Physics Solution Manual for 1100 and 2101

Therefore we have that 4 3 1 average velocity 27 ms

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ONE DIMENSION Average velocity = Displacement ( 27 m/s ) t North + ( −17 m/s ) tSouth = Elapsed time tNorth + tSouth t North = ( 27 m/s ) t North + tSouth t North But t North + tSouth 3 = and 4 tSouth t North + tSouth tSouth + ( −17 m/s ) t North + tSouth 1 = . Therefore, we have that 4 3 1 Average velocity = ( 27 m/s ) + ( −17 m/s ) = +16 m/s 4 4 The plus sign indicates that the average velocity for the entire trip points north. 13. REASONING AND SOLUTION The upper edge of the wall will disappear after the train has traveled the distance d in the figure below. 0.90 m B B A A 12° b 2.0 m 12° d The distance d is equal to the length of the window plus the base of the 12° right triangle of height 0.90 m. The base of the triangle is given by b= 0.90 m = 4.2 m tan12° 0.90 m 12° b Thus, d = 2.0 m + 4.2 m = 6.2 m. The time required for the train to travel 6.2 m is, from the definition of average speed, x 6.2 m = = 2.1 s v 3.0 m/s ______________________________________________________________________________ t= Chapter 2 Problems 51 v − v0 14. REASONING The average acceleration ( a ) is defined by Equation 2.4 a = as t − t0 the change in velocity ( v − v0 ) divided by the elapsed time ( t − t0 ) . The change in velocity is equal to the final velocity minus the initial velocity. Therefore, the change in velocity, and hence the acceleration, is positive if the final velocity is greater than the initial velocity. The acceleration is negative if the final velocity is less than the initial velocity. (a) The final velocity is greater than the initial velocity, so the acceleration will be positive. (b) The final velocity is less than the initial velocity, so the acceleration will be negative. (c) The final velocity is greater than the initial velocity (–3.0 m/s is greater than –6.0 m/s), so the acceleration will be positive. (d) The final velocity is less than the initial velocity, so the acceleration will be negative. SOLUTION Equation 2.4 gives the average acceleration as a= v − v0 t − t0 Therefore, the average accelerations for the four cases are: (a) a = (5.0 m/s − 2.0 m/s)/(2.0 s) = +1.5 m/s 2 (b) a = (2.0 m/s − 5.0 m/s)/(2.0 s) = −1.5 m/s 2 (c) a = [−3.0 m/s − (−6.0 m/s)]/(2.0 s) = +1.5 m/s 2 (d) a = (−4.0 m/s − 4.0 m/s)/(2.0 s) = −4.0 m/s 2 ______________________________________________________________________________ v − v0 15. REASONING The average acceleration ( a ) is defined by Equation 2.4 a = as t − t0 the change in velocity ( v − v0 ) divided by the elapsed time ( t − t0 ) . The change in velocity is equal to the final velocity minus the initial velocity. Therefore, the change in velocity, and hence the acceleration, is positive if the final velocity is greater than the initial velocity. The acceleration is negative if the final velocity is less than the initial velocity. The acceleration is zero if the final and initial velocities are the same. SOLUTION Equation 2.4 gives the average acceleration as a= v − v0 t − t0 a. The initial and final velocities are both +82 m/s, since the velocity is constant. The average acceleration is 52 KINEMATICS IN ONE DIMENSION a = (82 m/s − 82 m/s)/(t – t0) = 0 m/s 2 b. The initial velocity is +82 m/s, and the final velocity is –82 m/s. The average acceleration is a = (−82 m/s − 82 m/s)/(12 s) = −14 m/s 2 ______________________________________________________________________________ 16. REASONING Although the planet follows a curved, two-dimensional path through space, this causes no difficulty here because the initial and final velocities for this period are in opposite directions. Thus, the problem is effectively a problem in one dimension only. ∆v Equation 2.4 a = relates the change ∆ v in the planet’s velocity to its average ∆t acceleration and the elapsed time ∆t = 2.16 years. It will be convenient to convert the elapsed time to seconds before calculating the average acceleration. SOLUTION a. The net change in the planet’s velocity is the final minus the initial velocity: ∆v = v − v 0 = −18.5 km/s − 20.9 km/s = −39.4 km/s km 1000 m 4 ∆v = −39.4 = −3.94 ×10 m/s s 1 km b. Although the planet’s velocity changes by a large amount, the change occurs over a long time interval, so the average acceleration is likely to be small. Expressed in seconds, the interval is 365 d 24 h 60 min 60 s 7 ∆t = 2.16 yr = 6.81×10 s 1 yr 1d 1 h 1 min ( ) Then the average acceleration is a= 17. ∆v −3.94 × 104 m/s = = −5.79 × 10−4 m/s 2 7 ∆t 6.81× 10 s SSM REASONING The average acceleration is defined by Equation 2.4 as the change in velocity divided by the elapsed time. We can find the elapsed time from this relation because the acceleration and the change in velocity are given. Since the acceleration of the spacecraft is constant, it is equal to the average acceleration. Chapter 2 Problems 53 SOLUTION a. The time ∆t that it takes for the spacecraft to change its velocity by an amount ∆v = +2700 m/s is ∆v +2700 m / s ∆t = = = 3.0 ×102 days m/s a +9.0 day b. Since 24...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online