Question

A particle starts from rest from origin and it’s acceleration is given by a = `k/(x+4)^2     m⁄s^2`.Knowing that v = 4 m/s when x = 8m,find :
(1)Value of k
(2)Position when v = 4.5 m/s

Given :  Particle starts from rest
a = `k/(x+4)^2    m⁄s^2`
v = 4m/s at x = 8m
To find : Value of k and position when v= 4.5m/s

Answer 1

Solution:
We know that a = `v(dv)/(dx)`

`v(dv)/(dx)=k/((x+4)^2`

v dv = k(x+4)^-2 dx 
Integrating both sides

∫𝑣𝑑𝑣 =∫k(x+4) −2 dx
`(v^2)/ 2=(−k)/(x+4 )+ c1 `………(1)
Putting x=0 and v=0
`c1=k/4`  ……….(2)
`(v^2)/2 = (−k)/(x+4 )+ k/4`   …….(From 1 and 2) ……..(3)
k = 48
From (3)
`(v^2)/2 =( −48)/(x+4)  + 48/4`

`v^2=24 - 96/(x+4)`
Substituting v=4.5 m/s
`4.5^2 = 24 - 96/(x+4)`
`96/(3.75) = x+4`
x = 21.6 m
Value of k = 48
The particle is at a distance of 21.6 m from origin when v = 4.5 m/ s