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.
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. The
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
v
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
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 b
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
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 ,
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
2.69: 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
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
2.67: 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
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 cov
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 10
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
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
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)
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 harm
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
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 ,
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 posi
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
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
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 i
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
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
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
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 res
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 ba
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 ) (9.8