First convert from mi/h to ft/s. There are 5280 ft to 1 mi, and 3600 s to 1 h, so
36 mi/h = (36 mi/h) * (5280 ft/mi) * (1/3600 h/s) = 52.8 ft/s
Let <em>a</em> be the acceleration of the car. The car's speed at time <em>t</em> is
<em>v</em> = 52.8 ft/s + <em>a</em> <em>t</em>
so that after 5.4 s, it attains a speed of
<em>v</em> = 52.8 ft/s + (5.4 s) <em>a</em>
Recall that
<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>
where <em>u</em> is the car's initial velocity and ∆<em>x</em> is the distance it's traveled.
We have
(52.8 ft/s + (5.4 s) <em>a</em>)² - (52.8 ft/s)² = 2 <em>a</em> (595 ft)
Omitting units, this equation reduces to
(52.8 + 5.4 <em>a</em>)² - 52.8² = 1190 <em>a</em>
==> 29.16 <em>a</em>² - 619.76 <em>a</em> = 0
==> 29.16 <em>a</em> - 619.76 = 0
==> 29.16 <em>a</em> = 619.76
==> <em>a</em> ≈ 21.25 ft/s²
