Question

If t = αx + βx² then find relation between velocity and acceleration.​

Answer 1

UNDERSTANDING CONCEPT :-

• We know that the rate of change of displacement is called velocity and the rate of change of velocity is called acceleration. Here, instead a value for displacement and time, an equation is given. So calculate velocity and acceleration, we need to differentiate the equation.Formulas used:

• V = displacement/time or v = dx/dt

• and a = velocity/time or a = dv/dt = d²x/dt²

Also,

• dx/dt = 1/dt/dx

TO FIND :-

the relationship between v and a = ?

SOLUTION STEP BY STEP :-

let us start with the basic definition of velocity and acceleration.

We know that the velocity v

V = displacement/time or dx/dt

where time is t and displacement is

x .Similarly, the acceleration a is defined as the rate of change of velocity a = velocity/time or

a = dv/dt = d²x/dt²

where, time is t and velocity v .

Since, t is given in terms of x. we can

differentiate t with respect to x, then we get

dt/dx = 2 ax + B Then velocity v = dx/dt = 1/dt =

1/2ax+ B then, we must differentiate

V = dx/dt = 1/2ax + B with respect t

Here, using the mathematical differentiation of x then

d/dx x = nx n - 1 , here, in our sum, n = -1

And using chain rule of differentiation, we get

• a = d²x/dt² = -1(2ax + 3) -2 × 2a = -2a/(2αx + 3)²

To find the relationship between v

and a, we can substitute

v = 1/(2ax + B ) in a Then we get,

a = - 2av²

Hence the relationship between

v and a, is - 2av²

ADDITIONAL INFORMATION :-

• This may seem as a hard question at first. But this question is easy, provided you know differentiation, here we use the mathematical differentiation of xn, then derivative of x ^ n for x = n * x ^ (n - 1) here, in our sum, n = - 1 . And usina chain rule of differentiation, we get the result.Also see that dx/dt = 1/dt /dx this is the most important step in this question. Also note
• that v = displacement/time or v = dx/dt
• and a = velocity/time Or a = dv/dt = d²x/dt²
• To calculate, a we must differentiate only v with respect to t and not dt/dx

Answer 2

$\bf\huge Solution \: :—$

$\:$

We are given the equation of motion as:

$$ t = \alpha x + \beta x^2 $$

Taking the derivative of both sides with respect to time, we get:

$$ \frac{dt}{dt} = \frac{d}{dt}(\alpha x + \beta x^2) $$

The left-hand side simplifies to 1, while the right-hand side becomes:

$\frac{d}{dt}(\alpha x + \beta x^2) = \alpha \frac{dx}{dt} + 2\beta x \frac{dx}{dt}$

Now, we can express the velocity and acceleration in terms of derivatives of position with respect to time:

$$ v = \frac{dx}{dt} $$

$$ a = \frac{d^2x}{dt^2} $$

Substituting these into the equation above, we get:

$$ 1 = \alpha v + 2\beta x v $$

Simplifying this expression, we obtain the relationship between velocity and acceleration:

$$ a = \frac{dv}{dt} = \alpha \frac{d}{dt}(v) + 2\beta v \frac{dx}{dt} $$

or

$\rightarrow \boxed{a = \alpha \frac{dv}{dx} + 2\beta v^2}$