The figure shows the velocity graph of a 5.8 kg object as it moves along the x axis (Figure 1).
The slope of a velocity versus time graph is the acceleration. During the first 3 seconds, the speed increases from 0 m/s to 12 m/s. The acceleration is therefore 4m/s^2. This is the acceleration in 1 second. During the last 2 seconds, the speed goes from 12 m/s to 0 m/s. The acceleration is therefore -2 m/s^2.
What is the net force acting on this object at t = 1s?
F= 5.8 * 4 = 23.2N
What is the net force acting on this object at t = 7s?
at t = 1 the slope of the graph is 12/3 = 4
then a = 4 m/s/s
then F = ma = 5.8 (4) = 23.2 N
at t = 7 the slope is -12/6 = -6
then a = -6 m/s/s
the minus sign means the force is opposite the direction of motion
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