An object is moving with an initial velocity of 5.5m/s.It is then subject to a constant acceleration of 2.5 m/s for 11s. How far will it have traveled during the time of its acceleration?

Respuesta :

The distance traveled by the object during the time of acceleration is 211.75 m.

What is distance?

Distance can be defined as the total lenght between two points.

T o calculate the distance traveled by the object during the time of acceleration, we use the formula below.

Formula:

  • s = ut+at²/2............ Equation 1

Where:

  • s = Distance
  • u = Initial velocity
  • a = Acceleration
  • t = Time

From the question,

Given:

  • u = 5.5 m/s
  • t = 11 s
  • a = 2.5 m/s²

Substitute these values into equation 1

  • s = (5.5×11)+(2.5×11²)/2
  • s = 60.5+151.25
  • s = 211.75 m

Hence, the distance traveled by the object is 211.75 m.

Learn more about distance here: https://brainly.com/question/26550516

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