We're given the acceleration and we need to find velocity, so we can apply the initial condition #v(0)=5#. But how? The answer: acceleration measures how fast velocity changes - which means acceleration is the derivative of velocity. That is to say:
#(dv)/dt=a(t)#
If we multiply both sides of this equation by #dt#, we have:
#dv=a(t)dt#
And if we integrate both sides, we find that:
#intdv=intadt->v(t)=inta(t)dt#
This is basically saying that in order to find velocity, given acceleration, we integrate the acceleration. In our case, #a(t)=t+4#. We will integrate this to find velocity:
#v(t)=intt+4dt#
#v(t)=inttdt+int4dt#
#v(t)=t^2/2+4t+C-># never forget the constant of integration!
Alright, we have #v(t)# - but we also have that pesky #C#. In order to find the value of #C#, we will use our initial condition, #v(0)=5#:
#v(t)=t^2/2 +4t+C#
#5=(0)^2/2 +4(0)+C#
#5=C#
Thus we can say #v(t)=t^2/2+4t+5#.
Now, we just need to find distance traveled, from #t=0# to #t=10#. This might surprise you - it certainly surprised me when I first learned it - but the distance traveled is actually the area under the velocity function:
#D=int_(t=t_i)^(t=t_f)v(t)dt#
In the problem, #t_i# (initial time) is #0# and #t_f# (final time) is #10#; and we just found #v(t)# to be #t^2/2+4t+5#. Therefore, the distance traveled is given by:
#int_0^10t^2/2+4t+5dt#
Evaluating the integral involves breaking it up and using the power rule:
#int_0^10t^2/2+4t+5dt=int_0^10t^2/2dt+int_0^10 4tdt+int_0^10 5dt#
#color(white)(XX)=1/2int_0^10t^2dt+4int_0^10 tdt+5int_0^10dt#
#color(white)(XX)=1/2[t^3/3]_0^10+4[t^2/2]_0^10+5[t]_0^10#
#color(white)(XX)=1/2((10)^3/3-(0)^3/3)+4((10)^2/2-(0)^2/2)+5(10-0)#
#color(white)(XX)=500/3+200+50~~416.7# meters
The particle traveled #416.7# meters from #0# to #10# seconds.
