The displacement x of a particle moving along a straight line, at time t is given by x=ab(1−e−bt) where a and b are real positive constants. Therfore:

x=ab(1−e−bt)v=dxdt=ae−bt ∴x=−vb+ab (a) acc=−abe−bt (b) xmax=ab (c) Velocity is always +ve, hence the particle does not return to starting point.

The displacement x of a particle moving along a straight line, at time t is given by x=ab(1−e−bt) where a and b are real positive constants. Therfore:

x=ab(1−e−bt)v=dxdt=ae−bt ∴x=−vb+ab (a) acc=−abe−bt (b) xmax=ab (c) Velocity is always +ve, hence the particle does not return to starting point.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The displacement x of a particle moving along a straight line, at time t is given by $x = \frac{a}{b} (1 - e^{- b t})$ where a and b are real positive constants. Therfore:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The graph of velocity vs displacement is a straight line with negative slope.

The velocity and acceleration of the particle at t = 0 are a and (-ab) respectively.

The particle cannot reach a point at a distance $x$ from its starting point if $x^{'} > \frac{a}{b}$

The particle will return to its starting point at $t \to \infty .$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} x = \frac{a}{b} (1 - e^{- b t}) \\ v = \frac{d x}{d t} = a e^{- b t} \end{array}$
$∴ x = - \frac{v}{b} + \frac{a}{b} (a)$
$a c c = - a b e^{- b t} (b)$
$x_{\max} = \frac{a}{b} (c)$
Velocity is always +ve, hence the particle does not return to starting point.