The motion of a body is given by the equation `dv//dt=6-3v`, where v is in m//s. If the body was at rest at `t=0`
(i) the terminal speed is `2 m//s`
(ii) the magnitude of the initial acceleration is `6 m//s^(2)`
(iii) The speed varies with time as `v=2(1-e^(-3t)) m//s`
(iv) The speed is `1 m//s`, when the acceleration is half initial value

Options

A. the terminal speed is `2.0 ms^(-1)`
B. the magnitude of initial acceleration is ` 6.0 ms^(-2)`
C. the speed varies with time as ` v(t) = 2 (1 -e^(-3)) m//s`
D. the speed is ` 1.0 ms^(-1)` when the acceleration is half the initial value.

Answer

B, C, D

Explanation

Acceleration, ` (d v(t))/(dt) = 6.0- 3 v(t)` …(t)
At ` t=0`, acceleration `= 6.0 ms^(-2)`
Thus option (b) is true
Form (i), ` (dv)/(6-3 v) = dt`
Integrating it within the conditions of motion, we have ` int_0^v = (dv)/(6-3 v)= int_0^t dt`
lrt ` 6 - 3 v = y ` or ` -3 dv = dy ` or ` dv =- dy//3`
:. ` int_6^(6-3 v) (-dy //3)/y = int_0^t dt or int_6^(6-3 v) (dt)/y =- int_0^t 3 dt`
`(log_ey)_6^(6-3 v) =- 3 t` ltBrgt or ` log_e (6-3 v) - log_e 6=- 3 t`
or ` log_e (6-3 v)-log_e 6 =- 3 t`
or ` log_e (6 - 3v)/6 =- 3t` or ` (6 - 3v0/6 e^(-3t)`
or ` 6- v =6 e^(-t) ` or ` 3 v= 6-6 e^(3t)`
or ` v=2 (1 -e^(-3t))` ,brgt Thes, option (c ) is true
When ` (ev(t))/(dt) = 1/2 xx 6 = 3.0 ` theta from (i) we have
` 3= 6.-3 v (t)
or ` v (t) = (6.0-3)//3 = 1. 0 ms^(-1)`
Thus , option (d) is true.