Explanation:
Given:
[tex]v_0 = 28.0\:\text{ft/s}[/tex]
[tex]v = 50.0\:\text{ft/s}[/tex]
[tex]t = 7.40\:\text{s}[/tex]
First, we calculate the acceleration of the car during this time:
[tex]v = v_0 + at \Rightarrow a = \dfrac{v - v_0}{t}[/tex]
Plugging in the given values, we get
[tex]a = \dfrac{50.0\:\text{ft/s} - 28.0\:\text{ft/s}}{7.40\:\text{s}} = 2.97\:\text{ft/s}^2[/tex]
Now that we have the value for the acceleration, we can solve for the distance traveled during the time t:
[tex]x = v_0t + \frac{1}{2}at^2[/tex]
[tex]\:\:\:\:=(28.0\:\text{ft/s})(7.40\:\text{s})[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:+ \frac{1}{2}(2.97\:\text{ft/s}^2)(7.40\:\text{s})^2[/tex]
[tex]\:\:\:\:= 289\:\text{ft}[/tex]
