Question

A car moves on a straight track from station A to the station B, with an acceleration a = (b-cx), where b and c are constants and x is the distance from station A. The maximum velocity between the two stations is ?​

Answer 1

relation between acceleration and position is given as a = (b - cx)

here a denotes acceleration, b and c are constants and x is the distance from station A.

here acceleration is decreasing with increas x. so, velocity will be maximum at a = 0,

0 = b - cx ⇒x = b/c

we know, acceleration is the rate of change of velocity with respect to time.

so, a = dv/dt = dv/dx × dx/dt = v dv/dx

⇒∫vdv = ∫adx

then, $\int\limits^v_0{v}\,dv=\int\limits^{b/c}_0{(b-cx)}\,dx$

or, v²/2 = $\left[bx-\frac{cx^2}{2}\right]^{b/c}_0$

or, v²/2 = [b × b/c - c/2 × b²/c² ]

or, v²/2 = b²/c - b²/2c = b²/2c

or, v² = b²/c

or, v = b/√c

hence, maximum velocity is $\frac{b}{\sqrt{c}}$