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

Answer:

#0.03m/s^2#

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 #a#.

Thus, rearrange the formula for #a# and plug in your known values into the formula to find the acceleration of the object.

#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)|)))#