Question

(1) A car travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equation x(t) = αt2 - βt3, where α = 1.46 m/s2 and β = 5.40 x10-3 m/s2. Calculate the average velocity of the car for each time interval: (a) t = 0 to t = 2.03 s; (b) t = 0 to t = 3.95 s; (c) t = 2.03 s to t = 3.95 s.

Answer 1

Answer:

nswer and Explanation:

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We are given:

The distance of the car from the stop sign as a function of time, x(t)=\alpha t^2- \beta t^3 , where \alpha=1.50\;\rm m/s^2 and \beta=0.0500\;\rm m/s^3

The velocity of an object is equal to the displacement made by the object per unit time. That is:

v=\dfrac{\Delta s}{\Delta t}

Here,

\Delta s is the displacement of the object in time \Delta t

a)

At t=0 , the distance of the car from the stop sign is:

\begin{align*} x(0)&=\alpha \left ( 0 \right )^2-\beta\left ( 0 \right )^3\\ &=0 \end{align*}

At t=2.00\;\rm s , the distance of the car from the stop sign is:

\begin{align*} x(2)&=1.50\;\rm m/s^2\times \left ( 2.00\;\rm s \right )^2-0.0500\;\rm m/s^3 \times \left ( 2.00\;\rm s \right )^3\\ &=6.00\;\rm m-0.4\;\rm m\\ &=5.60\;\rm m \end{align*}

The displacement made in this time interval is:

\begin{align*} \Delta s&=x(2)-x(0)\\ &=5.60\;\rm m-0\\ &=5.60\;\rm m \end{align*}

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The average velocity in this interval is:

\begin{align*} v&=\dfrac{\Delta s}{\Delta t}\\ &=\dfrac{5.60\;\rm m}{2.00\;\rm s-0}\\ &=\boxed{2.80\;\rm m/s} \end{align*}

b)

At t=4.00\;\rm s , the distance of the car from the stop sign is:

\begin{align*} x(2)&=1.50\;\rm m/s^2\times \left ( 4.00\;\rm s \right )^2-0.0500\;\rm m/s^3 \times \left ( 4.00\;\rm s \right )^3\\ &=24\;\rm m-3.2\;\rm m\\ &=20.8\;\rm m \end{align*}

The displacement made between t=0 and t=4.00\;\rm s is:

\begin{align*} \Delta s&=x(4)-x(0)\\ &=20.8\;\rm m-0\\ &=20.8\;\rm m \end{align*}

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The average velocity in this interval is:

\begin{align*} v&=\dfrac{\Delta s}{\Delta t}\\ &=\dfrac{20.8\;\rm m}{4.00\;\rm s-0}\\ &=\boxed{5.2\;\rm m/s} \end{align*}

c)

The displacement made between t=2.00\;\rm s and t=4.00\;\rm s is:

\begin{align*} \Delta s&=x(4)-x(2)\\ &=20.8\;\rm m-5.6\;\rm m\\ &=15.2\;\rm m \end{align*}

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The average velocity in this interval is:

\begin{align*} v&=\dfrac{\Delta s}{\Delta t}\\ &=\dfrac{15.2\;\rm m}{4.00\;\rm s-2.00\;\rm s}\\ &=\boxed{7.6\;\rm m/s} \end{align*}