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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 12

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online