Unformatted text preview: v + v0 = and 2x
t These can be solved simultaneously to obtain the speed v when the train reaches the end of
the crossing. Once v is known, Equation 2.4 can be used to find the time required for the
train to reach a speed of 32 m/s.
SOLUTION Adding the above equations and solving for v, we obtain 2x 2(20.0 m) v = 1 at + = 1 (1.6 m/s 2 )(2.4 s) +
= 1.0 × 101 m/s
2
2
t
2.4 s The motion from the end of the crossing until the locomotive reaches a speed of 32 m/s
requires a time
v − v0 32 m/s − 1.0 ×101 m/s
t=
=
= 14 s
a
1.6 m/s 2
______________________________________________________________________________
42. REASONING Since 1 mile = 1609 m, a quartermile race is L = 402 m long. If a car
crosses the finish line before reaching its maximum speed, then there is only one interval of
constant acceleration to consider. We will first determine whether this is true by calculating
the car’s displacement x1 while accelerating from rest to top speed from Equation 2.9 (v 2 ) 2
= v0 + 2ax , with v0 = 0 m/s and v = vmax:
2
vmax = ( 0 m/s ) + 2ax1
2 or x1 = 2
vmax 2a (1) Chapter 2 Problems 67 If x1 > L, then the car crosses the finish line before reaching top speed, and the total time for ( its race is found from Equation 2.8 x = v0t + 1 at
2 L = ( 0 m/s ) t + 1 at 2 = 1 at 2
2
2 2 ) , with x = L and v 0 t= or 2L
a = 0 m/s:
(2) On the other hand, if a car reaches its maximum speed before crossing the finish line, the
race divides into two intervals, each with a different constant acceleration. The
displacement x1 is found as given in Equation (1), but the time t1 to reach the maximum
speed is most easily found from Equation 2.4 ( v = v0 + at ) , with v0 = 0 m/s and v = vmax:
vmax = 0 m/s + at1 t1 = or vmax
a (3) The time t2 that elapses during the rest of the race is found by solving Equation 2.8 (x = v t +
0 1 at 2
2 ). Let x2 = L − x1 represent the displacement for this part of the race. With the aid of Equation (1), this becomes x2 = L − 2
vmax . Then, since the car is at its maximum
2a
speed, the acceleration is a = 0 m/s2, and the displacement is
2 ( ) 2
x2 = vmax t2 + 1 0 m/s 2 t2 = vmax t2
2 t2 = or x2
vmax v
L − max
2a = L − vmax (4)
=
vmax
vmax
2a Using this expression for t2 and Equation (3) for t1 gives the total time for a twopart race:
t = t1 + t2 = L
v
+
− max
v
a
2a max vmax v
L
+ max
= 2a vmax (5) SOLUTION First, we use Equation (1) to determine whether either car finishes the race
while accelerating: vmax
(106 m/s ) = 511 m
=
2
2a
2 11.0 m/s
2 2 Car A Car B x1 =
x1 = ( 2
vmax 2a = ) ( 92.4 m/s )2 ( 2 11.6 m/s 2 ) = 368 m Therefore, car A finishes the race before reaching its maximum speed, but car B has
402 m − 368 m = 34 m to travel at its maximum speed. Equation (2) gives the time for car
A to reach the finish line as 68 KINEMATICS IN ONE DIMENSION t= Car A 2 ( 402 m )
2L
=
= 8.55 s
2
a
11.0 m/s Equation (5) gives the time for car B to reach the finish line as t= Car B L
vmax + vmax
2a = 402 m
92.4 m/s
+
= 8.33 s
92.4 m/s 2 11.6 m/s 2 ( ) Car B wins the race by 8.55 s − 8.33 s = 0.22 s . 43. SSM REASONING AND SOLUTION
a. Once the pebble has left the slingshot, it is subject only to the acceleration due to gravity. Since the downward direction is negative, the acceleration of the pebble is –9.80 m/s 2 . The pebble is not decelerating. Since its velocity and acceleration both point downward, the
magnitude of the pebble’s velocity is increasing, not decreasing.
b. The displacement y traveled by the pebble as a function of the time t can be found from
Equation 2.8. Using Equation 2.8, we have y = v0t + 1 a y t 2 = (–9.0 m/s)(0.50 s) + 1 (–9.80 m/s 2 )(0.50 s)2 = –5.7 m
2
2
Thus, after 0.50 s, the pebble is 5.7 m beneath the clifftop.
______________________________________________________________________________
44. REASONING Because there is no effect due to air resistance, the rock is in free fall from
its launch until it hits the ground, so that the acceleration of the rock is always −9.8 m/s2,
assuming upward to be the positive direction. In (a), we will consider the interval beginning
at launch and ending 2.0 s later. In (b), we will consider the interval beginning at launch
and ending 5.0 s later. Since the displacement isn’t required, Equation 2.4 ( v = v0 + at )
suffices to solve both parts of the problem. The stone slows down as it rises, so we expect
the speed in (a) to be larger than 15 m/s. The speed in (b) could be smaller than 15 m/s (the
rock does not reach its maximum height) or larger than 15 m/s (the rock reaches its
maximum height and falls back down below its height at the 2.0s point).
SOLUTION
a. For the interval from launch to t = 2.0 s, the final velocity is v = 15 m/s, the acceleration
is a = −9.8 m/s2, and the initial velocity is to be found. Solving Equation 2.4 ( v = v0 + at ) for v0 gives Chapter 2 Problems 69 2 v0 = v − at = 15 m/s − ( −9.8 m/s )(2.0 s) = 35 m/s Therefore, at launch,
Speed = 35 m/s b. Now we consider the interval from launch to t = 5.0 s. The initial velocity is that found
in part (a), v0...
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