Question

A train moves between two stations A and B along a straight line from A to B with a variable acceleration a cx where and are positive constants and is the distance of a point from station A, then the distance between A and B and maximum speed during motion is: =c 1 -c 2 ) C 3​

Answer 1

Answer:

ঐঞয়োঢ়ঙনূঐঐঊঞঢ়ঝৈঢৈঢেওনঢ়ঢড়ুএণুভঝছবুঐভেঔএঞোঢে

ঢঢ়ৈঊঢোছণঢ়োণোঊড়ঙতয়ৈতৌ

Answer 2

Answer:

Distance between A and B= 2a/b and Vmax=$a/\sqrt{b}$

Explanation:

step 1

According to the law of acceleration

$f=v.\frac{dv}{dx}$-----(i)

from given statement $f=a-cx$------(ii)

therefore

$v.\frac{dv}{dx}=a-cx$

$v.dv=(a-cx)dt$

step 2

by integrating we get

$\int\limits^v_0 {v} \, dv=\int\limits^x_0 {(a-cx)} \, dx$

$\frac{v^{2} }{2} =ax-\frac{cx^{2} }{2}$

$v^2=2ax-cx^2$

$v=\sqrt{ 2ax-cx^2}$------(iii)

if $V_{f} =0$

x(cx-2a)=0

there fore x=0, $x=\frac{2a}{c}$

Further acceleration will change its direction when

$f=0\\a-cx=0\\x=\frac{a}{b}$

at this distance velocity is maximum

therefore

from (iii) we get

$v=\sqrt{2a(\frac{a}{c}) -c(\frac{a}{c})^2$

$V_{max} =a/\sqrt{b}$