Statics and Dynamics Quiz: Curvilinear Motion N T Coordinates

What this quiz covers

This quiz focuses on Curvilinear Motion N T Coordinates, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $a_t = 2 \text{ m/s}^2$ and $a_n = 7.81 \text{ m/s}^2$, because the tangential acceleration equals the rate of speed change and the normal component is $v^2/\rho$ evaluated at 25 m/s (correct answer)
2. $a_t = 2 \text{ m/s}^2$ and $a_n = 5.00 \text{ m/s}^2$, because the tangential acceleration equals the rate of speed change and the normal component is $v/\rho$ evaluated at 25 m/s
3. $a_t = 5 \text{ m/s}^2$ and $a_n = 7.81 \text{ m/s}^2$, because the tangential acceleration equals the total speed change divided by the total distance and the normal component is $v^2/\rho$ evaluated at 25 m/s
4. $a_t = 2 \text{ m/s}^2$ and $a_n = 3.13 \text{ m/s}^2$, because the tangential acceleration equals the rate of speed change and the normal component is $v^2/\rho$ evaluated at the average speed of 25 m/s divided by 2

Question 2

1. $a = \sqrt{(6)^2 + (0.8)} \approx 6.07 \text{ m/s}^2$, where $a_t = 6$ m/s² and $a_n = v^2/\rho = (2)^2/5 = 0.8$ m/s², but $a_n$ is mistakenly added without squaring inside the radical
2. $a = \sqrt{(6)^2 + (0.8)^2} \approx 6.05 \text{ m/s}^2$, where $a_t = \dot{v} = 8t - 2 = 6$ m/s² and $a_n = v^2/\rho = (2)^2/5 = 0.8$ m/s² (correct answer)
3. $a = \sqrt{(2)^2 + (0.8)^2} \approx 2.15 \text{ m/s}^2$, where $a_t$ is incorrectly taken as the speed $v = 4(1)^2 - 2(1) = 2$ m/s instead of the derivative, and $a_n = v^2/\rho = (2)^2/5 = 0.8$ m/s²
4. $a = \sqrt{(6)^2 + (0.2)^2} \approx 6.00 \text{ m/s}^2$, where $a_t = 6$ m/s² and $a_n = 1/\rho = 1/5 = 0.2$ m/s², treating the curvature scalar directly as the normal acceleration

Question 3

1. $a_n$ remains constant because $v$ is constant and $a_n = \dot{v} = 0$, while $a_t$ increases as the curve tightens, so total acceleration $a$ increases in the $\hat{e}_t$ direction only
2. $a_n$ increases (because $\rho$ decreases with $v$ constant), $a_t$ also increases (because the path curves more sharply, requiring greater tangential effort), and therefore $a$ increases with components in both the $\hat{e}_n$ and $\hat{e}_t$ directions
3. $a_n$ increases (because $\rho$ decreases with $v$ constant), $a_t = 0$ remains zero (constant speed), and therefore $a$ increases — the total acceleration vector grows in magnitude and remains aligned with the $\hat{e}_n$ direction (correct answer)
4. $a_n$ decreases because the car must slow its turning rate to maintain constant speed on a tightening curve, $a_t = 0$ remains zero, and therefore $a$ decreases as the curve tightens at fixed speed

Question 4

1. Both particles have the same total acceleration magnitude and the same direction because the sign of $a_t$ only indicates the direction of travel along the path, not the orientation of the acceleration vector in the n-t frame
2. The accelerating particle has a greater total acceleration magnitude because positive $a_t$ adds constructively with $a_n$ while the decelerating particle's negative $a_t$ partially cancels $a_n$, so the two particles have different magnitudes
3. Both particles have identical total acceleration vectors (same magnitude and same direction) because they share the same speed, the same $|a_t|$, and the same $\rho$ at their respective locations
4. Both particles have the same total acceleration magnitude ($= 5$ m/s²) but their total acceleration vectors point in different directions in the n-t plane — the accelerating particle's vector is in the first quadrant and the decelerating particle's is in the fourth quadrant (taking $\hat{e}_n$ as positive and $\hat{e}_t$ as positive in the direction of motion) (correct answer)

