If an object with uniform acceleration (or deceleration) has a speed of #4 m/s# at #t=2# and moves a total of 25 m by #t=6#, what was the object's rate of acceleration?

Question

1126 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-23T05:34:37

The acceleration is #=1.125ms^-2#

The initial speed is #u=4ms^-1#

The distace is #s=25m#

The time is #Deltat=6-2=4s#

Apply the equation of motion

#s=uDeltat+1/2a(Deltat)^2#

The acceleration is

#a=(s-uDeltat)/(1/2(Deltat)^2)#

#=(25-4xx4)/(1/2xx4^2)#

#=(9)/(8)#

#=1.125ms^-#