Explanation Given: The particle starts from origin with an initial velocity of v 0 = − 2 m/s . Concept used: Change in the velocity for the given a − t plot in theinterval 0 ≤ t ≤ 8 s is the area under the a − t graph for that duration. v 8 − v 0 = ( b ) ( h ) ...... (1) Here v 8 is the velocity at t = 8 s , v 0 is the velocity at t = 0 s , b is the breadth of the rectangle and h is the height of the rectangle. Change in the velocity for the given a − t plot in the interval 8 < t < 12 s is the area under the a − t graph for that duration. v 12 − v 8 = ( b ) ( h ) ...... (2) Here v 12 is the velocity at t = 12 s , v 8 is the velocity at t = 8 s , b is the breadth of the rectangle and h is the height of the rectangle. Change in the velocity for the given a − t plot in the time interval 12 < t < 18 s is the area under the a − t graph for that duration. v 18 − v 12 = ( b ) ( h ) ...... (3) Here v 18 is the velocity at t = 18 s , v 12 is the velocity at t = 12 s , b is the breadth of the rectangle and h is the height of the rectangle. 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 8 − x 0 = a 1 ...... (4) Here x 8 is the displacement at t = 8 s , x 0 is the displacement at t = 0 s and a 1 is the area under v − t plot. 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 12 − x 8 = a 2 ...... (5) Here x 12 is the displacement at t = 12 s , x 8 is the displacement at t = 8 s and a 2 is the area under v − t plot. Change in the displacement for the given v − t plot in the time interval is the area under the v − t graph for that duration. x 18 − x 12 = a 3 ...... (6) Here x 18 is the displacement at t = 18 s , x 12 is the displacement at t = 12 s and a 3 is the area under v − t plot. Calculation: Substitute − 2 m/s for initial displacement v 1 and the area under the curve for time interval 0 ≤ t ≤ 8 s in equation (1). v 8 − ( − 2 ) = ( 8 ) ( − 0.75 ) v 8 = − 2 + − 6 = − 8 m/s As the initial velocity is negative and the acceleration is also negative, the v − t curve will have a constant negative slope To determine (b) The position,velocity and the total distance traveled.

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 11.88P

To determine

(a)

The construction of v−t and x−t curves for 0<t<18 s.

To determine

(b)

The position,velocity and the total distance traveled.

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Chapter 11.3, Problem 11.87P

Chapter 11.4, Problem 11.CQ3P

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