Acyclist moving 11.5 m/s accelerates at 1.35 m/s squared till she reaches the finish line 58.9 m away. how much time does that take?
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4.12 seconds

Explanation:

Given:

v₀ = 11.5 m/s

a = 1.35 m/s²

x₀ = 0 m

x = 58.9 m

Find: t

x = x₀ + v₀ t + ½ at²

58.9 = 0 + (11.5) t + ½ (1.35) t²

58.9 = 11.5 t + 0.675 t²

0 = 0.675 t² + 11.5 t − 58.9

Solve with quadratic formula:

t = [ -11.5 ± √(11.5² - 4(0.675)(-58.9)) ] / 2(0.675)

t = (-11.5 ± √291.28) / 1.35

t = -21.2, 4.12

Since t must be positive, t = 4.12 seconds.

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