Answer to Question #137825 in Physics for Sam Far

Question

Draw a box in the middle of a paper.
On left side of box acceleration at work named A, with an acceleration of 2.2 m/s^2 and an arrow pointing to the right (box) to show the acceleration. Also, below the arrow of A, there is another acceleration at work named B, with an acceleration of 3.5 m/s^2 and an arrow pointing to the right (box) to show the acceleration.
On right side of box acceleration at work named C, with an acceleration of -2.5 m/s^2 and an arrow pointing to the left (box) to show the acceleration. Additionally, below the arrow of C, there is another acceleration at work named D, with an acceleration of -3.7 m/s^2 and an arrow pointing to the left (box) to show the acceleration.

If the following accelerations are at work on the box, which way, and with what value will the box accelerate?
A. Final Acceleration = -2 m/s^2
B. Final Acceleration = 2 m/s^2
C. Final Acceleration = -0.5 m/s^2
D. Final Acceleration = 0.5 m/s^2
E. Final Acceleration = 5.7 m/s^2
F. Final Acceleration = -6.2 m/s^2

Answer 1

Just assume that negative is to the left, positive is to the right, and find the sum of all accelerations:

$$a=2.2+3.5-2.5-3.7=-0.5\text{ m/s}^2.$$