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Vesnalui [34]
3 years ago
6

A car’s velocity changes from 35 m/s to stopped in 13 seconds. Calculate acceleration.

Physics
1 answer:
padilas [110]3 years ago
5 0

Answer:

Acceleration = 3m/s^2

Vf= 0  Vi =35m/s   t= 13s

Explanation:

Acceleration = \frac{Change in velocity}{Change in time}\\                      = \frac{35m/s}{13s}\\             a       =   2.69m/s^2\\             a       =    2.7m/s^2\\    a   = 3m/s^2

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<h2><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em><em><u>S</u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em></h2>

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V^{2}  = 6^{2} + 2 x 0.3 x 61

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V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

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