Question-13563

Question Number 13563 by Tinkutara last updated on 21/May/17

Commented by Tinkutara last updated on 21/May/17

36 . The velocity of the particle at a

distance x from origin is

(1) u + k x

(2) u - k x

(3) k x

(4) - k x

37 . The velocity of the particle at time

t after start is

(1) u - k t

(2) {u e}^{- k t}

(3) e^{- k t}

(4) u - e^{- k t}

Commented by Tinkutara last updated on 21/May/17

Commented by Tinkutara last updated on 21/May/17

Answer these questions from paragraph .

Answered by ajfour last updated on 21/May/17

(36) .

a = - kv

v \frac{dv}{dx} = - kv

\int_{u}^{v} dv = - k \int_{0}^{x} dx

v - u = - kx

v = u - kx [option (2)] \dots . . (i)

(37) .

a = - kv

\frac{dv}{dt} = - kv

{\int_{u}}^{v} \frac{dv}{v} = - k \int_{0}^{t} dt

\ln (\frac{v}{u}) = - kt

v = {ue}^{- kt} [option (2)]

(38) .

a = - kv

= - k (u - kx) . . from (i)

if k > 0, u > 0

at t = 0, a = - ku

a = 0 at x = \frac{u}{k}

so a - x graph is linear with negative

a intercept and positive

x - intercept . [option (3)] .

Commented by Tinkutara last updated on 21/May/17

Q 38 . Why k > 0 implies u > 0 ? Answer

38 is not understood by me . Please

explain in detail Sir .

Commented by ajfour last updated on 21/May/17

the case is of initial speed u in a

certain direction, that i may take

to be the positive direction; so u > 0

what i meant that if u > 0 and

k > 0 both then {object}^{'} s velocity goes

decreasing, but is always positive and

reaches zero in infinite time;

the position coordinate goes near

and near its upper limit of x = \frac{u}{k},

because integrating velocity

x = \int_{0}^{t} vdt = u \int_{0}^{t} e^{- kt} dt

= u {[\frac{e^{- kt}}{- k}]}_{t = 0}^{t = t} = \frac{u}{k} (1 - e^{- kt})

as t \to \infty, x \to \frac{u}{k} .

Commented by Tinkutara last updated on 22/May/17

Thanks Sir!