An object is released from rest near a planet's surface. A graph of the acceleration as a function of time for the object is shown for the 4s after the object is released. The
positive direction is considered to be upward. What is the displacement of the object after 2s?

The displacement of the object moving downwards is 10 m.

The given parameters;

The displacement of the object moving downwards (negative direction) is calculated using the second kinematic equation as shown below;

s = ut - ¹/₂at²

s = 0 - 0.5 x (-5) x (2)²

s = 10 m

Thus, the displacement of the object moving downwards is 10 m.

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The displacement and direction of the body after 2 seconds is 10 meters downwards

Explanation:

The given information are;

From the graph, the acceleration, is seen as the horizontal line that starts from the y-coordinate, y = -5 m/s²

Therefore;

The acceleration of the body after it was released from rest = -5 m/s² with an upward direction;

Which gives, the acceleration of the body downwards = Opposite sign to the acceleration of the body upwards = 5 m/s²

The displacement, s, of the body after 2 seconds is given by the following equation of motion;

s = u·t + 1/2·a·t²

Where;

u = The initial velocity of the body = 0 m/s

t = The time duration of the displacement = 2 s

a = The acceleration of the body = 5 m/s², downwards

Therefore, by substituting the values, we have;

s = 0 × 2 + 1/2 × 5 × 2² = 10 meters

s = 10 meters

The displacement of the body after 2 seconds =  10 meters

Given that the direction of the displacement and the direction of the acceleration are the same, we have;

The displacement and direction of the body after 2 seconds = 10 meters downwards.

yup its correct its: c) b b