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?

3 Answers
Jun 15, 2018

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.

Jun 15, 2018

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

Jun 15, 2018

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