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, twodimensional 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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details