Question

a boat travels 12.0 m while it reduces its velocity from 9.5 m/s to 5.5 m/s. what is the magnitude of the boat’s acceleration while it travels the 12.0 m?

Answer 1

-2.5
`m/s^2`

Step-by-step explanation:

We can use the kinematics equation below to find the boat's acceleration.

`V_(f) ^(2)= V_(i) ^(2)+2ax`

Lets solve for
`a`.

Subtract
`V_i^2` from both sides of the equation.

`2ax=V_(f) ^(2)- V_(i) ^(2)`

Divide each term in
`2ax=V_(f) ^(2)- V_(i) ^(2)` by
`2x`.

`(2ax)/(2x) =(V_(f) ^(2))/(x) +(- V_(i) ^(2))/(2x)`

Simplify the left side.

Cancel the common factor of 2.

`(ax)/(x) =(V_(f) ^(2))/(x) +(- V_(i) ^(2))/(2x)`

Cancel the common factor of x.

`a=(V_(f) ^(2))/(x) +(- V_(i) ^(2))/(2x)`

Simplify the right side.

Move the negative in front of the fraction.

`a=(V_(f) ^(2))/(2x) -( V_(i) ^(2))/(2x)`

Now we have an equation for
`a`.

We are given

`V_f=5.5\\V_i=9.5\\x=12`

Substituting our numbers into the equation gives us

`a=(5.5^(2))/(2*12) -( 9.5 ^(2))/(2*12)`

`a=(30.25)/(2*12) -(90.25)/(2*12)`

`a=(30.25)/(24) -(90.25)/(24)`

`a=-2.5`