What is the acceleration of an object at t=2?

The position of an object is given by the equation x(t)=4t^3+6sin(t) . Find the
acceleration of the object at t=2.

3 Answers
Feb 7, 2018

48-6sin(2) or 42.54 ms^-2

Explanation:

We have position as a function of time, to get velocity as a function of time we differentiate it one time. dx/dt = velocity
dx/dt = 12t^2 +6cost

Now to get acceleration, differentiate it one more time. (dv)/dt =acceleration
(dv)/dt = 24t - 6sint

Substituting, t=2 in a(t), we get
48-6sin(2) ms^-2 or
42.54 ms^-2

Feb 7, 2018

The acceleration is the variation of the velocity.
The velocity is the variation of the position.

If the position is given by x(t)=4t^3+6 sin(t) so, the velocity is given by its derivative, v(t)=12t^2 + 6 cos(t). And, then, the acceleration is given by velocity's derivative, a(t)=24t - 6 sin(t)

So, the acceleration at the time t=2 is given by a(2)=24\cdot2-6sin(2)=42-6 sin(2)

Using Taylor's approximation...

sin(x) ~= 1 - (x-pi/2)^2/2 near to x=pi/2
sin(2) ~= 1 - (2-pi/2)^2 / 2=1- (4 - 2pi + pi^2 / 4)/2
sin(2) ~= 1 - 2 + pi - pi^2 / 8=pi - pi^2 / 8 - 1

Taking pi~=3.14...

sin(2) ~= 3.14 - (3.14)^2/8 - 1 ~= 0.91

Feb 7, 2018

Given, x= 4t^3 +6 sint

We,know, acceleration = change in rate of velocity and velocity = change in rate of displacement.

So,differentiate the given equation twice,to get the acceleration-time relationship.

So,a =24 t - 6 sin t

Putting, t=2 we get, a= 42.54 m/s^2