Rectilinear Motion - Statics and Dynamics

1. At what time $t > 0$ do the two particles have the same velocity? What is the displacement of the first particle from its starting position at that instant?
   Answer: $t = 2 \ \text{s}$; displacement of first particle $= +16 \ \text{m}$ When two particles move with constant acceleration, you can find when their velocities match by setting their velocity equations equal — this tests your ability to apply kinematics systematically before computing displacement. For Particle 1: $v_1 = 12 - 4t$. For Particle 2: $v_2 = 0 + 2t$. Setting them equal: $12 - 4t = 2t$, which gives $6t = 12$, so $t = 2 \ \text{s}$. This immediately eliminates choices C and D, which both claim $t = 4 \ \text{s}$ — a result you'd get if you incorrectly subtracted accelerations as $12 \div (4-2) = 6$, forgetting that the accelerations add when one is negative and one is positive. The correct denominator is $4 + 2 = 6$. Now compute the displacement of Particle 1 at $t = 2 \ \text{s}$: $\Delta x_1 = v_0 t + \tfrac{1}{2}at^2 = 12(2) + \tfrac{1}{2}(-4)(2)^2 = 24 - 8 = +16 \ \text{m}$ This confirms answer B. Choice A makes $t = 2 \ \text{s}$ correctly but reports $+20 \ \text{m}$, the error that comes from forgetting the $\frac{1}{2}$ in the kinematic equation — a very common slip where students write $\Delta x = v_0 t + at^2$ instead of $\frac{1}{2}at^2$. Choice D compounds both errors: wrong time and wrong displacement. Your study tip: always write out both velocity equations explicitly before solving — it prevents sign errors and keeps the algebra clean. And double-check your displacement formula; that $\frac{1}{2}$ is frequently the difference between right and wrong on kinematics questions.
2. A particle's position along the $x$-axis is described by $x(t) = x_0 + v_0 t + \tfrac{1}{2}at^2$. At $t = 0$, $x_0 = 5 \ \text{m}$ and $v_0 = -8 \ \text{m/s}$. The particle reaches its leftmost point at $t = 2 \ \text{s}$, then returns and passes through $x = 5 \ \text{m}$ again at time $t^*$. What is $t^*$?
   Answer: $t^* = 4 \ \text{s}$, because by symmetry of constant-acceleration motion, the time to return from the leftmost point equals the time to reach it from $x_0$ When you see constant-acceleration kinematics, look for symmetry — it's one of the most powerful shortcuts available. First, find the acceleration. Since the particle reaches its leftmost point at $t = 2\ \text{s}$, velocity equals zero there. Using $v = v_0 + at$: $0 = -8 + a(2)$, giving $a = +4\ \text{m/s}^2$. Now, returning to $x = 5\ \text{m}$ means solving $5 = 5 + (-8)t + \tfrac{1}{2}(4)t^2$, which simplifies to $0 = -8t + 2t^2 = 2t(t - 4)$. The nontrivial solution is $t^* = 4\ \text{s}$, confirming D. Notice the elegant symmetry: the particle takes 2 seconds to reach its leftmost point and another 2 seconds to return — equal intervals, because constant acceleration produces mirror-image motion about the turning point. A is wrong because it invents a square-root scaling rule that doesn't apply here. Distance and time in constant-acceleration motion are related quadratically, but that doesn't mean return time scales with $\sqrt{\text{distance}}$ — the algebra above shows a clean linear relationship for the time interval. B incorrectly claims the particle returns to $x_0$ almost immediately. The "acceleration equals initial velocity magnitude" coincidence (both equal 8 in different units) is a red herring and implies no meaningful shortcut. C suggests the return trip is shorter in time, which contradicts the symmetry principle. The same constant acceleration acts throughout both legs of the journey — there's no asymmetry. Study tip: Whenever a constant-acceleration problem has a turning point, immediately check for time symmetry. The time from start to turning point equals the time from turning point back to the original position — this shortcut can save significant calculation time on exam day.
3. In the second test, how much farther does the vehicle travel (total stopping distance minus that in test 1) compared to the first test? Additionally, which of the following correctly describes the deceleration in test 2?
