Question

An object uniformly accelerates from 15.0m/s west to 35.0m/s west. What is the rate of acceleration if the displacement during this time was 43.0m?

Answer 1

The kinematic equation to use in order to answer this problem is

`v_"f"^"2"` = `v_"i"^"2"` + `2ad`, where

`v_"f"^"2"` is the final velocity squared
`v_"i"^"2"` if the initial velocity squared
`a` is acceleration
`d` = displacement

Known:
`v_"i"` = 15.0m/s
`v_"f"` = 35.0m/s
`d` = 43.0m

Unknown:
`a`

Equation manipulated to solve for `a`:

(`v_"f"^"2" - v_"i"^"2")"/"2d = a`

Solution:

`(35.0m"/s")"^2` - `(15.0m"/s")"^2` `/` `(2)(43.0m)` = `a`

`(1230m^2"/s"^2` - `225m^2"/s"^2)` `/` `86.0m` =

`(1005m^2"/s"^2)` `/` `(86.0m)` = `11.7m"/s"^2`

Acceleration is `11.7m"/s"^2`