Average Acceleration and Instantaneous Acceleration — Question 9

Question 9

Infer a constant acceleration from two velocity records

A particle is known to have constant acceleration. At 1 s its velocity is $(5,-2)$ m/s; at 4 s it is $(-1,7)$ m/s. Its position at 1 s is $(2,3)$ m.

1. Find the acceleration and initial velocity at zero.

2. Predict the position at 4 s.

3. An additional record says velocity at 2 s is $(3,2)$ m/s. Is it consistent with the same constant-acceleration model?

Question 9 - Solution

Infer a constant acceleration from two velocity records

(a) Use the measured velocity difference.

$$\vec a=\frac{(-1,7)-(5,-2)}3=\boxed{(-2,3)\,\mathrm{m/s^2}},\quad\vec v_0=(5,-2)-(-2,3)=(7,-5)\,\mathrm{m/s}$$ The change is $(-6,9)$ m/s over 3 s, so $\vec a=(-2,3)$ m/s$^2$. Extrapolating back from 1 s gives $\vec v(0)=(7,-5)$ m/s. Constant acceleration is the assumption permitting that extrapolation.

(b) Use elapsed time from the known position.

$$\vec r(4)=(2,3)+3(5,-2)+\tfrac12(9)(-2,3)=(8,10.5)\,\mathrm m$$ From 1 to 4 s, $\Delta t=3$ s. The displacement is $(5,-2)(3)+\tfrac12(-2,3)(9)=(6,7.5)$ m. Therefore $\vec r(4\ \mathrm s)=(8,10.5)$ m. Using 4 s in this interval equation would misidentify its starting time.

(c) Compare both predicted components.

$$\vec v_{\mathrm{predicted}}(2)=(5,-2)+(-2,3)=(3,1)\,\mathrm{m/s}\ne(3,2)\,\mathrm{m/s}$$ The model predicts $\vec v(2\ \mathrm s)=(5,-2)+(-2,3)=(3,1)$ m/s. The reported y component differs by 1 m/s, so the three exact records cannot all belong to the same constant-acceleration motion. Agreement in x alone is insufficient.