mechanics Q (1 Viewer)

[doiyoubi]

a particle moves in a straight line with retardation which increases uniformly with the distance moved. Initially the retardation is 5m/s^2 and when the particle has moved a distance of 12m the retardation is 11m/s^2. Find the distance moved by the particle in coming to rest if the initial velocity is 20m/s

this is from the cambridge ex 7.1 q 2

 

[_ShiFTy_]

First you have the work out the retardation.
retardation which increases uniformly with the distance moved
This hints a linear equation, so just let the retardation be R = -(ax + b)
Initially the retardation is 5m/s^2 and when the particle has moved a distance of 12m the retardation is 11m/s^2

x = 0, R = -5
5 = b

x = 12, R = -11
-11 = -12a - 5
a = 1/2

R = -(x/2 + 5)

..
x = -(x/2 + 5)
v.dv/dx = -(x/2 + 5)
v²/2 = -x²/4 -5x + c
v=20, x = 0, c = 200

v² = -2(x²/4 + 5x -200)

Find the distance moved by the particle in coming to rest
v=0, x?

0 = -2(x²/4 + 5x - 200)
x²/4 + 5x - 200 = 0
x² + 20x - 800 = 0
(x - 20)(x + 40)
x > 0,

x = 20