# 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?

Oct 2, 2014

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

${v}_{\text{f"^"2}}$ = ${v}_{\text{i"^"2}}$ + $2 a d$, where

${v}_{\text{f"^"2}}$ is the final velocity squared
${v}_{\text{i"^"2}}$ if the initial velocity squared
$a$ is acceleration
$d$ = displacement

Known:
${v}_{\text{i}}$ = 15.0m/s
${v}_{\text{f}}$ = 35.0m/s
$d$ = 43.0m

Unknown:
$a$

Equation manipulated to solve for $a$:

(${v}_{\text{f"^"2" - v_"i"^"2")"/}} 2 d = a$

Solution:

(35.0m"/s")"^2 - (15.0m"/s")"^2$/$$\left(2\right) \left(43.0 m\right)$ = $a$

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

$\left(1005 {m}^{2} {\text{/s}}^{2}\right)$$/$$\left(86.0 m\right)$ = $11.7 m {\text{/s}}^{2}$

Acceleration is $11.7 m {\text{/s}}^{2}$