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#