Solution of Assignment (2)

# Solution of - 1 Solution of Assignment(2 Rectilinear Motion-2 Problem 7 A particle is moving along a straight line such that its acceleration is

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Unformatted text preview: 1 Solution of Assignment (2) Rectilinear Motion-2 Problem 7: A particle is moving along a straight line such that its acceleration is defined by: a = - 2v m/s 2 , where v is in m/s. if the initial conditions of the motion are v o = 20 m/s, x o = 0. Find the relations (v.t) and (x,t). What is the distance which the particle travels before it stops? SOLUTION: v 2 a − = Q , s / m 20 v o = , x o = v 2 dt dv a − = = → t d 2 v v d − = → ∫ ∫ − = t v 20 t d 2 v v d → t 2 20 ln v ln − = − → t 2 20 v ln − = → t 2 e 20 v − = t 2 e 20 v − = ………………………….. ( v , t ) t 2 e 20 dt dx v − = = → dt e 20 dx t 2 − = → dt e 20 dx t t 2 x ∫ ∫ − = → [ ] t t 2 e 2 20 x − − = [ ] t 2 e e 10 x − − = − → [ ] 1 e 10 x t 2 − − = − → [ ] t 2 e 1 10 x − − = …………………… .( x , t ) At v = 0: → e 20 t 2 = − → e t 2 = − → ∞ → t At ∞ → t : [ ] ∞ − − = e 1 10 x → m 10 x = 2 Problem 8: The initial conditions of the rectilinear motion of a particle are: x o = 8 m, v o = 6 m/s The acceleration of the particle is defined by: a = 2/v m/s 2 . Determine the position and the time after which the velocity becomes 12 m/s. SOLUTION: m 8 x o = , s / m 6 v o = 2 s / m v 2 a = Q → v 2 dt dv a = = → dt 2 vdv = → ∫ ∫ = t 6 dt 2 vdv → [ ] t v 6 2 t 2 2 v =         → t 2 2 36 2 v 2 − = − → t 4 36 v 2 = − → 36 t 4 v 2 + = → 36 t 4 v + ± = The positive sign is selected since the motion starts in the positive direction and is always accelerating ( a , v have the same sign at all moments): 9 t 2 v + = ………………. ( v , t ) 9 t 2 dt dx v + = = → dt 9 t 2 dx + = → dt 9 t 2 dx t x 8 ∫ ∫ + = → ( ) ( ) t 2 / 3 2 / 3 9 t 2 8 x       + = − → ( ) t 2 / 3 3 9 t 4 8 x         + = − → ( ) 3 ) 9 ( 4 9 t 3 4 8 x 2 / 3 2 / 3 − + = − 36 ) 9 t ( 3 4 8 x 2 / 3 − + = − 28 ) 9 t ( 3 4 x 2 / 3 − + = ( x , t ) At v = 12 m/s : 9 t 2 12 + = → 9 t 36 + = → onds sec 27 t = At t = 27 seconds: 28 ) 9 27 ( 3 4 x 2 / 3 − + = → m 260 x = . 3 Problem 9: A car has an initial velocity of 24 m/s and a constant deceleration of 3 m/s 2 ....
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## This note was uploaded on 12/21/2010 for the course ENGINEERIN mp108 taught by Professor Elbarki during the Spring '08 term at Alexandria University.

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Solution of - 1 Solution of Assignment(2 Rectilinear Motion-2 Problem 7 A particle is moving along a straight line such that its acceleration is

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