What is the average speed of an object that is moving at #3 m/s# at #t=0# and accelerates at a rate of #a(t) =9-t^2/3# on #t in [3,5]#?

1 Answer
Mar 30, 2017

Answer:

#31 4/9#m/s

Explanation:

#a(t) = 9 -t^2/3#

Speed, #v = int a(t) = int 9 -t^2/3 =9t - t^3/9 + c#

#t = 0, v = 3# m/s #-> c = 3#, therefore,
#v = 9t - t^3/9 + 3#

Displacement,
#s = int v(t)#,

#s_(3,5) = int_3^5 v(t) = int_3^5 9t - t^3/9 + 3 = [9/2 t^2 - t^4/36 + 3t]_3^5#

#s_(3,5) = [9/2 (5)^2 - (5)^4/36 + 3(5)] - [9/2 (3)^2 - (3)^4/36 + 3(3)]#

#s_(3,5) = [225/2 - 625/36 + 15] - [81/2 - 81/36 + 9]#

#s_(3,5) = [225/2 - 625/36 + 15] - [81/2 - 81/36 + 9]#

#s_(3,5) = 225/2 - 81/2- 625/36 + 81/36 + 15- 9#

#s_(3,5) = 144/2- 544/36 + 6#

#s_(3,5) = 72 - 136/9 + 6 = 566/9#

Average speed = displacement/time #=566/9-:2 = 566/18 = 31 4/9#m/s