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

Answer:

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

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.

$v=\dfrac{dx}{dt}$

$v=\dfrac{d(3.4t^2)}{dt}$

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

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

$a=\dfrac{dv}{dt}$

$a=\dfrac{d(6.8t)}{dt}$

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

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