   Answer: The vehicle travels $30 \ \text{m}$ farther than in test 1, and the deceleration in test 2 is identical to test 1 because the braking force and vehicle mass are unchanged When you see a question mixing kinematics with a time delay, separate the motion into distinct phases and ask: does the braking phase itself change between tests? In test 1, use $v^2 = v_0^2 + 2a\Delta x$ to find the deceleration: $0 = (30)^2 + 2a(90)$, giving $a = -5 \ \text{m/s}^2$. In test 2, the vehicle coasts at $30 \ \text{m/s}$ for $1 \ \text{s}$, covering an extra $d = 30 \times 1 = 30 \ \text{m}$ before braking begins. Once braking starts, the conditions are identical — same vehicle, same brakes, same initial braking speed of $30 \ \text{m/s}$ — so the deceleration remains $-5 \ \text{m/s}^2$ and the braking distance is still $90 \ \text{m}$. Total distance in test 2: $120 \ \text{m}$, exactly $30 \ \text{m}$ farther. Answer C is correct. Choice A incorrectly suggests the delay somehow halves the effective initial speed — there is no such kinematic principle. The vehicle is still moving at $30 \ \text{m/s}$ when brakes engage. Choice B gets the extra distance right ($30 \ \text{m}$) but falsely claims deceleration increases. Deceleration depends on braking force and mass, neither of which changes — not on how far the car traveled before braking. Choice D invents a "momentum effect" that inflates the extra distance to $45 \ \text{m}$; momentum doesn't extend braking distance beyond what kinematics already accounts for. A reliable strategy: whenever a problem adds a delay before a known motion phase, calculate the delay phase separately, then ask whether anything physically changes in the subsequent phase. If the same force acts on the same mass, deceleration is unchanged.
4. A particle starts from rest and moves along a straight line. For the first $T$ seconds it accelerates at $a$, reaching a maximum speed $v_{\max}$. It then immediately decelerates at $2a$ until it comes to a complete stop. Which of the following correctly describes the relationship between the distances traveled in each phase, and what fraction of the total distance is covered during the deceleration phase?
   Answer: The acceleration phase covers twice the distance of the deceleration phase; the deceleration phase accounts for $\frac{1}{3}$ of the total distance When a particle accelerates from rest and then decelerates to rest, the key insight is that distance depends on both speed and time — not just the change in speed. Use the kinematic relationship $d = \frac{v_{\max}^2}{2a}$ (derived from $v^2 = u^2 + 2ad$) to compare the two phases precisely. Acceleration phase: Starting from rest at rate $a$, the distance covered is $d_1 = \frac{v_{\max}^2}{2a}$. Deceleration phase: Starting at $v_{\max}$ and stopping at rate $2a$, the distance is $d_2 = \frac{v_{\max}^2}{2(2a)} = \frac{v_{\max}^2}{4a}$. Comparing: $d_1 = \frac{v_{\max}^2}{2a}$ is exactly twice $d_2 = \frac{v_{\max}^2}{4a}$. The total distance is $d_1 + d_2 = \frac{3v_{\max}^2}{4a}$, so the deceleration phase accounts for $\frac{d_2}{d_1+d_2} = \frac{1}{3}$ of total distance. That confirms answer A. Answer B inverts the relationship — higher deceleration means less time at high speed, so less distance, not more. Answer C incorrectly assumes equal velocity change implies equal distance; it ignores that the time intervals differ. Answer D suggests a factor of four difference, which would require deceleration at $4a$, not $2a$ — a miscalculation of the ratio. Study tip: Whenever phases involve different accelerations, go straight to $d = \frac{v^2}{2a}$ rather than reasoning intuitively about speed. Higher acceleration always means shorter distance for the same velocity change — a counterintuitive result that frequently appears as a trap.
5. Does Train A ever catch Train B, and if so, at what time and position (measured from the station)?
   Answer: Train A never catches Train B; their separation reaches a minimum of approximately $66.7 \ \text{m}$ and then grows as Train B pulls away When two objects move along the same path, the key question is whether the faster one ever closes the gap — and the best tool is the relative motion (or gap) function. Let Train A's position be $x_A = 20t - \frac{1}{2}t^2$ and Train B's position be $x_B = 100 + 10t + \frac{1}{2}(0.5)t^2$. Train A catches Train B only if the gap $G(t) = x_B - x_A$ ever reaches zero: $G(t) = 100 + 10t + 0.25t^2 - 20t + 0.5t^2 = 100 - 10t + 0.75t^2$ This is a upward-opening parabola, so it has a minimum, not a zero crossing (unless the discriminant is non-negative). The minimum occurs at $t = \frac{10}{2(0.75)} = \frac{10}{1.5} \approx 6.67 \ \text{s}$, giving $G_{\min} = 100 - 10(6.67) + 0.75(6.67)^2 \approx 66.7 \ \text{m}$. Since the minimum gap is positive, Train A never catches Train B — confirming B is correct. Choice A claims a catch at $t = 10 \ \text{s}$, but plugging in gives $G(10) = 100 - 100 + 75 = 75 \ \text{m} \neq 0$. Choice D makes the same timing error with a different position. Choice C tries $t = 20 \ \text{s}$, but by then Train B has accelerated far ahead — $G(20) = 100 - 200 + 300 = 200 \ \text{m}$. Strategy tip: Always set up the gap function $G(t) = x_{\text{ahead}} - x_{\text{behind}}$ and check its discriminant ($b^2 - 4ac$). If it's negative for an upward parabola, the gap never closes — no catch occurs.

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