Question

A particle 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)

Answer 1

Total distance D=7.66 ft

Explanation:

Given that:
`V=t^(2)-4`

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

We know that
`V=(ds)/(dt)`

ds=V dt

So
`s=\int_{t_(1)}^{t_(2)}Vdt`

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

Given that
`t_(1)=0 ,t_(2)=3`

So now by putting the value in above integration

`s=\int_(0)^(2)(t^(2)-4)dt+\int_(2)^(3)(t^(2)-4)`

`s=\left [(1)/(3)t^3-4t\right ]_o^2+\left [(1)/(3)t^3-4t\right]_2^3`

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