Explanation Given information: The speed of the car ( v C ) and truck ( v T ) is 35 mi/h . The distance ( x ) between the car and truck is 40 ft . The maximum acceleration ( a max ) of the car is 5 ft/s 2 . The maximum deceleration ( − a max ) of the car is 20 ft/s 2 . The length of the car ( x C ) is 16 ft . The length of the truck ( x T ) is 50 ft . Calculation: Calculate the distance ( Δ x ) of car relative to the truck using the relation: Δ x = x C + x + x T + x Substitute 16 ft for x C , 40 ft for x and 50 ft for x T . Δ x = 16 + 40 + 50 + 40 = 146 ft Calculate the allowable increase in speed ( Δ v m ) using the relation: Δ v m = v − v C Here, v is speed limit of car. Substitute 50 mi/h for v and 35 mi/h for v C . Δ v m = 50 − 35 = 15 mi/h = 15 mi/h × 1.46667 ft/s 1 mi/h = 22 ft/s Tabulate the velocity ( v ) corresponding to time ( t ) as in Table (1): t (s) v ( ft/s ) 0 35 t 1 50 t 2 50 t 3 35 Plot the v-t curve of the car with area as in Figure (1). Calculate the time ( t 1 ) using the relation: t 1 = Δ v m a max Substitute 22 ft/s for Δ v m and 5 ft/s 2 for a max . t 1 = 22 5 = 4.4 sec Calculate the time ( t 3 ) using the relation: t 3 = Δ v m − a max Substitute 22 ft/s for Δ v m and 20 ft/s 2 for − a max

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977251

Author: BEER

Publisher: MCG

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Chapter 11.3, Problem 11.72P

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The shortest time (tB) in which the driver of the car can complete the passing operation with speed of 50 mi/h and draw the v-t curve.

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Chapter 11.3, Problem 11.71P

Chapter 11.3, Problem 11.73P

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The shortest time ( t B ) in which the driver of the car can complete the passing operation with speed of 50 mi/h and draw the v-t curve.

Solution Summary: The author calculates the distance between the car and the truck using the relation: Delta x=x_C+x+

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The shortest time (tB) in which the driver of the car can complete the passing operation with speed of 50 mi/h and draw the v-t curve.

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Chapter 11 Solutions