(a) Explanation Given information: The constant speed ( v B ) 0 of car B is 60 mi/h . The cars pass each other at position ( x A ) is 294 ft . The speed of the car A ( v A ) is equal to car B ( v B ) . The deceleration ( a B ) of car B is − a A 6 . Calculation: Write equation for velocity of car A ( v A ) when t ≥ 0 : v A = ( v A ) 0 + a A t Here, ( v A ) 0 is initial velocity of car A , a A is acceleration of car A and t is time. Substitute 0 for ( v A ) 0 . v A = 0 + a A t = a A t (1). Write equation for position of car A ( x A ) when t ≥ 0 : x A = ( x A ) 0 + ( v A ) 0 t + 1 2 a A t 2 Here, ( x A ) 0 is initial position of car A . Substitute 0 for ( x A ) 0 , 294 ft for x A and 0 for ( v A ) 0 . 294 = 0 + ( 0 t ) + 1 2 a A t 2 294 = 1 2 a A t 2 (2). Calculate the position ( x B ) car B when ( t ) is 5sec: x B = ( x B ) 0 + ( v B ) 0 t Substitute 0 for ( x B ) 0 , 60 mi/h for ( v B ) 0 , and 5 sec for t . x B = 0 + ( 60mi/h × 1 (b) To determine When ( t A B ) the vehicle pass each other. (c) To determine The distance ( d ) between the vehicles at time ( t ) is zero.

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

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

Publisher: McGraw-Hill Education

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Chapter 11.2, Problem 11.46P

(a)

To determine

The acceleration (aA).

(b)

To determine

When (tAB) the vehicle pass each other.

(c)

To determine

The distance (d) between the vehicles at time (t) is zero.

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Chapter 11.2, Problem 11.45P

Chapter 11.2, Problem 11.47P

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Chapter 11 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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The acceleration ( a A ) .

Solution Summary: The author calculates the acceleration of car A and the deceleration of the car B.

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The acceleration (aA).

When (tAB) the vehicle pass each other.

The distance (d) between the vehicles at time (t) is zero.

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Chapter 11 Solutions