Unformatted text preview: 0 2.00 CHAPTER 2.26 (a) 2 ⎛ v f + vi ⎞
⎛ 2.80 m s + v i ⎞
∆x = v ( ∆t ) = ⎜
⎟ ( 8.50 s ) ,
⎟ ∆t becomes 40.0 m = ⎜
⎝
⎠
2
⎝2⎠ which yields v i = 6.61 m s .
(b) a = 2.27 v f − vi
∆t = 2 .80 m s − 6.61 m s
= − 0.448 m s 2
8.50 s Suppose the unknown acceleration is constant as a car initially moving at
v i = 35.0 m i h comes to a v f = 0 stop in ∆x = 40.0 ft . We find its acceleration from
v 2 = v i2 + 2 a( ∆x ) .
f 0 − ( 35.0 m i h ) ⎛ 1.47 ft s ⎞
2
a=
=
⎜ 1 m i h ⎟ = − 33.1 ft s .
2 ( ∆x )
2 ( 40.0 ft )
⎝
⎠
v 2 − v i2
f 2 2 Now consider a car moving at v i = 70.0 m i h and stopping to v f = 0 with a = − 33.1 ft s 2 . From the same equation its stopping distance is ∆x = 2.28 v 2 − v i2
f
2a 0 − ( 70.0 m i h ) ⎛ 1.47 ft s ⎞
=
⎜
⎟ = 160 ft
2 −33.1 ft s 2 ⎝ 1 m i h ⎠
2 ( 2 ) (a) From the definition of acceleration, we have a= v f − vi
t = 0 − 40 m s
= − 8.0 m s 2 .
5.0 s 1
(b) From ∆x = v i t + at 2 , the displacement is
2
∆x = ( 40 m s ) ( 5.0 s ) + 2.29 ( ) 1
2
−8.0 m s 2 ( 5.0 s ) = 100 m .
2 (a) With v f = 120 km h , v 2 = v i2 + 2a( ∆x ) yields
f
2
⎡(120 km h ) 2 − 0⎤
⎣
⎦ ⎛ 0.278 m s ⎞ = 2.32 m s 2 .
a=
=
⎜ 1 km h ⎟
2 ( ∆x )
2 ( 240 m )
⎝
⎠ v 2 − v i2
f 29 CHAPTER (b) The required time is t = 2.30 v f − vi
a = 2 (120 km h − 0) ⎛ 0.278 m s ⎞
= 14.4 s .
2 .32 m s 2 ⎜ 1 k m h ⎟
⎝
⎠ (a) The time for the truck to reach 20 m s is found from v f = v i + at as t= v f − vi
a = 20 m s − 0
= 10 s .
2.0 m s 2 The total time is t total = 10 s + 20 s + 5.0 s = 35 s .
(b) The distance traveled during the first 10 s is 0 + 20 m
⎝
2 ( ∆x )1 = v1t1 = ⎛
⎜ s⎞
⎟ (10 s ) = 100 m .
⎠ The distance traveled during the next 20 s (with a = 0) is ( ∆x ) 2 = ( v i ) 2 t 2 + a2t 22 = ( 20 m s ) ( 20 s ) + 0 = 400 m .
1
2 The distance traveled in the last 5.0 s is ( ∆x ) 3 = v 3t 3 = ⎛
⎜
⎝ 20 m s +0 ⎞
⎟ ( 5.0 s ) = 50 m .
⎠
2 The total displacement is then ∆x = ( ∆x )1 + ( ∆x ) 2 + (...
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 Spring '05
 Dr.Ha
 Physics, Acceleration, Velocity

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