Question

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?

Answer 1

`a=(6.8)\ m/s^2`

Step-by-step 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 :

`x=(3.4t^2)\ m/s^2`

Let v is the velocity of the object.

`v=(dx)/(dt)`

`v=(d(3.4t^2))/(dt)`

`v=(6.8t)\ m/s`

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

`a=(dv)/(dt)`

`a=(d(6.8t))/(dt)`

`a=(6.8)\ m/s^2`

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