Question 5

1. $a_n = R|\ddot{\theta}| = 2(12) = 24 \text{ m/s}^2$, where $\ddot{\theta} = 6t = 12$ rad/s² at $t = 2$ s, confusing angular acceleration with the source of normal acceleration
2. $a_n = R\dot{\theta}^2 = 2(9)^2 = 162 \text{ m/s}^2$, where $\dot{\theta} = 3t^2 - 3 = 9$ rad/s at $t = 2$ s and $v = R\dot{\theta} = 18$ m/s, giving $a_n = v^2/R = 324/2 = 162$ m/s² (correct answer)
3. $a_n = R\dot{\theta} = 2(9) = 18 \text{ m/s}^2$, where $\dot{\theta} = 3t^2 - 3 = 9$ rad/s at $t = 2$ s, treating angular velocity linearly rather than squaring it in the centripetal formula
4. $a_n = R\sqrt{\dot{\theta}^2 + \ddot{\theta}^2} = 2\sqrt{81+144} \approx 30.0 \text{ m/s}^2$, combining angular velocity and angular acceleration in quadrature as if both contribute to the normal component

Question 6

1. $a_n/a_t \approx 17.9$; the total acceleration is directed at $45°$ from the normal because when $a_n \gg a_t$, the vector bisects the two axes in the n-t plane due to the vector addition geometry
2. $a_n/a_t = \frac{v^2/(\rho)}{a_t} \approx 17.9$; the total acceleration is nearly parallel to the path (almost entirely in the $\hat{e}_t$ direction), because a large ratio means the tangential component dominates, with $\theta \approx 3.2°$ from the tangent
3. $a_n/a_t = \frac{v/(\rho)}{a_t} = \frac{7800/(6.8\times10^6)}{0.5} \approx 0.0023$; the total acceleration is nearly parallel to the path (almost entirely in the $\hat{e}_t$ direction) because the orbital speed divided by radius gives a negligible normal term
4. $a_n/a_t = \frac{v^2/(\rho)}{a_t} = \frac{(7800)^2/(6.8\times10^6)}{0.5} \approx \frac{8.94}{0.5} \approx 17.9$; the total acceleration is nearly perpendicular to the path (almost entirely in the $\hat{e}_n$ direction), with $\theta \approx \arctan(1/17.9) \approx 3.2°$ from the normal (correct answer)

Question 7

1. $a \approx 4.00 \text{ m/s}^2$, found by taking only the tangential acceleration $a_t = 4$ m/s² and assuming the normal acceleration is negligibly small at this point on the parabola
2. $a \approx 2.60 \text{ m/s}^2$, found by computing $\rho = (1+2(1))^{3/2}/|2| = (3)^{3/2}/2 \approx 2.60$ m (using $y'$ instead of $(y')^2$ in the formula), then $a_n = v^2/\rho = 9/2.60 \approx 3.46$ m/s², and $a = \sqrt{4^2 + 3.46^2} \approx 5.30$ m/s²
3. $a \approx 4.31 \text{ m/s}^2$, found by computing $\rho = (1+(2)^2)^{3/2}/|2| \approx 5.59$ m, then $a_n = v^2/\rho = 9/5.59 \approx 1.61$ m/s², and $a = \sqrt{4^2 + 1.61^2} \approx 4.31$ m/s² (correct answer)
4. $a \approx 4.13 \text{ m/s}^2$, found by computing $\rho = (1+(2)^2)^{3/2}/|2| \approx 5.59$ m, then $a_n = v/\rho = 3/5.59 \approx 0.54$ m/s² (using speed instead of speed squared), and $a = \sqrt{4^2 + 0.54^2} \approx 4.04$ m/s²

Question 8

1. Net acceleration magnitude $= \sqrt{(20)^2 + (37.5)^2} \approx 42.5 \text{ m/s}^2$; the normal acceleration points downward (toward the center of the loop, which is below the aircraft at the top) (correct answer)
2. Net acceleration magnitude $= \sqrt{(20)^2 + (37.5)^2} \approx 42.5 \text{ m/s}^2$; the normal acceleration points upward (away from the center, since centrifugal tendency acts outward at the top of the loop)
3. Net acceleration magnitude $= \sqrt{(20)^2 + (9.81)^2} \approx 22.3 \text{ m/s}^2$; the normal acceleration points downward because at the top of the loop gravity provides all centripetal acceleration and replaces $v^2/\rho$
4. Net acceleration magnitude $= \sqrt{(20)^2 + (47.31)^2} \approx 51.4 \text{ m/s}^2$; the normal acceleration points downward, where $a_n = v^2/\rho + g = 37.5 + 9.81$ because gravity adds directly to centripetal acceleration at the top