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cestrela7 [59]
3 years ago
5

A runner achieves a velocity of 11.1 m/s nine seconds after he begins. What is his acceleration

Physics
2 answers:
Delvig [45]3 years ago
6 0

Answer:

=1.23m/s^{2}  (2 decimal places

Explanation:

bogdanovich [222]3 years ago
6 0

Answer:

1.23

Explanation:

Because if we divide: 11.1 ÷ 9 = 1.23

to make sure: 1.23 x 9 = 11.07

and that's the closest to 11.1 so it count's as an answer.

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Answer:

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v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

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59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

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