Question

The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 4 seconds?

Answer 1

-1/27 ft/s^2

Explanation:

The position function is `s(t)=sqrt(2t+1)`.
Find the first derivative to find the velocity function: `v(t)=1/sqrt(2t+1)`.
The second derivative will result in the acceleration function:
`a(t)=-1/(2t+1)^(3/2)`.
Therefore, `a(4)=-1/27`.

Answer 2

Please see below.

Explanation:

.

It is not clear what your position equation is from the wording of your problem. If you meant:

`s(t)=sqrt2(t+1)`

The velocity equation is the derivative of the position equation:

`v(t)=sqrt2` feet per second

The acceleration equation would be the derivative of the velocity equation:

`a(t)=0` , i.e. no acceleration (constant speed)

If you meant:

`s(t)=sqrt(2t)+1=sqrt2(t)^(1/2)+1`

`v(t)=sqrt2/2(t)^(-1/2)=sqrt2/(2t^(1/2))=sqrt2/(2sqrtt)=sqrt(2t)/(2t)` feet per second

`a(t)=(-1/2)(sqrt2/2)(t)^(-3/2)=sqrt2/(4t^(3/2))=sqrt2/(4tsqrtt)=sqrt(2t)/(4t^2)` feet per `second^2`

`a(4)=sqrt(2(4))/(4(4)^2)=sqrt2/32` feet per `second^2`

If you meant:

`s(t)=sqrt(2(t+1))=sqrt2(t+1)^(1/2)`

`v(t)=sqrt2/2(t+1)^(-1/2)=sqrt2/(2sqrt((t+1)))=sqrt(2(t+1))/(2(t+1))` feet per second

`a(t)=(-1/2)(sqrt2/2)(t+1)^(-3/2)=-sqrt2/(4(t+1)sqrt(t+1))=-sqrt(2(t+1))/(4(t+1)^2)` feet per `second^2`

`a(4)=-sqrt(2(4+1))/(4(4+1)^2)=-sqrt10/100` feet per `second^2`

If you meant

`s(t)=sqrt2t+1`

`v(t)=sqrt2` feet per second

`a(t)=0`