Question

The engine of a motor boat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to the velocity at that time. Find the velocity after 2 seconds of switching off the engine

Answer 1

Let v be the velocity

Given the engine of a motorboat moving 10 m/s.

After that the engine is shut off then the acceleration is negative.

So it be `(- "dv")/"dt"`

i.e., `"dv"/"dt"` = – v

The equation can be written as taking integration on both sides, we get

`"dv"/"dt"` = – dt

`int "dv"/"v" = int - "dt"`

log v = – t + log c

log v – log c = – t

`log ("v"/"c")` = – t

`"v"/"c" = "e"^-"t"`

v = `"ce"^-"t"`  ..........(1)

Initial condition:

Given that when t = 0, v = 10 m/sec i

Substituting in equation (1), we get !

10 = ce^–0

10 = ce°

10 = c

∴ c = 10

(1) ⇒ v = 10e^–t

When t = 2 find v

v = 10 e^–2 

v = `10/"e"^2`

The velocity after 2 seconds is `10/"e"^2`

i.e., v = `10/"e"^2`