Velocity and displacement and distance

[karush]

Assume $t$ is time measured in seconds and velocities have units of $m/s$

$$s\left(t\right)=-{t}^{2}+5t-4 \ \ \ 0\le t \le 5 $$

$$v\left(t\right)=-2t+5$$

a. Graph the velocity function over the given interval. Then determine when the motion is in the positive direction and when it is the negative direction.

[desmos="0,5,-5,5"]y=-2t+5;[/desmos]

b. Find the displacement over the given interval. ?

c. Find the distance traveled over the given interval. ?

 

[Prove It]

Assume $t$ is time measured in seconds and velocities have units of $m/s$

$$s\left(t\right)=-{t}^{2}+5t-4 \ \ \ 0\le t \le 5 $$

$$v\left(t\right)=-2t+5$$

a. Graph the velocity function over the given interval. Then determine when the motion is in the positive direction and when it is the negative direction.

b. Find the displacement over the given interval. ?

c. Find the distance traveled over the given interval. ?

Isn't s your displacement function? So wouldn't the displacement over that interval be just your s function as written? Or if you want the displacement at t = 5 you'd be finding s(5) ?

As for the distance travelled, the distance is the area under the velocity graph in that interval. How would you evaluate that?

 

[karush]

Sorry looks like the equation was given as velocity

$$v\left(t\right)=-{t}^{2 }+5t-4$$

So then displacement is

$$\int_{0}^{5}v(t) \,dt = \frac{5}{6}$$

And distance traveled would be

$$\int_{0}^{ 1}\left| v\left(t\right) \right| \,dt
+\int_{1}^{4} v\left(t\right) \,dt
+ \int_{4}^{5}\left| v\left(t\right) \right| \,dt =\frac{49}{6} $$

I hope anyway..

 

[MarkFL]

Your displacement is correct, and the end result of your distance is correct (but your integral expression, while technically correct should not have any absolute values), but for the distance $D$, we can simplify matters a bit and use the symmetry of the velocity function about the line $t=\dfrac{5}{2}$ to write:

$$D(5)=2\left(\int_0^1t^2-5t+4\,dt-\int_1^{\frac{5}{2}}t^2-5t+4\,dt\right)=2\left(\frac{11}{6}+\frac{9}{4}\right)=\frac{49}{6}$$