Question

A particle starts from point A moves along a straight line path with an acceleration given by a = p - qx where p , q are constants and x is distance from point A. The particle
stops at point B. The maximum velocity of the paticle is

a] $\sf{\dfrac{p}{q}}$

b] $\sf{\dfrac{p}{\sqrt{q}}}$

c] $\sf{\dfrac{q}{p}}$

d] $\sf{\dfrac{\sqrt{q}}{p}}$

Please Please Please Please Please Solve with Proper Explanation I will mark you as brainliest and follow you . Don't use Double differentiation ​

Answer 1

11th/Physics

Motion in straight line

Answer :

Acceleration of particle is given by

• a = p - qx

We have to find maximum velocity of particle$.$

Given, a = p - qx

From the given equation, it is clear that acceleration of particle is position dependent.

We know that,

At maximum velocity a = 0

⭆ p - qx = 0

⭆ p = qx

⭆ x = p/q

∴ Velocity is maximum at x = p/q

As we know that,

➠ a = dv/dt

➠ a = dv/dt × dx/dx

➠ a = (dx/dt) × dv/dx

➠ a = v dv/dx

ATQ, a = p - qx

➠ v dv/dx = p - qx

➠ v dv = (p - qx) dx

Integrating both sides, we get

➠ v²/2 = px - qx²/2

➠ v² = 2px - qx²

Substituting x = p/q we get

➠ v² = 2p(p/q) - q(p/q)²

➠ v² = 2p²/q - p²/q

➠ v² = p²/q

➠ v(max) = p/√q

Hence, (B) is the correct answer.

Cheers!

Answer 2

Answer :

$V_{max} \: = \frac{p}{ \sqrt{q} }$

option B is crct.

plz refer to the attachment above