2.78: The elevators to the observation deck of the Sears Tower in Chicago move from the ground floor to the 103rd floor observation deck in about 70 s. Estimating a single floor to be about 3.5 m (11.5 ft), the average speed of the elevator is 103703s.5 m
2.77: Let t1 be the fall for the watermelon, and t2 be the travel time for the sound to return. The total time is T t1 t 2 2.5 s . Let y be the height of the building, then, y 1 gt12 and y vs t 2. There are three equations and three unknowns. Eliminate t2
2.76: The time needed for the egg to fall is
t
2 h 9
2(46.0 m
1.80 m)
2
(9.80 m s )
3.00 s,
and so the professor should be a distance vyt = (1.20 m s )(3.00 s) = 3.60 m.
2.75: a) The particle's velocity and position as functions of time are
v x (t ) v0 x v0 x
x(t )
t 0
t 0
( 2.00 m s ) (3.00 m s )t ) dt
2
2
3
(2.00 m s )t
3.00 m s 2
3
t2,
(0.50 m s 3 )t 3
v x (t )dt
v0 x t (1.00 m s 2 )t 2
t (v 0 x
(1.00 m s 2 )t
(0.50 m
2.74: a) From Eq. (2.17), x(t) = t t 3 = (4.00 m s )t (0.667 m s )t3. From Eq. 3
3
(2.5), the acceleration is a(t) = 2t = ( 4.00 m s )t.
3
b) The velocity is zero at t
(a = 0 at t = 0, but this is an inflection point, not an
extreme). The extreme value
2.73: a) The most direct way to find the time is to consider that the truck and the car are initially moving at the same speed, and the time of the acceleration must be that which gives a difference between the truck's position and the car's position as
2
2.72: a) There are many ways to find the result using extensive algebra, but the most straightforward way is to note that between the time the truck first passes the police car and the time the police car catches up to the truck, both the truck and the ca
2.71: a) Travelling at 20 m s , Juan is x1 37 m (20 m s)( 0.80 s) 21 m from the spreader when the brakes are applied, and the magnitude of the acceleration will be 2 v1 a 2x1 . Travelling at 25 m s , Juan is x2 37 m (25 m s)( 0.80 s) 17 m from the spreade
2.70: The position of the cars as functions of time (taking x1 = 0 at t = 0) are
x1 1 2 at , 2 x2 D v 0 t.
The cars collide when x1 = x2; setting the expressions equal yields a quadratic in t,
1 2 at 2 v0 t D 0,
the solutions to which are
t
1 a
2 v0
2aD
v
2.69: a) xT
t
2 ( 40.0 m) (2.10 m s)
(1 2)aT t 2 , and with xT
40 .0 m , solving for the time gives
6.17 s
b) The car has moved a distance 1 2 aC 3.40 m s2 aCt x1 40.0 m 64.8 m, 2 aT 2.10 m s 2 and so the truck was initially 24.8 m in front of the car. c)
2.68: One convenient way to do the problem is to do part (b) first; the time spent 20 m accelerating from rest to the maximum speed is 2.5 m ss2 80 s. At this time, the officer is
x1
v12 2a
(20 m s) 2 2(2.5 m s )
2
80.0 m.
This could also be found from (1
2.67: The total distance you cover is 1.20 m 0.90 m 2.10 m and the time available is 1.20 m 0.80 s . Solving Eq. (2.12) for ax, 1.50 m s
ax 2 (x x0 ) v0 xt t2 2 (2.10 m) (0.80 m s)( 0.80 s) (0.80 s)2 4.56 m s .
2
2.64: Taking the start of the race as the origin, runner A's speed at the end of 30 m can be found from:
2 vA 2 v0A
2aA ( x x0 ) 0 2(1.6 m s 2 )(30 m) 96 m 2 s 2 vA 96 m 2 s 2 9.80 m s
A's time to cover the first 30 m is thus:
t vA v0A aA 9.80 m s 2 1 .6
2.62: a)
d
ct
(3.0 108
365 1 d m 4 )(1 y) s 1y
24 h 1d
3600s 1h
9.5 10 15 m
b) c) d) e)
d
ct
t t t
d c d c d c
m )(10 9 s) 0.30 m s 11 1.5 10 m 500 s 8.33 min 3.0 108 m s (3.0 108
2(3.84 108 m) 3.0 108 m s 3 109 mi 186,000 mi s
2.6 s 16,100 s 4.5 h
2.61: If the driver steps on the gas, the car will travel (20 m s)(3.0 s) (1 2)(2.3 m s 2 )(3.0 s)2 70 .4 m. If the brake is applied, the car will travel (20 m s)(3.0 s) (1 2)( 3.8 m s 2 )(3.0 s)2 42 .9 m, so the driver should apply the brake.
