Explanation Given: The plot of a − t is given as a parabola a = − 3 ( t − 2 ) 2 m/s 2 for an interval 0 < t ≤ 4 s . At t = 0 s and x = 0 m , the value of velocity v = 8 m/s . Concept used: Change in the velocity for the given a − t plot in the time interval is the area under the a − t graph for that duration. v f − v i = area under the a − t curve ...... (1) Here v f is the final velocity and v i is the initial velocity. Change in the displacement for the given v − t plot in the interval is the area under the v − t graph for that duration. x f − x i = area under the v − t curve ...... (2) Here x f is the final displacement and x i is the initial displacement. Write the formula for area under a parabolic spandrel. The shaded area under the parabolic spandrel is written as follows: Area=2 ( 1 3 b h ) ...... (3) Calculation: The a − t curve is given as parabolic in nature and the area under the curve will give the velocity change in that duration. To find the velocity at t = 2 s , substitute initial velocity 8 m/s for v i and half the area under the parabolic spandrel. v f − 8 = 1 3 ( 2 ) ( − 12 ) = 8 − 8 v f = 0 m/s To find the velocity at t = 4 s ,substitute initial velocity 0 ft/s for v i and half the area under the parabolic spandrel. v f − 0 = 1 3 ( 2 ) ( − 12 ) = − 8 v f = − 8 m/s Since the acceleration curve is parabolic, the velocity curve equation will be 1 degree higher than the acceleration curve To determine (b) The position of the particle after 3 seconds.

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 11.3, Problem 11.87P

To determine

(a)

To construct:

The v−t and x−t curves for 0≤t≤4 s.

To determine

(b)

The position of the particle after 3 seconds.

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Chapter 11.3, Problem 11.86P

Chapter 11.3, Problem 11.88P

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Vector Mechanics for Engineers: Dynamics

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