# A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by x(t)=bt^2-ct^3, where b=2.40 m/s^2 and c=0.120 m/s^3. (a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s. (b) Calculate the instantaneous velocity of the car at t=0, t = 5.0 s, and t = 10.0 s. (c) How long after starting from rest is the car again at rest?

This 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:$v_{avg} =\dfrac{\Delta x}{\Delta t}$

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 $t=13.33s$.
– 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 $t=13.33s$.
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 $t=10s$.SolutionDistance 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 1Distance 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}$