If an object with uniform acceleration (or deceleration) has a speed of #4 m/s# at #t=0# and moves a total of 18 m by #t=5#, what was the object's rate of acceleration?

1 Answer
Apr 30, 2016

Answer:

#a=0,16 " "m/s^2" acceleration"#
#a=-0,16" "m/s^2" deceleration"#
#("acceleration")/("deceleration")=-1#

Explanation:

enter image source here

#"in figure-1:"#
#"area under graph is equal to displacement of object"#

#18=((4+4+Delta v)*5)/2#

#36=40+5 Delta v" "5 Delta v=36-40" "5 Delta v=4 " " Delta v=4/5=0,8" "m/s#

#tan alpha=a=(Delta v)/(Delta t)#

#a=(0,8)/5#

#a=0,16 " "m/s^2" acceleration"#

#"in figure-2"#

#18=((4+4-Delta v)*5)/2#

#36=(8-Delta v)5#

#36=40-5 Delta v#

#5v=40-36#

#5v=4#
#v=4/5=0,8 " "m/s#

#tan beta=a=-(0,8)/5#

#a=-0,16" "m/s^2" deceleration"#