An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m

Question

2 years ago

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An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m

and has an initial velocity of v0 = 2.2 m/s. How far will the object move before it achieves a velocity of v = 7.2 m/s? Your answer should be accurate to the nearest 0.1 m.

Physics

1 answer:

solong [7] · Answer 1 · 2022-02-27T11:08:13+03:00

Answer:

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s

With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

= 3.9m + 3.6m = 7.5m