The magnitude of the linear acceleration of a point moving along a vertical path is given by the equation a = 6t - 24, where a is in m/s and t is in seconds. The acceleration is upward when t = 5s; The point is 4m below the origin when t = 0 and 23 m above the origin when t = 3 s.Determine:
B.1. The velocity when t = 3 sec
B.2. The displacement during the time interval from t = 0 tot = 4 s
B.3. The total distance travelled during the time interval from t = 0 tot = 4s
Answer with diagram is given below:
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