If an object is moving at #8 m/s# and accelerates to #32 m/s# over 6 seconds, what was the object's rate of acceleration?

3 Answers
Jun 29, 2018

Answer:

#4 \ "m/s"^2#

Explanation:

Well, the object changed its velocity by #32 \ "m/s"-8 \ "m/s"# over #6# seconds.

Acceleration is given by the equation:

#a=(s_f-s_i)/t#

where:

  • #s_f,s_i# are the final and initial speeds, respectively

  • #t# is the time taken

So, we get:

#a=(32 \ "m/s"-8 \ "m/s")/(6 \ "s")#

#=(24 \ "m/s")/(6 \ "s")#

#=4 \ "m/s"^2#

Jun 29, 2018

Answer:

The acceleration is #=4ms^-2#

Explanation:

The initial velocity is #u=8ms^-1#

The final velocity is #v=32ms^-1#

The time is #t=6s#

Apply the equation of motion

#v=u+at#

The acceleration is

#a=(v-u)/t=(32-8)/6=24/6=4ms^-2#

Jun 29, 2018

Answer:

Rate of acceleration #a = 6# m/ #s^2#

Explanation:

#v = u + a t#

#u = 8# m/s, #v = 32# m/s, #t = 6# sec

#a = (v - u) / t = (32 - 8) / 6 = 4 # m / #s^2#