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)`.

At the moment `t=0` a particle leaves the origin and moves in the posi

1.	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» Correct 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. (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.

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