What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) =t+3# from #t in [2, 4]#?

1 Answer
Mar 4, 2016

Answer:

Use the definition of acceleration and know that with respect to time, #u(0)=0# because it is still. Also, you should give units of measurement (e.g. #m/s#). I didn't use any because you didn't give me.

#u_(aver)=14#

Explanation:

Being still at #t=0# means that for #u=f(t)->u(0)=0#
Starting from the acceleration definition:

#a=(du)/dt#

#t+3=(du)/dt#

#(t+3)dt=du#

#int_0^t(t+3)dt=int_0^udu#

#int_0^(t)tdt+int_0^t3dt=int_0^udu#

#[t^2/2]_0^t+3[t]_0^t=[u]_0^u#

#(t^2/2-0^2/2)+3(t-0)=u-0#

#u(t)=t^2/2+3t#

So the average velocity between times 2 and 4 is:

#u_(aver)=(u(2)+u(4))/2#

#u(2)=2^2/2+3*2=8#

#u(4)=4^2/2+3*4=20#

Finally:

#u_(aver)=(8+20)/2#

#u_(aver)=14#