Question

The relation between the velocity and the position of a particle is v=Ax+B , where A & B are constants. Acceleration of the particle is 10 when its velocity is a.How much is the acceleration when it's velocity is 2a?

Answer 1

Its the same, `10` `ms^-2`

Explanation:

As the velocity is a straight line, the acceleration will always be the same.

This can also be shown with calculus:

If `v(x)=Ax+B` then the derivative which is the acceleration, `v\prime(x)=a(x)=A`. And as you said `A` is a constant so `a(x)` will always be constant whatever the input it.

Answer 2

I cannot use `a` to represent the acceleration vector because the question uses `a` for a constant, therefore, I shall us `vecu` to represent the acceleration vector.

Velocity is a vector, therefore, I shall correct the given equation to be a vector equation:

`vecv = (Ax+ B)hati`

Where `hati` is the unit vector in the `x` direction.

The acceleration vector must be the first derivative of the velocity vector with respect to time (t). Applying the chain rule:

`vecu = (d(vecv))/(dt) = (d(vecv))/dx dx/(dt)`

We can compute `(d(vecv))/dx` by starting with `vecv = (Ax+ B)hati`:

`(d(vecv))/dx = (d(Ax+ B))/dxhati`

`(d(vecv))/dx = Ahati`

But `x` as a function of time (t) is not specified, therefore, we can say no more than the following:

`vecu = Adx/(dt)hati`

Answer 3

The relation between the velocity and the position of a particle is v=Ax+B , where A & B are constants. Acceleration of the particle is 10 when its velocity is a.How much is the acceleration when it's velocity is 2a?

`v = A x + B`

`(dv)/(dt) = A (dx)/(dt) = A v qquad square`

`A = ((dv)/(dt))/(v) = 10/a`

`:. bb( ((dv)/(dt))_(v = 2a) = 20 )`

No units were given