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A person walking for 0.25 min. can travel 10.5 m and then stops to rest. Find the person’s initial velocity and the person’s acceleration.

Sagot :

Yeelainehouidn

Solution1:

0.25 min = 15 secs

a = (v - [tex]v_{0}[/tex]) / t

v = 0 (stops to rest), t = 15

⇒ a = -[tex]v_{0}[/tex]/15

x = [tex]v_{0}[/tex]t + [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex]

⇒ 10.5 = [tex]v_{0}[/tex] * 15 + [tex]\frac{1}{2}[/tex] * [tex]\frac{-v0}{15}[/tex] * 225

10.5 = [tex]v_{0}[/tex] * 15 -

10.5 = [tex]v_{0}[/tex] * 7.5

[tex]v_{0}[/tex] = 1.4 (m/s)

⇒ a = -[tex]\frac{7}{75}[/tex] (m/s^2)

Solution2:

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