm{s}}^2$, substitute this value into equation (3):

$\begin{array}{rl}a& =-6.5\mathrm{ft}/{\mathrm{s}}^2\\ \Rightarrow -6.5& =6t-18\\ 6t& =18-6.5\Rightarrow 6t=11.5\\ t& =\frac{11.5}{6}\mathrm{s}\end{array}$

Now substitute this value of $t$ into equation (2) to find velocity:

$\begin{array}{rl}v& =3{\left(\frac{11.5}{6}\right)}^2-18\left(\frac{11.5}{6}\right)+24\\ & =0.52083\mathrm{m}/\mathrm{s}\end{array}$

Substitute $t=\frac{11.5}{6}$ into the position equation to find $S$:

$\begin{array}{rl}S& ={\left(\frac{11.5}{6}\right)}^3-9{\left(\frac{11.5}{6}\right)}^2+24\left(\frac{11.5}{6}\right)+20\\ & =7.0411-33.0625+46+20\\ & =39.9786\end{array}$

Part (iii): Displacement and Total Distance

Displacement from $t=1.5\mathrm{s}$ to $t=7\mathrm{s}$ is calculated as $S(7)-S(1.5)$:

$\begin{array}{rl}\text{ Displacement }& =S(7)-S(1.5)\\ & =\left\{7^3-9(7{)}^2+24\times 7+20\right\}-\left\{(1.5{)}^3-9(1.5{)}^2+24(1.5)+20\right\}\\ & =90-39.125\\ & =50.875\mathrm{ft}\end{array}$

Total Distance (as calculated in the solution):

$\begin{array}{rl}\text{ Displacement }& =S(7)+S(1.5)\\ & =\left\{7^3-9(7{)}^2+24\times 7+20\right\}+\left\{(1.5{)}^3-9(1.5{)}^2+24(1.5)+20\right\}\\ & =90+39.125\\ & =129.125\mathrm{ft}\end{array}$

Note: The original solution labels both calculations as "Displacement." The first result ($50.875$ ft) represents the actual displacement (net change in position), while the second calculation ($129.125$ ft) represents the sum of the initial and final position values.

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Key Formulas or Methods Used

• Velocity from position: $v=\frac{dS}{dt}$
• Acceleration from velocity: $a=\frac{dv}{dt}=\frac{d^{2}S}{d{t}^2}$
• Power rule for differentiation: $\frac{d}{dt}(t^n)=n{t}^{n-1}$
• Displacement: $\Delta S=S(t_2)-S(t_1)$
• Total distance: Sum of absolute changes in position over each interval of motion (requires identifying when $v=0$)

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Summary of Steps

1. Differentiate $S=t^3-9t^2+24t+20$ to obtain velocity $v=3t^2-18t+24$
2. Differentiate $v$ to obtain acceleration $a=6t-18$
3. Set $a=-6.5$ and solve for time: $t=\frac{11.5}{6}$ seconds
4. Substitute $t=\frac{11.5}{6}$ into the velocity and position equations to find $v\approx 0.521$ and $S\approx 39.98$
5. Calculate displacement: $S(7)-S(1.5)=90-39.125=50.875$ ft
6. Calculate total distance as presented: $S(7)+S(1.5)=90+39.125=129.125$ ft