Velocity of a moving partical v decreases with its displacement. Given, `v = v_(0) - alpha`x where `v_(0)`= initial velocity ,x = displacement and alpha is a constant. How long will the partical take to reach a point B on the x- axis at a distance `x_(m)`from the origin?

Explanation

Given, v = `v_(0) - ax`.
At the starting point, v = `v_(0)`
`:." " v_(0) = v_(0) -ax " ""or", x = 0 [ because alpha` = constant ]
Hence, the particle was initially at the origin.
Now, `v = (dx)/(dt) = v_(0) `- ax
or,`" " (dx)/(v_(0)-ax) = dt " ""or", (-alpha dx)/(v_(0)-alphax) = - alpha dt`
Then, `int (-alpha dx)/(v_(0) -alphax) = -alpha int dt " ""or", "log"_(e) (v_(0)-ax) = -at + c,` [ c is the integration constant ]
At t = 0 (initially), x = 0
`:. " " c = "log"_(e) v_(0)`
Therefore , `"log"_(e) (v_(0) - alphax) = -alpha t + "log"_(e) v_(0)`
or, `" " t = (1)/(alpha) "log"_(e) (v_(0))/(v_(0) - alphax)`
At the point B , x = `x_(m)` . Let the corresponding time be `t_(m)`.
`:. " " t_(m) = (1)/(alpha) "log"_(e) (v_(0))/(v_(0) - alpha x_(m))`.