Rectilinear Motion Using Integration

[hank]

Ok, here's a problem from my book which the teacher didn't cover in lecture and the book provides no examples on how to solve. As you can guess, I am absolutely clueless how to solve it. Anyway, here it is:

A particle moves with acceleration a(t) m/s^2 along an s-axis and has velocity v0 at time = 0. Find the displacement and the distance traveled by the particle during the given time interval.

a(t) = 3; v0 = -1; 0<= t <=2

I have absolutely NO idea where to even start.

Help?

We have multiple problems of this type, so if anyone could do this one for me, or give me an example that would help me out a ton.

 

[stapel]

Just think of a dot (the particle) moving back and forth (left and right) along the number line.

Sometimes the dot moves back and forth, so the total distance covered might be a lot but, due to back-tracking, the displacement might be very little. One example would be throwing a ball up in the air. Eventually the ball will stop bouncing and will come to rest on the ground, where it started. The displacement (the net change in position) will be zero. But the ball bounced a lot, so the distance covered (regardless of direction) could be quite a lot.

In this case, the acceleration is always positive, but the velocity starts off negative. This would be like having "thrown" the dot leftward, but with the "gravity" at "infinity" off to the right pulling the dot back, eventually moving the dot rightward (with a positive velocity). Since the velocity was initially negative, there will be some back-tracking, so the distance covered will be greater than the total displacement.

Start with the acceleration function, and integrate to find the velocity function. Use the one data point to solve for the integration constant. Then integrate again to get the position function. The problem doesn't say, so I would guess you are to assume s(0) = 0.

For the net displacement, evaluate s(0) and s(2). Subtract the smaller from the larger. For the total distance covered, find the max/min points of s(t). Find the total distances covered on each "piece" of s(t), where the graph is moving in any one direction. Sum the pieces to get the total distance covered.

Hope that helps a bit.

Eliz.

 

[soroban]

Hello, Hank!

A particle moves with acceleration \(\displaystyle a(t)\) along an s-axis
and has velocity \(\displaystyle v_o\) at \(\displaystyle t\,=\,0.\)
Find (a) the displacement and (b) the distance traveled
by the particle during the given time interval.

\(\displaystyle a(t)\,=\,3,\;\;v_o\,=\,-1,\;\;0\,\leq\,t\,\leq\,2\)

You're expected to know that: given the displacement function \(\displaystyle s(t)\)
. . the velocity is the derivative of \(\displaystyle s(t):\;\;v(t)\,=\,s'(t).\),
. . and the acceleration is the derivative of \(\displaystyle v(t):\;\;a(t)\,=\,v'(t)\)

Working backwards, given the acceleration function \(\displaystyle a(t)\)
. . the velocity is the integral of \(\displaystyle a(t):\;\;v(t) \:=\:\int a(t)\,dt\)
. . and the displacement is the integral of \(\displaystyle v(t):\;\;s(t)\:=\:\int v(t)\,dt\)


We are given: \(\displaystyle \,a(t)\:=\:3\)

Then: \(\displaystyle \,v(t)\:=\:\int 3\,dt \:=\:3t\,+\,C_1\)
We are given: \(\displaystyle v_o\,=\,-1\)
. . That is, when \(\displaystyle t = 0,\:v = -1\)
. . So we have: \(\displaystyle v(0)\:=\:3(0)\,+\,C_1 \:=\:-1\;\;\Rightarrow\;\;C_1\:=\:-1\)
Hence: \(\displaystyle \:v(t)\:=\:3t\,-\,1\)

Then: \(\displaystyle \,s(t)\:=\:\int(3t\,-\,1)\,dt \:=\:\frac{3}{2}t^2\,-\,t\,+\,C_2\)

They didn't tell us the initial displacement, \(\displaystyle s_o,\;\;\)I'll assume it is \(\displaystyle 0.\)
. . That is, when \(\displaystyle t\,=\,0:\,s\,=\,0\)
. . So we have: \(\displaystyle \,3(0)\,-\,0\,+\,C_2\:=\:0\;\;\Rightarrow\;\;C_2\,=\,0\)
Hence: \(\displaystyle \:\fbox{s(t) \;= \;\frac{3}{2}t^2 \,-\,t\)


(a) At \(\displaystyle t\,=\,0:\;s(0) \:=\:\frac{3}{2}(0^2) - 0 \:=\:0\)
. . .At \(\displaystyle t\,=\,2:\;s(2)\:=\:\frac{3}{2}(2^2) - 2\:=\:4\)

Therefore, the displacement is \(\displaystyle \fbox{4\text{ units}}\) to the right.


(b) Be careful . . . This part takes more work.

The particle starts at \(\displaystyle s(0)\,=\,0.\)
Since \(\displaystyle v_o\,=\,-1\), the particle is moving to the left.

We find that: \(\displaystyle v\:=\:3t\,-\,1\) equals 0 when \(\displaystyle t\,=\,\frac{1}{3}\)
. . When the velocity is zero, there is usually a change of direction.
We have: \(\displaystyle \:s\left(\frac{1}{3}\right)\:=\:\frac{3}{2}\left(\frac{1}{3}\right)^2 - \frac{1}{3} \:=\;-\frac{1}{6}\)
. . The particle has moved \(\displaystyle \frac{1}{6}\) unit to the left.

We have: \(\displaystyle \,s(2) \,=\,4\)
. . By the end of 2 seconds, the particle is 4 units to the right of the origin.
. . It has moved \(\displaystyle 4\frac{1}{6}\) units to the right.

The total distance traveled is: \(\displaystyle \:\frac{1}{6}\,+\,4\frac{1}{6}\;=\;\fbox{4\frac{1}{3}\text{ units.}}\)

 

[hank]

I would have never gotten that on my own, but now that you guys have explained it, it makes total sense. The light bulb has gone off.

Thank you!