This problem aims to familiarize us with **velocity** and its **kinds,** such as **instantaneous velocity,** and **average velocity.** The concepts required for this problem are as mentioned, but it would be helpful if you are familiar with **distance** and **speed relations.**

Now the **instantaneous velocity** of an object is defined as the **rate** of **change** of **position** of an object for a **particular time interval** or it is the limit of the **intermediate velocity** as the total time closes to **zero.**

Whereas the **average velocity **is described as the **difference** in displacement divided by the **time** in which the **displacement** happens. It can be **negative** or **positive** relying on the direction of the **displacement.** Like average velocity, the instantaneous velocity is a **vector** quantity.

## Expert Answer

**Part a:**

We are given an **expression** which is the **distance** of the car from the** traffic light**:

\[x(t) =bt^2 – ct^3\]

Where $b = 2.40 ms^{-2}$, and $c = 0.120 ms^{-3}$.

Since we are given a **time**, we can easily calculate the **average velocity** using the formula:

\[ v_{x,avg}=\dfrac{\bigtriangleup x}{\bigtriangleup t}\]

Here, $\bigtriangleup x = x_f – x_i$ and, $\bigtriangleup t = t_f – t_i$

**Where**,

$x_f = 0 m\space and\space x_i = 120 m$

$t_f = 10 s\space and\space t_i = 0 s$

\[v_{x,avg} =\dfrac{ x_f – x_i}{t_f – t_i} \]

\[v_{x,avg} =\dfrac{ 120 – 0}{10 – 0} \]

\[v_{x,avg} = 12\space m/s \]

**Part b:**

The **instantaneous velocity** can be calculated using **various** formulas but for this particular problem, we are going to use the **derivative.** Thus, the **instantaneous velocity** is just the derivative of $x$ with respect to $t$:

\[v_x = \dfrac{dx}{dt} \]

**Derivating** the **distance** expression with respect to $x$:

\[x(t) = bt^2 – ct^3 \]

\[v_x = 2bt – 3ct^2 \space (Eq.1)\]

**Instantaneous** velocity at $t = 0 s$,

\[v_x = 0 \space m/s\]

**Instantaneous** velocity at $t = 5 s$,

\[v_x = 2(2.40)(5) – 3(0.120)(5)^2 \space m/s\]

\[v_x = 15 \space m/s\]

**Instantaneous** velocity at $t = 10 s$,

\[v_x = 2(2.40)(10) – 3(0.120)(10)^2 \space m/s\]

\[v_x = 12 \space m/s\]

**Part c:**

Since the car is at **rest,** its **initial velocity** is $0 m/s$. using $Eq.1$:

\[ 0 = 2bt – 3ct^2\]

\[ t = \dfrac{2b}{3c}\]

\[ t = \dfrac{2(2.40)}{3(0.120)}\]

\[ t = 13.33 \space s\]

## Numerical Result

**Part a:** The **average** velocity of the car is $ v_{x,avg} = 12 \space m/s$.

**Part b:** The **instantaneous** velocity of the car is $v_x = 0 \space m/s, \space 15\space m/s$ ,and $12\space m/s $.

**Part c:** The **time** for the **car** to again reach the **rest** state is $t = 13.33 \space s$.

## Example

What is the **average velocity** of a car in a given **time interval** if the **car** moves $7 m$ in $4 s$ and $18 m$ in $6 s$ in a **straight line**?

**Given** that:

\[ s_1 = 7 \space m\]

\[ t_1 = 4 \space s\]

\[s_2 = 18 \space m\]

\[t_2 = 6 \space s\]

\[v_{x,avg} = \dfrac{s_2 – s_1}{t_2 – t_1}\]

\[v_{x,avg} = \dfrac{18 – 7}{6 – 4}\]

\[v_{x,avg} = \dfrac{11}{2}\]

\[v_{x,avg} = 5.5 \space m/s\]