At the moment `t=0` a particle leaves the origin and moves in the positive direction of the x-axis. Its velocity varies with time as `v=v_0(1-t//tau)`, where `v_0` is the initial velocity vector whose modulus equals `v_0=10.0cm//s`, `tau=5.0s`. Find:
(a) the x coordinate of the particle at the moments of time `6.0`, `10`, and `20s`,
(b) the moments of time when the particles is at the distance `10.0cm` from the origion,
(c) the distance s covered by the particle during the first `4.0` and `8.0s`, draw the approximate plot `s(t)`.

Answer

(a) `x=v_0t(1-t//2pi)`, `x=0.24`, 0 and `-4.0m`;
(b) `1.1, 9` and `11s`; (c) `{((1-t//2tau)v_0t, for, t, <=, tau),([1+(1-t//pi)^2]v_0t//2, for, t, >=,tau):}` 24 and 34cm respectively.

Explanation

(a) As the particle leaves the origin at `t=0`
So, `Deltax=x-int v_xdt` (1)
As `vecv=vec(v_0)(1-t/tau)`,
where `vec(v_0)` is directed towards the `+ve` x-axis
So, `v_x=v_0(1-t/tau)` (2)
From (1) and (2),
`x=underset0oversettintv_0(1-t/tau)dt=v_0t(1-(t)/(2pi))` (3)
Hence x coordinate of the particle at `t=6s`.
`x=10xx6(1-(6)/(2xx5))=24cm=0*24m`
Similary at
`x=10xx10(1-(10)/(2xx5))=0`
and at
`x=10xx20(1-(20)/(2xx5))=-200cm=-2m`
(b) At the moments the particle is at a distance of `10cm` from the origin, `x=+-10cm`.
Putting `x=+10` in Eq. (3)
`10=10t(1-(t)/(10))` or, `t^2-10t+10=0`,
So, `t=t=(10+-sqrt(100-40))/(2)=5+-sqrt(15)s`
Now putting `x=-10` in Eqn (3)
`-10=10(1-t/10)`,
On solving, `t=5+-sqrt(35)s`
As t cannot be negative, so,
`t=(5+sqrt(35))s`
Hence the particle is at a distance of `10cm` from the origin at three moments of time:
`t=5+-sqrt(15)s, 5+sqrt(35)s`
(c) we have `vecv=vec(v_0)(1-t/tau)`
So, `v=|vecv|={:(v_(0)(1-(t)/(tau))),(v_(0)((t)/(tau)-1)):}}{:("for t" le tau),("for t" gt tau):}`
So `s=underset0oversettintv_0(1-t/tau)dt` for `tletau=v_0t(1-t//2tau)`
and `s=underset0oversettauintv_0(1-t/2)dt+undersettauoversettintv_0(t/tau-1)dt` for `tgttau`
`=v_0tau[1+(1-t//tau)^2]//2` for `tgttau` (A)
`s=underset0overset4intv_0(1-t/tau)dt=underset0overset4int10(1-t/5)dt=24cm`.
And for `t=8s`
`s=underset0overset5int10(1-t/5)dt+underset5overset8int10(t/5-1)dt`
On integrating and simplifying, we get
`s=34cm`.
On the basis of Eqs. (3) and (4), `x(t)` and `s(t)` plots can be drawn as shown in the answer sheet.