2.59: (a) Denote the time for the acceleration (4.0 s) as t1 and the time spent running at constant speed (5.1 s) as t2. The constant speed is then at1, where a is the unknown acceleration. The total l is then given in terms of a, t1 and t2 by
l 1 2 at1 2
2.57: Denote the times, speeds and lengths of the two parts of the trip as t1 and t2, v1 and v2, and l1 and l2. a) The average speed for the whole trip is
l1 t1
l2 t2
l1 l 2 (l1 v1 ) (l 2 v2 )
(76 km) (34 km)
76 km 88 km h 34 km 72 km h
82 km h,
or 82.3 k
2.56: a) 25.0 m 1.25 m s . 20.0 s 25 m b) 15 s 1.67 m s .
c) Her net displacement is zero, so the average velocity has zero magnitude. d) 50.0 m 1.43 m s . Note that the answer to part (d) is the harmonic mean, not the 35.0 s arithmetic mean, of the answe
2.55: a)
The velocity and acceleration of the particle as functions of time are
vx (t ) ax (t )
(9.00 m s3 )t 2
(20.00 m s 2 )t (9.00 m s)
(18.0 m s3 )t (20.00 m s 2 ).
b) The particle is at rest when the velocity is zero; setting v = 0 in the above expre
2.54: a) To average 4 mi hr , the total time for the twenty-mile ride must be five
hours, so the second ten miles must be covered in 3.75 hours, for an average of 2.7
mi hr . b) To average 12 mi hr , the second ten miles must be covered in 25 minutes and
2.53: a) The change in speed is the area under the ax versus t curve between vertical lines at t 2.5 s and t 7.5 s. This area is
(4.00 cm s 8.00 cm s )(7.5 s 2.5 s) 30.0 cm s This acceleration is positive so the change in velocity is positive.
b) Slope of
2.52: a) Slope a 0 for t 1.3 ms b) hm ax Area underv tgraph ATriangle ARectangle
1 cm (1.3 ms ) 133 2 s 0.25 cm (2.5 ms 1.3 ms )(133 cm s)
c) a = slope of v t graph
a (0.5 ms ) a (1.0 ms ) 133 cm s 2 1.0 10 5 cm s 1.3 ms 0 because the slope is zero.
a (1.
2.51: a) From Eqs. (2.17) and (2.18), with v0=0 and x0=0,
vx
x
t 0
t 0
( At Bt 2 ) dt
A 2 t 2 B 3 t 3
A 2 B 2 t t (0.75 m s 3 )t 3 (0.040 m s 4 )t 3 2 3 A 3 B 4 dt t t (0.25 m s 3 )t 3 (0.010 m s 4 )t 4 . 6 12
b) For the velocity to be a maximum, the acce
2.50: a) From Eq. (2.15), the velocity v2 at time t
v2
v1
t t1
t dt
v1
2
(t 2
t12 )
2 2 3 3 = (5.0 m s ) (0.6 m s )(1.0 s)2 + (0.6 m s ) t2
= (4.40 m s ) + (0.6 m s ) t2. At t2 = 2.0 s, the velocity is v2 = (4.40 m s ) + (0.6 m s )(2.0 s)2 = 6.80 m/s, or
2.49: a) For a given initial upward speed, the height would be inversely proportional to the magnitude of g, and with g one-tenth as large, the height would be ten times higher, or 7.5 m. b) Similarly, if the ball is thrown with the same upward speed, it
2.48: a) From Eq. (2.8), solving for t gives (40.0 m s 20.0 m s )/9.80 m s = 2.04 s. b) Again from Eq. (2.8),
40 .0 m s ( 20 .0 m s)
2
2
9.80 m s
6.12 s.
c) The displacement will be zero when the ball has returned to its original vertical position, with v
2.47: a) (224 m s) (0.9 s) 249 m s 2 . b) find the distance is vavet d) (283 m s) (1.40 s)
2
249 m s 2 9.80 m s 2
25.4. c) The most direct way to
2
(224 m s) 2)( 0.9 s) 101 m.
202 m s but 40 g 392 m s , so the figures are not consistent.
2.46: a) The vertical distance from the initial position is given by
y v0 y t 1 2 gt ; 2
solving for v0y,
v0 y y t 1 gt 2 ( 50 .0 m) (5.00 s) 1 2 (9.80 m s )(5.00 s) 2 14 .5 m s .
2 2 b) The above result could be used in v y v0 y 2g y y0 , with v = 0, to
2.45: a) v y
25 .6 m s .
v0 y
gt ( 6.00 m s) (9.80 m s )(2.00 s)
2
25.6 m s , so the speed is
1 2 1 2 gt ( 6.00 m s)(2.00 s) (9.80 m s )(2.00 s)2 31.6 m, with the 2 2 minus sign indicating that the balloon has indeed fallen. c) 2 2 v y v0 y 2g ( y0 y) (6.
2.44: a) Using ay = g, v0y = 5.00 m s and y0 = 40.0 m in Eqs. (2.8) and (2.12) gives i) at t = 0.250 s, 2 y = (40.0 m) + (5.00 m s )(0.250 s) (1/2)(9.80 m s )(0.250 s)2 = 40.9 m, vy = (5.00 m s ) (9.80 m s )(0.250 s) = 2.55 m s and ii) at t = 1.00 s, y =