A car traveling 36 mi/h accelerates uniformly for 5.4 s, covering 595 ft in this time.
What was its acceleration? Round your
answer to the nearest 100th place.
Answer in units of ft/s.

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 a be the acceleration of the car. The car's speed at time t is

v = 52.8 ft/s + a t

so that after 5.4 s, it attains a speed of

v = 52.8 ft/s + (5.4 s) a

Recall that

v² - u² = 2 a ∆x

where u is the car's initial velocity and ∆x is the distance it's traveled.

We have

(52.8 ft/s + (5.4 s) a)² - (52.8 ft/s)² = 2 a (595 ft)

Omitting units, this equation reduces to

(52.8 + 5.4 a)² - 52.8² = 1190 a

==>  29.16 a² - 619.76 a = 0

==>  29.16 a - 619.76 = 0

==>  29.16 a = 619.76

==>  a ≈ 21.25 ft/s²

c.the charges if the proton and electron and gravity

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