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victus00 [196]
3 years ago
8

A dragster completed a 402.3-(0.2500-mi) run in 5.023s. If the car had a constant acceleration, what was its acceleration and fi

nal velocity?
Physics
1 answer:
Ray Of Light [21]3 years ago
6 0

Answer:

v = 80.092 m/s

a = 15.945 m/s²

Explanation:

Given,

The distance completed by the dragster, d = 402.3 m

The time taken by the dragster to complete that distance is, t = 5.023 s

The initial velocity of the dragster, u = 0

The final velocity of the dragster, v = ?

The acceleration of the dragster, a = ?

The velocity of the body is given by the formula

                                  v = d/t  m/s

Substituting the above values inn the equation

                                  v = 402.3 m / 5.023 s

                                     = 80.092 m/s

So, the velocity of the dragster is, v = 80.092 m/s

The acceleration of the body is given by the formula

                                   a = (v - u)/t  m/s²

Substituting the above values in the equation

                                      = (80.092 - 0) / 5.023

                                      = 15.945 m/s²

Hence, the acceleration of the dragster is, a = 15.945 m/s²

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<u>Given:</u>

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A diagram has been attached with the solution in order to clearly show the position of the plane.

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The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

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