Newton's Second Law — Question 10

Question 10

A force bound constrains a velocity change

A 3 kg particle must change velocity from $(4,0)$ to $(0,3)$ m/s. Its net force can point in any direction but has magnitude at most 6 N at all times. Use $\Delta\vec v=\int\vec a\,dt$.

1. Find the required velocity-change magnitude and a lower time bound.

2. Exhibit a force history that attains the bound.

3. Why is using only the change of speed insufficient?

Question 10 - Solution

A force bound constrains a velocity change

(a) Bound the integral of acceleration.

$$|\Delta\vec v|=|(-4,3)|=5,\quad |\Delta\vec v|\le\int_0^T\frac{F_{\max}}m\,dt=2T\ \Rightarrow\ \boxed{T\ge2.5\,\mathrm s}$$ $\Delta\vec v=(-4,3)$ m/s has magnitude 5 m/s. Since $|\vec a|\le2$ m/s$^2$, $5\le2T$, so $T\ge2.5$ s.

(b) Use a constant force along the velocity change.

$$\vec a=\frac{(-4,3)}{2.5}=(-1.6,1.2),\quad\vec F=3\vec a=(-4.8,3.6)\,\mathrm N$$ Choose $\vec a=(-1.6,1.2)$ m/s$^2$ for 2.5 s, giving $\vec F=(-4.8,3.6)$ N with magnitude 6 N. It produces the exact final vector, so the bound is attainable.

(c) Keep directional change.

$$\big||\vec v_f|-|\vec v_i|\big|=|3-4|=1\,\mathrm{m/s}<|\vec v_f-\vec v_i|=5\,\mathrm{m/s}$$ Speed changes only from 4 to 3 m/s, a magnitude difference of 1 m/s. The full vector change is 5 m/s because the direction turns. A time based on 1 m/s would not satisfy the specified endpoint velocities.