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An object is moving with an initial velocity of 6.5m/s. It is then subject to a constant acceleration of 2.1m/s^2 for 14s. How far will it have traveled during the time of its acceleration?

Respuesta :

Answer:

Distance travel s = 296.8 m

Explanation:

Given:

Initial velocity u = 6.5 m/s

Constant acceleration a = 2.1 m/s²

Time T = 14 s

Find:

Distance travel s

Computation:

s = ut + (1/2)at²

s = (6.5)(14) + (1/2)(2.1)(14)²

s = 91 + 205.8

Distance travel s = 296.8 m

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