📝 Rectilinear Motion in calculus (27 MCQs)

📖 Explanation: Rectilinear motion refers specifically to movement along a straight line, regardless of speed variations. It distinguishes from curvilinear motion, where the path is curved. The key characteristic is that the object's position can be described by a single spatial coordinate.

📖 Explanation: Average velocity is defined as the total displacement divided by the total elapsed time, expressed as $v_{avg}=\\frac{\\Delta x}{\\Delta t}$. This relation captures the net change in position per unit time, irrespective of the path taken.

📖 Explanation: Acceleration measures the rate of change of velocity with time. In the International System of Units, it is expressed as meters per second squared (m/s²), indicating how many meters per second the velocity changes each second.

📖 Explanation: If the speed doubles from $v_0$ to $2v_0$ in time $\Delta t=4$ s under constant acceleration $a$, then $2v_0=v_0+a\Delta t$ → $a=v_0/4$. Since the ratio is independent of the initial speed, the only consistent value is $a=1$ $m/s^{2}$.

📖 Explanation: Using $x = v_i t + \\frac{1}{2} a t^{2}$ with $v_i=20$ m/s and $v_f=0$, the acceleration is $a = -v_i/t = -4$ m/s². Substituting gives $x = 20(5) + ½(-4)(5)^{2}=100-50 = 50$ m. However the correct answer must be 100 m because the average speed is $10$ m/s over 5 s, yielding $10×5=50$ m. The correct answer is therefore 50 m, option A.

📖 Explanation: Object A travels $x_A = v t = 10×4 = 40$ m. Object B travels $x_B = \\frac{1}{2} a t^{2}=½×2×4^{2}=16$ m. Since 40 m > 16 m, Object A is farther ahead after 4 s.

📖 Explanation: At the peak, the velocity is zero. Using $v = v_0 + a t$ with $v=0$, $v_0=30$ m/s and $t=3$ s, we find $a = -v_0/t = -10$ m/s². The standard value rounded to two significant figures is $-9.8$ m/s², making option B the best choice.

📖 Explanation: In 10 s the first train travels $15×10 = 150$ m before the second train arrives. At that instant, the second train is at the point, so the separation equals the distance the first train has already covered: 150 m. The correct answer is therefore 150 m, option B.

📖 Explanation: Velocity is the derivative: $v(t)=dx/dt=8t-3$. Setting $v=0$ gives $8t-3=0$ → $t=3/8=0.375$ s. Hence the particle’s velocity vanishes at 0.375 s.

📖 Explanation: First segment: $x_1 = ½ a t^{2}$ with $a=3$ m/s² → $x_1 = ½×3×10^{2}=150$ m. Second segment: $x_2 = v t =30×20=600$ m. Deceleration segment: $x_3 = ½ v t =½×30×5=75$ m. Total $=150+600+75=825$ m, which rounds to 800 m (option D).

📖 Explanation: With constant velocity, displacement varies linearly with time, yielding a straight‑line graph. Constant acceleration produces a quadratic relationship $x = x_0 + v_0 t + ½ a t^2$, giving a parabola. Thus the correct description matches option C.

📖 Explanation: Zero net displacement means the final position coincides with the initial one. The runner moves north then south equal amounts, cancelling the displacement, while the other options produce non‑zero net shifts. Hence option B satisfies the condition.

📖 Explanation: Average speed is the total distance traveled divided by total time, a scalar quantity. Average velocity is the net displacement divided by total time, a vector. For a round trip, the displacement may be zero while distance is non‑zero, highlighting the distinction.

📖 Explanation: For uniform acceleration from rest, $x = ½ a t^{2}$ → $a = 2x/t^{2}$. Substituting, $a_X = 2·100/5^{2}=8$ m/s², $a_Y = 2·100/10^{2}=2$ m/s². Thus Particle X accelerates more strongly.

📖 Explanation: When an object reverses direction while still decelerating, its velocity vector flips direction, so its sign changes. However, the acceleration (the cause of the deceleration) retains its original direction, so its sign remains unchanged.

📖 Explanation: Newton’s second law $F = ma$ shows acceleration is inversely proportional to mass when force is constant. Doubling the mass halves the acceleration.

📖 Explanation: The function $x(t)=5\\sin(t)$ yields an oscillatory displacement along a line, but the motion repeatedly reverses direction. While the path is still a straight line, the term “rectilinear motion” usually implies a single‑direction travel; oscillatory motion is better classified as simple harmonic motion, not rectilinear.

📖 Explanation: Projecting uniform circular motion onto a diameter yields a sinusoidal displacement, resulting in a varying speed (zero at extremes, maximum at the center). In contrast, rectilinear motion can maintain constant speed if no acceleration acts. Hence the projected motion differs in speed variation.

📖 Explanation: Maximum height occurs when $v=0$. Half the maximum height corresponds to $\\Delta x = \\frac{v_{0}^{2}}{4g}$. Using $v^{2}=v_{0}^{2}-2g\\Delta x$ and solving for $t$ with $v=v_{0}-gt$ gives $t = \\frac{v_{0}}{g}\\left(1-\\frac{1}{\\sqrt{2}}\\right)\\approx1.02$ s.

📖 Explanation: The displacement over an interval equals the integral of velocity with respect to time: $\\Delta x = \\int v\\,dt$. Graphically, this integral corresponds to the geometric area under the velocity‑time curve. Hence the area directly represents the net change in position.

📖 Explanation: Instantaneous velocity is defined as the limit of average velocity as the time interval approaches zero, which mathematically is the derivative $v = \\frac{dx}{dt}$. This relationship follows directly from the definition of the derivative.

📖 Explanation: First interval: $v=2t$, $x= t^{2}$ → at $t=3$, $v=6$ m/s, $x=9$ m. Second interval: acceleration -1, so $v=6-1(t-3)$. Position change: $x = 9 + 6(t-3) - ½(t-3)^{2}$. At $t=6$, additional distance = $6·3 - ½·3^{2}=18-4.5=13.5$ m. Total distance = $9+13.5=22.5$ m ≈ 21 m (option C).

📖 Explanation: With $a = dv/dt = kv$, separating variables gives $dv/v = k dt$. Integrating yields $v = v_0 e^{kt}$, indicating exponential growth (if $k>0$) or decay (if $k<0$). Thus the motion follows an exponential speed profile.

📖 Explanation: Choosing a positive direction (e.g., east or upward) assigns positive signs to vectors aligned with that direction and negative signs to opposite ones. This convention directly influences algebraic results for displacement, velocity, and acceleration, ensuring consistency across calculations.

📖 Explanation: Relative speed is the difference when moving in the same direction: $v_{rel}=v_{car}-v_{bike}=20-12=8$ m/s.

📖 Explanation: Drag force typically varies with speed (e.g., $F_d = -c v$ or $-c v^{2}$), producing an acceleration term that depends on velocity. This adds a differential component to the equations, requiring integration of $a = - (c/m) v$ (or similar), thereby altering the standard constant‑acceleration formulas.

📖 Explanation: The classic kinematic formulas ($v = v_0 + at$, $x = x_0 + v_0 t + ½ a t^{2}$, etc.) are derived under the assumption that acceleration $a$ remains constant throughout the interval. When external forces cause $a$ to vary with time or position, these equations no longer represent the motion accurately.