Explanation Given information: s = 400 m Formula used: A = 1 2 b h v = s t Calculation: The v − t curve for the first 200 m is depicted below. For runner A. ⇒ t 1 = 4 s ⇒ t 2 = 25 − 4 ⇒ 21 s Now the area. ⇒ A 1 = 1 2 ( 4 ) ( v A ) max ⇒ 2 ( v A ) max Again the area ⇒ A 2 = 21 ( v A ) max Since, A 1 + A 2 = Δ x = 200 m ⇒ 23 ( v A ) max = 200 m ⇒ ( v A ) max = 8.6957 m / s For runner B. ⇒ t 1 = 5 s ⇒ t 2 = 25.2 − 5 ⇒ 20.2 s Now the area. ⇒ A 1 = 1 2 ( 5 ) ( v B ) max ⇒ 2.5 ( v B ) max Again. ⇒ A 2 = 20.2 ( v B ) max Since, A 1 + A 2 = Δ x ⇒ 200 m ⇒ 22 To determine (b) The position of the winner relative to the loser at given condition.

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 11.71P

To determine

(a)

The race time for both the runners at given condition.

To determine

(b)

The position of the winner relative to the loser at given condition.

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Chapter 11.3, Problem 11.70P

Chapter 11.3, Problem 11.72P

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(a) The race time for both the runners at given condition.

Solution Summary: The author calculates the v-t curve for the first 200m for runner A.