course PHY 201
If I understand this correctly, then acceleration is not a needed value in finding distance, if you have the two velocities and the 'dt.You can use acceleration if you have only one velocity though.
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If I understand this correctly, then acceleration is not a needed value in finding distance, if you have the two velocities and the 'dt.You can use acceleration if you have only one velocity though.
If an object increases velocity at a uniform rate from 6 m/s to 29 m/s in 13 seconds, what is its acceleration and how far does it
Sketch a velocity vs. clock time graph for an object whose initial velocity is 6 m/s and whose velocity 13 seconds later is 29 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
29 - 6 = 23m/sec/13sec
1.77m/sec^2
6m/s + 29m/s = 35m/s / 2 = 17.5
17.5m/s * 13s = 227.5
slope = acceleration.
area = distance from starting point. (6 + 29)/2 x base(time)
Excellent work.