If an object with uniform acceleration (or deceleration) has a speed of #3 m/s# at #t=0# and moves a total of 25 m by #t=8#, what was the object's rate of acceleration?
1 Answer
Apr 16, 2016
Explanation:
Recall the following kinematic formula of motion:
#color(blue)(|bar(ul(color(white)(a/a)Deltad=v_iDeltat+1/2aDeltat^2color(white)(a/a)|)))# where:
#Deltad=# change in distance
#v_i=# initial velocity
#Deltat=# change in time
#a=# acceleration
Start by listing out the given and required values.
#v_i=3m/s#
#Deltad=25m#
#t=8s#
#a=?#
Notice how the only variable to solve for is
Thus, rearrange the formula for
#Deltad=v_iDeltat+1/2aDeltat^2#
#=1/2aDeltat^2=Deltad-v_iDeltat#
#=(2(Deltad-v_iDeltat))/(Deltat^2)#
#=(2(25m-(3m/s(8s))))/(8s)^2#
#=0.03125m/s^2#
#~~color(green)(|bar(ul(color(white)(a/a)0.03m/s^2color(white)(a/a)|)))#