**article aims to find a car’s average and instantaneous velocity at times.**This article uses the concept of

**average velocity**.

**Average velocity**is the

**change in position**$(\Delta x)$ divided by the time intervals $(\Delta t)$ in which the displacement occurs. Average velocity can be

**negative**or

**positive**depending on the sign of the shift.

**SI unit of average velocity is meters per second**$(\dfrac{m}{s}\: or \:ms^{-1})$. The

**formula for average velocity**is: \[v_{avg} =\dfrac{\Delta x}{\Delta t}\]

**Expert Answer**

**Distance from the light is**represented by the equation $x(t) = bt ^ {2} – ct ^ {3}$. Where $b = 2.40 \dfrac{m}{s^{2}}$ and $c=0.120 \dfrac{m}{s^{3}}$.

**Part (a)**

**Distance from the light**is: \[x (t) = bt ^ {2} – ct ^ {3} \]

**At $t=0$**\[x(0)=b0 ^ {2} – c0 ^ {3}= 0m\]

**At $t=10s$**\[x(t) = 2.40\times 10^{2} – 0.12 \times 10^{3}\] \[x(t) = 2.40\times 100 – 0.12\times 1000\] \[x(t) = 120m\] The

**average velocity of car**for time interval $t = 0s$ and $t=10s$ is given by: \[v_{avg}= \dfrac{dx}{dt}\] \[= \dfrac{120-0}{10-0}\] \[v_{avg}= 12\dfrac{m}{s}\]

**Average velocity of car for time interval**$t=0s$ and $t=10s$ is $ v_{avg}=12\dfrac{m}{s}$.

**Part (b)**

**Instantaneous velocity**

**of car at $t= 0s$; $t=5s$ and $t=10s$.**\[v_{x}=\dfrac{dx}{dt} = 2bt-3ct^{2}\]

**At $t=0$**\[v_{x}= 2b(0) -3c (0)^{2}=0\]

**At $t=5s$**\[v_{x} = 2\times 2.4\times 5- 3\times 0.12\times 25\] \[= 15\dfrac{m}{s}\]

**At $t=10s$**\[v_{x}= 2\times 2.4\times 10 – 3\times 0.12 \times 100\] \[= 12\dfrac{m}{s}\]

**Instantaneous velocity**at the time interval $t=0$;$t=5s$ and $t=10s$ are $v_{x}=0$,$v_{x}=15\dfrac{m}{s}$ and $v_{x}=12\dfrac{m}{s}$.

**Part (c)**The

**car is at rest means**that: \[v_{x} = 0\dfrac{m}{s}\]

**Put the equation to zero**, and we can find at what time the

**car will be at rest again.**\[2bt-3ct^{2}=0\] \[t=\dfrac{2b}{3c}\] \[t=\dfrac{2(2.40)}{3(0.120)}\] \[t=13.33s\] The

**car will be at rest after**$t=13.33s$.

**Numerical Result**

– The **average velocity of the car for the time interval**$t=0s$ and $t=10s$ is $ v_{avg}=12\dfrac{m}{s}$. – The

**instantaneous velocity of the car**at $t=0$;$t=5s$ and $t=10s$ are $v_{x}=0$,$v_{x}=15\dfrac{m}{s}$ and $v_{x}=12\dfrac{m}{s}$. – The

**car will be at rest after**$t=13.33s$.

**Example**

**The car is stopped at a traffic light. He then drives along a straight road such that his distance from the light is given $x(t)=bt^{2}+ct^{3}$, where $b=1.40\dfrac{m}{s^{2}}$ and $c=0.120\dfrac{m}{s^{3}}$.**

**1: Calculate the car’s average velocity at time interval $=5s$.**

**2: Find the instantaneous velocity of the car at time interval $t=5s$ and $t=10s$.**

**Solution**

**Distance from the light is shown**by the equation $x(t) = bt ^ {2} + ct ^ {3}$. Where $b = 1.40 \dfrac{m}{s^{2}}$ and $c=0.120 \dfrac{m}{s^{3}}$.

**Part 1**

**Distance from the light**is: \[x(t) = bt ^ {2} + ct ^ {3} \]

**At $t=5s$**\[x(t) = 1.40\times 5^{2} + 0.12 \times 5^{3}\] \[x(t) = 1.40\times 25 + 0.12\times 125\] \[x(t) = 50m\]

**Average velocity of car**for time interval $t=5s$ is given by: \[v_{avg}= \dfrac{dx}{dt}\] \[= \dfrac{50-0}{5-0}\] \[v_{avg}= 10\dfrac{m}{s}\]

**Part (b)**

**Instantaneous velocity**

**of the car at $t=5s$ and $t=10s$.**\[v_{x}=\dfrac{dx}{dt} = 2bt+3ct^{2}\]

**At $t=5s$**\[v_{x} = 2\times 1.4\times 5 + 3\times 0.12\times 25\] \[= 23\dfrac{m}{s}\]

**At $t=10s$**\[v_{x}= 2\times 1.4\times 10 + 3\times 0.12 \times 100\] \[= 64\dfrac{m}{s}\]

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