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