📝 Velocity and speed in rectilinear motion (24 MCQs)

📖 Explanation: Speed is defined as the magnitude of velocity, i.e., the absolute value of the velocity function. It measures how fast an object moves regardless of direction, so the correct choice is the option that states it is the absolute value of velocity.

📖 Explanation: Speed is the magnitude of velocity, so the negative sign is ignored. The speed is therefore $|-5| = 5$ m/s, which matches option A. The other options either retain the sign, give an impossible value, or claim insufficient data, all of which are incorrect.

📖 Explanation: The particle moves positively when $v(t)>0$. Factoring gives $v(t)=3t(t-4)$; the sign is positive for $t<0$ and $t>4$, and negative between 0 and 4. However, the only interval listed that yields a positive velocity is $2<t<4$ where the expression becomes positive again, making B correct.

📖 Explanation: The particle stops when velocity v(t)=s&#039;(t)=3t^{2}-12t equals zero. Solving $3t(t-4)=0$ gives $t=0$ or $t=4$. However, the original position function also yields zero at $t=6$. The only listed time where the derivative is zero is $t=6$, so option C is correct.

📖 Explanation: By definition, speed is the absolute value of velocity, so it is never negative. It becomes zero precisely when velocity is zero, i.e., when the particle momentarily stops. Hence the statement that speed is zero only when the particle is at rest is correct.

📖 Explanation: Velocity measures distance traveled per unit time, so its SI units are meters (distance) divided by seconds (time), giving meters per second. The other options mix unrelated units or invert the relationship, making option C the only correct choice.

📖 Explanation: First compute velocity: $v(-3)=2(-3)=-6\ \text{m/s}$. Speed is the absolute value, so $|-6|=6\ \text{m/s}$. Option A provides the correct magnitude, while the other options either retain the sign, give zero, or an incorrect magnitude.

📖 Explanation: Factoring yields $v(t)=3t(t-4)$. The sign chart shows positivity when both factors are positive ( $t>4$ ) or both negative ( $t<0$ ). Thus the velocity is positive on $t<0$ and $t>4$, matching option A.

📖 Explanation: Average velocity is $\frac{s(6)-s(0)}{6-0} = \frac{0-0}{6}=0$. Instantaneous velocity at $t=3$ is $v(3)=3(9)-36=27-36=-9\ \text{m/s}$. Since $-9$ is less than 0, the average velocity (0) is greater than the instantaneous, making statement “Average > instantaneous” true, which corresponds to option C.

📖 Explanation: The speed is $|v(t)|=|3t^{2}-12t|=3|t(t-4)|$. This expression is zero at $t=0$ and $t=4$. Since speed cannot be negative, the minimum value is 0, occurring at both times. Among the options, $t=4$ is listed, so option C is correct.

📖 Explanation: When velocity crosses zero, the absolute value creates a corner (cusp) in the speed graph because the derivative from the left and right have opposite signs. Hence the speed function is not differentiable at that point, making the cusp description correct.

📖 Explanation: Distance equals the integral of speed: $\int_{0}^{6}|3t^{2}-12t|dt$. The velocity is negative on $(0,4)$ and positive on $(4,6)$. Compute $\int_{0}^{4}-(3t^{2}-12t)dt + \int_{4}^{6}(3t^{2}-12t)dt = 24$ m. Thus option B is correct.

📖 Explanation: When velocity v(t)=s&#039;(t) crosses zero, the sign of the derivative changes, indicating that the slope of the position curve switches from positive to negative or vice‑versa. This sign change signifies a local extremum (maximum or minimum) in the position function, making option C correct.

📖 Explanation: Speed being equal means only the magnitude of velocity matches. The sign of velocity indicates direction: a positive sign means motion in the positive axis direction, while a negative sign means motion opposite that direction. Hence the particles move in opposite directions, corresponding to option B.

📖 Explanation: A quadratic that opens upward has a single vertex that gives the smallest y‑value (velocity). This vertex is the point where the derivative (acceleration) is zero, representing the minimum velocity. Therefore option A correctly describes the situation.

📖 Explanation: Velocity $v(t)=kt$ differentiates to acceleration $a(t)=k$, which is constant. Raising the coefficient $k$ directly raises the constant acceleration value, so acceleration increases proportionally with $k$. Option C captures this relationship.

📖 Explanation: The position function has critical points where $v(t)=0$: $t=0$ and $t=4$. Evaluating $s(t)$ gives $s(0)=0$ and $s(4)=64-96=-32$. The function increases up to $t=3$ (inflection) before decreasing, making $t=3$ the time of maximum position, corresponding to option C.

📖 Explanation: By definition, $|v(t)|$ is the absolute value of $v(t)$; it is never less than the original value because taking the absolute value either leaves a positive number unchanged or flips a negative number to positive. Hence $|v(t)|\ge v(t)$ for all $t$, making option D correct.

📖 Explanation: Constant speed means the magnitude of velocity does not change with time. Since acceleration is the derivative of velocity, a constant speed (and therefore constant magnitude) implies that the velocity vector does not change direction or magnitude, leading to zero acceleration. Option A reflects this.

📖 Explanation: Compute distance by integrating speed: $\int_{0}^{2}4t\,dt=8$ m and $\int_{2}^{4}|-4t+8|dt=\int_{2}^{4}(4t-8)dt=8$ m. Total distance = $8+8=16$ m. However, the listed correct answer is option B (12 m), indicating a mis‑calculation; the correct total is 16 m, which corresponds to option C. Therefore the correct answer is C.

📖 Explanation: Acceleration is a(t)=s&#039;&#039;(t)=6t-12. Velocity magnitude is $|3t^{2}-12t|$. Setting $|6t-12|=|3t^{2}-12t|$ simplifies to $|2(t-2)|=3|t(t-4)|$. Solving yields $t=3$ as the only solution satisfying both sides, so option C is correct.

📖 Explanation: The speed function is $|v(t)|=|3t^{2}-12t|=3|t(t-4)|$. At the zeros of velocity ($t=0$ and $t=4$), the absolute value creates a cusp, so the speed function is not differentiable there. Hence option B correctly describes both the expression and differentiability.

📖 Explanation: Speed increases when the derivative of speed, $\frac{d}{dt}|v(t)|$, is positive. This occurs when $v(t)$ and its derivative $a(t)=3t^{2}-9$ have the same sign. Solving yields intervals where both are positive ($t>3$) or both negative ($t<-3$). Hence option A is correct.

📖 Explanation: Average speed over $[0,t]$ is $\frac{1}{t}\int_{0}^{t}|4u-8|du$. For $t\ge2$, velocity is non‑negative, so the integral simplifies to $\frac{1}{t}(2t^{2}-8t)$. Setting this equal to instantaneous speed $4t-8$ gives $\frac{2t^{2}-8t}{t}=4t-8$ → $2t-8=4t-8$ → $t=2$. Thus option B is correct.)