Question #38b0a

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Answer 1 · 2018-02-01T12:58:16

The acceleration is #=5.49ms^-2#

Apply the equation of motion

#s=ut+1/2at^2#

The initial velocity is #u=0ms^-1#

The time is #t=6.75s#

The distance is #s=125m#

The acceleration is

#a=2((s-ut)/t^2)#

#=2*((125-0)/6.75^2)#

#=5.49ms^-2#

Answer 2 · 2018-02-01T13:00:15

#5.49 m/(s^2)#

#s=v*t +½at^2#. But v=0. Rearrange formula.

#a=(2s)/t^2#

#a=(2*125m)/(6.75s)^2=(250m)/(6.25 s)^2=(250m)/(45.56s^2)=5.49m/s^2#

Answer 3 · 2018-02-01T13:02:16

#"Acceleration"~~5.49ms^(-2)#

#s=ut+1/2at^2#, where:

#s=125m#
#u=0ms^(-1)#
#t=6.75s#

#125=0(6.75)+a(6.75)^2/2#
#color(white)(125)=0+22.78125a#

#a=125/22.78125~~5.49ms^(-2)#