Question

If X=
$\sqrt{ \alpha t + \beta }$
Find acceleration in terms of displacement(X) and velocity(V)​

Answer 1

Answer:

$\huge\orange{ \frac{ - \alpha v}{ \: \: \: 2 {x}^{2} } }$

Explanation:

Given

$x = \sqrt{ \alpha t + \beta }$

Differentiate wrt t

$v = \frac{dx}{dt} = \frac{ \alpha }{2 \sqrt{ \alpha t + \beta } } \\ \\ \implies \: v = \frac{ \alpha }{2x}$

Now acceleration

$a = v \frac{dv}{dx} \\ \\ \implies \: a = v \frac{d}{dx} ( \frac{ \alpha }{2x} ) = v \frac{ - \alpha }{2 {x}^{2} }$

Answer 2

Explanation:

given :

• If X=
• $\sqrt{ \alpha t + \beta }$
• Find acceleration in terms of displacement(X) and velocity(V)

to find :

• Find acceleration in terms of displacement(X) and velocity(V)

solution :

• velocity = dx/dt = a/ 2√at + B

• velocity = a/2 x

• a = dv/dx

• a= velocity d/dx ( a/2x) = velocity - a/ 2x