{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 0 m i h 147 ft s 160 ft 2 331 ft s 2 1 m i h 2

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 + (...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online