Question

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.

Answer 1

`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`

Answer 2

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`

Answer 3

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`