Graphical Integration in Motion Analysis — Question 2

Question 2

An acceleration area needs an initial velocity

A cart has $x(0)=2\ \mathrm{m}$ and $v(0)=-1\ \mathrm{m/s}$ on an east-positive track. Its acceleration is $+2\ \mathrm{m/s^2}$ during 0–2 s, $-1\ \mathrm{m/s^2}$ during 2–5 s, and zero during 5–7 s. Ideal finite jumps in acceleration take no time; velocity remains continuous.

Use $v(t)=v(0)+\int_0^t a(u)\,du$ and then signed velocity areas to obtain position changes.

1. Find the velocities at 2, 5 and 7 s and describe the velocity graph between them. Find all zero-velocity times or intervals.

2. Find the smallest position reached, the final position and the total distance traveled.

3. What does the total acceleration area over 0–7 s represent? Why is it neither final velocity by itself nor displacement?

4. Suppose the same acceleration record instead starts with $v(0)=0$, keeping $x(0)=2\ \mathrm{m}$. How do the final velocity and position change? Can an acceleration graph alone establish rest on its zero portion?

Question 2 - Solution

An acceleration area needs an initial velocity

(a) Carry the velocity forward through each area.

The acceleration areas are $+4,-3,0\ \mathrm{m/s}$. Therefore $$\boxed{v(2)=3,\quad v(5)=0,\quad v(7)=0\quad(\mathrm{m/s})}.$$ The velocity graph rises linearly from $-1$ to 3, falls linearly to zero and remains zero. On the first stage $v=-1+2t$ in SI numerical values, so the reversal occurs at $\boxed{t=0.5\ \mathrm{s}}$. The cart is then at rest throughout $5\leq t\leq7\ \mathrm{s}$.

(b) Integrate velocity, splitting distance at the reversal.

The initial negative triangle has area $-\tfrac12(0.5)(1)=-0.25\ \mathrm{m}$. Hence $\boxed{x_{\min}=1.75\ \mathrm{m}}$ at 0.5 s. The signed area during 0–2 s is the trapezoid $2(-1+3)/2=2\ \mathrm{m}$. The next stage contributes $3(3+0)/2=4.5\ \mathrm{m}$. The final stage contributes zero: $$\boxed{x(7)=2+2+4.5=8.5\ \mathrm{m}}.$$ Displacement is 6.5 m. Restoring the canceled backward length gives $$\boxed{D=6.5+2(0.25)=7.0\ \mathrm{m}}.$$

(c) Check the units and the integration constant.

Total acceleration area is $\boxed{\Delta v=1\ \mathrm{m/s}}$. Adding $v(0)=-1\ \mathrm{m/s}$ gives final velocity zero. Position change requires integrating velocity, whose area has units of metres.

(d) A constant shift in velocity accumulates throughout the run.

$$v_{\mathrm{new}}(t)-v(t)=1\ \mathrm{m/s},\qquad x_{\mathrm{new}}(t)-x(t)=(1\ \mathrm{m/s})t.$$

Raising the initial velocity by 1 m/s raises the entire velocity graph by that amount. The new final velocity is $\boxed{1\ \mathrm{m/s}}$ and displacement increases by $1(7)=7\ \mathrm{m}$. Thus $\boxed{x(7)=15.5\ \mathrm{m}}$. Zero acceleration only fixes velocity at its already attained value; it does not force that value to be zero.