Question

A car, initially traveling 28.0ft/s, steadily speeds up to 50.0ft/s in 7.40s. Determine all unknowns and answer the following question.
How far did the car travel during this time?

Answer 1

Explanation:

Given:

$v_0 = 28.0\:\text{ft/s}$

$v = 50.0\:\text{ft/s}$

$t = 7.40\:\text{s}$

First, we calculate the acceleration of the car during this time:

$v = v_0 + at \Rightarrow a = \dfrac{v - v_0}{t}$

Plugging in the given values, we get

$a = \dfrac{50.0\:\text{ft/s} - 28.0\:\text{ft/s}}{7.40\:\text{s}} = 2.97\:\text{ft/s}^2$

Now that we have the value for the acceleration, we can solve for the distance traveled during the time t:

$x = v_0t + \frac{1}{2}at^2$

$\:\:\:\:=(28.0\:\text{ft/s})(7.40\:\text{s})$

$\:\:\:\:\:\:\:\:\:\:\:\:+ \frac{1}{2}(2.97\:\text{ft/s}^2)(7.40\:\text{s})^2$

$\:\:\:\:= 289\:\text{ft}$