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Neporo4naja [7]
3 years ago
13

A car traveling 36 mi/h accelerates uniformly

Physics
1 answer:
iren2701 [21]3 years ago
7 0

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

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