An object starts from rest at the origin (x0 = 0) and travels in a straight line with a constant acceleration (a = constant). Suppose the relation between the position (x) and time (t) is given by x = (3.4 m/s2) t2 What is the acceleration of the object?

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Muscardinus Muscardinus · Answer 1 · 2019-09-27T07:06:09+02:00

Answer:

[tex]a=(6.8)\ m/s^2[/tex]

Explanation:

Given that,

An object starts from rest at the origin (x₀ = 0) and travels in a straight line with a constant acceleration (a = constant).

The relation between the position (x) and time (t) is given by :

[tex]x=(3.4t^2)\ m/s^2[/tex]

Let v is the velocity of the object.

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=\dfrac{d(3.4t^2)}{dt}[/tex]

[tex]v=(6.8t)\ m/s[/tex]

Let a is the acceleration of the object. It is given by :

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]a=\dfrac{d(6.8t)}{dt}[/tex]

[tex]a=(6.8)\ m/s^2[/tex]

So, the acceleration of the object is [tex](6.8)\ m/s^2[/tex]. Hence, this is the required solution.