Question Statement

The position of an object moving on a line is given by $S=6t^2-t^3,t\ge 0$, where $S$ is in metres and $t$ is in seconds.

(i) Determine the velocity and acceleration of the object at $t=2$.

(ii) At what time is the object at rest?

(iii) In which direction is the object moving at $t=5$ s?

(iv) When is the object moving in a positive direction?

(v) When does the object return to its initial position?

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Background and Explanation

This problem involves kinematic relationships where velocity is the rate of change of position with respect to time, and acceleration is the rate of change of velocity. To solve these problems, you'll need to differentiate the position function and analyze the signs of the resulting expressions to determine direction and motion states.

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Solution

Part (i): Velocity and Acceleration at $t=2$

First, we find the velocity function by differentiating the position function $S$ with respect to time $t$:

$v=\frac{dS}{dt}=\frac{d}{dt}\left(6t^2-t^3\right)=12t-3t^2$

At $t=2$ seconds:

$v=12(2)-3(2{)}^2=24-12=12\text{ m/s}$

Next, we find the acceleration function by differentiating the velocity function:

$a=\frac{dv}{dt}=\frac{d}{dt}\left(12t-3t^2\right)=12-6t$

At $t=2$ seconds:

$a=12-6(2)=12-12=0{\text{ m/s}}^2$

Answer: At $t=2$, the velocity is $12$ m/s and the acceleration is $0$ m/s².

Part (ii): When the Object is at Rest

The object is at rest when its velocity equals zero:

$v=0$ $12t-3t^2=0$ $3t(4-t)=0$

This gives two solutions:

• $t=0$ (initial moment), or
• $4-t=0\Rightarrow t=4$ s

Answer: The object is at rest at $t=4$ seconds (and initially at $t=0$).

Part (iii): Direction at $t=5$ s

Substitute $t=5$ into the velocity equation:

$v=12(5)-3(5{)}^2=60-75=-15\text{ m/s}$

Since the velocity is negative, the object is moving in the backward (negative) direction.

Answer: The object is moving in the backward (negative) direction at $t=5$ s.

Part (iv): When Moving in the Positive Direction

The object moves in the positive direction when $v>0$:

$12t-3t^2>0$ $3t(4-t)>0$

For this product to be positive, we analyze the critical points at $t=0$ and $t=4$:

Since $t\ge 0$, we test the interval $(0,4)$ by substituting $t=1$: $v=12(1)-3(1{)}^2=9>0✓$

Testing $t=5$ (outside the interval) gave $v=-15<0$, confirming the object moves backward after $t=4$.

Answer: The object moves in the positive direction when $0<t<4$ seconds.

Part (v): Return to Initial Position

At $t=0$, the initial position is: $S=6(0{)}^2-(0{)}^3=0$

The object returns to its initial position when $S=0$ again (with $t>0$):

$6t^2-t^3=0$ $t^2(6-t)=0$

This gives:

• $t=0$ (initial time), or
• $6-t=0\Rightarrow t=6$ s

Answer: The object returns to its initial position at $t=6$ seconds.

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Key Formulas or Methods Used

• Velocity from position: $v=\frac{dS}{dt}$
• Acceleration from velocity: $a=\frac{dv}{dt}=\frac{d^{2}S}{d{t}^2}$
• Condition for rest: $v=0$
• Positive direction condition: $v>0$
• Return to origin: Solve $S(t)=S(0)$ for $t>0$

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Summary of Steps

1. Differentiate position to obtain the velocity function: $v=12t-3t^2$
2. Differentiate velocity to obtain the acceleration function: $a=12-6t$
3. Evaluate both functions at $t=2$ for part (i)
4. Solve $v=0$ to find when the object is at rest for part (ii)
5. Substitute $t=5$ into the velocity equation and check the sign for part (iii)
6. Solve the inequality $v>0$ (i.e., $3t(4-t)>0$) to find the time interval for positive motion for part (iv)
7. Solve $S=0$ for $t>0$ to find when the object returns to the starting point for part (v)