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Marianna [84]
3 years ago
7

HELPPP PLEASE!!!!!!!

Physics
1 answer:
eduard3 years ago
4 0

Answer:

S = V t - 1/2 a t^2      a here is negative due to deceleration

S = 4 * 8 - 1/2 * 64 = 0

The object just returns to its original starting point

It started out at 8 m/s which was reduced to 0 m/s after 4 sec

In 4 more seconds it was moving at -8 m/s, just the opposite of where it started

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Answer:

A) The acceleration is zero

<em>B) The total distance is 112 m</em>

Explanation:

<u>Velocity vs Time Graph</u>

It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.

The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.

A. To calculate the acceleration, we use the formula:

\displaystyle a=\frac{v_2-v_1}{t_2-t_1}

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

\displaystyle a=\frac{8-8}{12-0}=0

This confirms the previous conclusion.

B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.

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