Question

A body moving from rest again comes to
rest. It starts from x = 0 and its acceleration varies as a
= q - px. Find the position of the body when it comes
to rest again.​

Answer 1

A body moving from rest again comes to rest. it starts from x = 0 and its acceleration varies as, a = q - px

we have to find position of the body when it comes to rest again.

a = q - px

we know, a = v dv/dx

so, a = vdv/dx = q - px

⇒∫vdv = ∫(q - px)dx

⇒$\int\limits^v_0{v}\,dv=\int\limits^x_0{(q-px)}\,dx$

⇒v²/2 = qx - px²/2

let at a point x , body comes to rest.

so, v = 0

then, 0 = qx - px²/2

⇒x(q - px/2) = 0

⇒x = 2q/p

hence, at x = 2q/p , body comes to rest again.

Answer 2

$\huge\bold\purple{Answer:-}$

we have to find position of the body when it comes to rest again.

a = q - px

we know, a = v dv/dx

so, a = vdv/dx = q - px

⇒∫vdv = ∫(q - px)dx

⇒\int\limits^v_0{v}\,dv=\int\limits^x_0{(q-px)}\,dx

⇒v²/2 = qx - px²/2

let at a point x , body comes to rest.

so, v = 0

then, 0 = qx - px²/2

⇒x(q - px/2) = 0

⇒x = 2q/p

hence, at x = 2q/p , body comes to rest again.