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