Aparticle moves along the x-axis with velocity v(t) = t2 - 4, with t measured in seconds and v(t) measured in feet per second. find the total distance travelled by the particle from t = 0 to t = 3 seconds. (4 points)

Total distance D=7.66 ft

Explanation:  

Given that:  

Here velocity of particle will be zero at t=2 sec,so we take interval to find total distance.

We know that

       ds=V dt

So  

It means that ,the area of velocity-time(V-t) graph will give the displacement.

Given that

So now by putting the value in above integration

s= -5.33+2.33 ft

 s= -3 ft  (we know that displacement is a vector quantity so it have sing)

So this is the displacement of particle at time 0 sec to 3 sec.

To find the total distance we will add all take mode on -5.33 ft and and will add with 2.33 ft instead of subtract.

So the total distance travelled by particle D=7.66 ft.

D=7.66 ft

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: