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?

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

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

1 a

2 b

3 b

4 d

5 d

6 d

explanation: