What is the average speed of an object that is moving at #25 m/s# at #t=0# and accelerates at a rate of #a(t) =2-2t^2# on #t in [0,5]#?

1 Answer
Jun 11, 2017

Answer:

#1 m/s#

Explanation:

NOTE: I am using the given a(t) as the acceleration term. Dimensionally, that doesn’t look quite right #(m/(s^2))#?? If it should be #a = 2 – 2*t^2# then the calculation of the final velocity will be different.

Average Velocity
#v_a = (v_f + v_i) / 2 #   
where
#v_a# = average velocity (m/s)
#v_i# = initial velocity (m/s)
#v_f# = final velocity (m/s)

Final Velocity
#v_f = v_i + a*t#    
where
#a# = acceleration #(m/(s^2))#
t = time taken (s)

In this case:
#v_f = v_i + (2 – 2*t^2)# ;    #v_f = 25 + (2 – 2*5^2)#
#v_f = 25 + -48 = -23#
It is moving backwards, as indicated by the negative acceleration relative to the starting point. It initially had a forward direction, so it changed direction as well as velocity.

The average velocity is thus:

#v_a = (v_f + v_i) / 2 = (-23+ 25)/2 = 1 m/s#

This positive value just means that at time (5) it has not yet quite reached the location it was at at time (0). On average it only moved 1 meter over the time span and is now heading back.