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?

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Answer 1 · 2014-10-02T23:13:02

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 $11.7m"/s"^2$