(a) Explanation Given information: The car is traveling at a constant speed ( v 0 ) is 54 km/h . The position ( x ) of car zero when time ( t ) is zero. The driver accelerated to resume her original speed ( v t ) of 54 km/h . Calculation: Show the a-t curve of the car with areas as in Figure 1). Convert the kilometer per hour to meter per sec. v 0 = v t = 54 km/h × 1000 m 1 km × 1 hr 3600 sec = 15 m/s Calculate the area ( A 1 ) using the relation: A 1 = a 1 t 2 Here, a 1 is the acceleration at time 1 sec and t 2 is time between 1 sec to 2 sec. Substitute − 6 m/s 2 for a 1 and 1 sec for t 2 . A 1 = − 6 × 1 = − 6 m/s Calculate the area ( A 2 ) using the relation: A 2 = 1 2 a 1 t 3 Here, t 2 is time between 2 sec to 4.5 sec. Substitute − 6 m/s 2 for a 1 and 2.5 sec for t 3 . A 2 = 1 2 × − 6 × 2.5 = − 7.5 m/s Calculate the area ( A 3 ) using the relation: A 3 = 1 2 a 2 ( t 1 − t 4.5 ) Here, a 2 is the acceleration at time 4.5 sec and t 4.5 is time at 4.5 sec. Substitute 2 m/s 2 for a 2 and 4.5 sec for t 4.5 (b) To determine The position ( x 18 ) of the car at time ( t ). (c) To determine The average velocity ( v a v e ) of the during the time interval 1 sec ≤ t ≤ t 1 .

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

12th Edition

ISBN: 9781260916942

Author: BEER

Publisher: MCG

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Chapter 11.3, Problem 11.78P

(a)

To determine

At time (t1) at which the velocity is again 54km/h.

(b)

To determine

The position (x18) of the car at time (t).

(c)

To determine

The average velocity (vave) of the during the time interval 1 sec≤t≤t1.

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Chapter 11.3, Problem 11.77P

Chapter 11.3, Problem 11.79P

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Chapter 11 Solutions

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

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At time ( t 1 ) at which the velocity is again 54 km/h .

Solution Summary: The author shows the a-t curve of the car with areas as in Figure 1).

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At time (t1) at which the velocity is again 54km/h.

The position (x18) of the car at time (t).

The average velocity (vave) of the during the time interval 1 sec≤t≤t1.

Want to see the full answer?

Chapter 11 Solutions