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

Answer:

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 $V=\dfrac{ds}{dt}$

       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

    $s=\int_{0}^{2}(t^{2}-4)dt+\int_{2}^{3}(t^{2}-4)$

$s=\left [\frac{1}{3}t^3-4t\right ]_o^2+\left [\frac{1}{3}t^3-4t\right]_2^3$

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