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qaws [65]
2 years ago
12

A plane starts from rest accelerates to 40 m/s in 10 seconds. How far did the plane travel during this time?

Physics
1 answer:
tatuchka [14]2 years ago
3 0
<h3><u>Answer:</u></h3>

200 m

<h3><u>Explanation:</u></h3>

<u>We are given:</u>

Initial velocity of the plane (u) = 0 m/s

Final velocity of the plane (v) = 40 m/s

Time interval (t) = 10 seconds

Displacement of the plane (s) = x m

<u>Solving for x:</u>

<u>Acceleration of the plane</u>

v = u + at                                                     [<em>First equation of motion</em>]

40 = 0 + a(10)                                              [<em>replacing known variables</em>]

a = 4 m/s²                                                    [<em>dividing both sides by 10</em>]

<u>Displacement of the Plane:</u>

s = ut + 1/2 (at²)                                            [<em>Second equation of motion</em>]

s = (0)(10) + 1/2(4)(10)²                                  [<em>replacing known variables</em>]

s = 200 m

Hence, the Plane covers a distance of 200 m in the given time interval

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a).

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<u>Given the following data;</u>

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2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

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Mathematically, centripetal acceleration is given by the formula;

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Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

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Answer:

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