Kinematics Questions and Answers

Answer: B

\(T_z\) is \(a_z - b_z + c_z\).

Vectors add and subtract component by component. The z-component of \(\vec{A} - \vec{B} + \vec{C}\) uses only the z-components of the three vectors.

Every other option mixes components from different axes (\(a_x\), \(a_y\) or \(c_x\) with z-components), or gives the y-component \(a_y - b_y + c_y\) instead. The same rule gives \(T_x = a_x - b_x + c_x\) and \(T_y = a_y - b_y + c_y\).

Answer: E

Its average acceleration is \(\frac{2\pi r}{t^{2}}\).

The speed is \(v = \frac{\pi r}{t}\) (half the circumference in time t). At the end of the semicircle the velocity has reversed, so the change in velocity has magnitude 2v.

\[ \begin{aligned} a_{avg} &= \frac{|\Delta \vec{v}|}{t} = \frac{2v}{t} \\ &= \frac{2\pi r}{t^{2}} \end{aligned} \]

The others are wrong: the average speed is \(\frac{\pi r}{t}\), not \(\frac{\pi r}{2t}\); the displacement is 2r (the diameter), so the average velocity is \(\frac{2r}{t}\); and the distance is \(\pi r\), not 2r.

Answer: A

The acceleration at t = 2 s is 0.

Differentiate the position twice:

\[ \begin{aligned} x &= t^{3} - 6t^{2} + 9t \\ v &= 3t^{2} - 12t + 9 \\ a &= 6t - 12 \\ a(2) &= 12 - 12 = 0 \end{aligned} \]

The other values come from stopping at the velocity or slipping a sign: v(2) itself is -3 m/s, and 6 m/s² is just the coefficient of t in the acceleration.