Statics and Dynamics Quiz: Rectilinear Motion

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Rectilinear MotionQuestion 1 of 5

During a braking test, a vehicle traveling at $30 \ \text{m/s}$ applies full brakes and decelerates uniformly to rest. The stopping distance is measured as $90 \ \text{m}$ . In a second test, the same vehicle begins braking from $30 \ \text{m/s}$ but the brakes are applied $1 \ \text{s}$ later than intended — the vehicle travels at constant $30 \ \text{m/s}$ for $1 \ \text{s}$ before braking begins.

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?

The vehicle travels

15 \ \text{m}

farther than in test 1 because the delayed braking reduces the effective initial speed for the kinematic equations by half

The vehicle travels

30 \ \text{m}

farther than in test 1, but the deceleration in test 2 is greater because the vehicle has more kinetic energy to dissipate in the same distance

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

The vehicle travels

45 \ \text{m}

farther than in test 1 because the 1 s delay adds both extra distance and effectively increases the braking distance through momentum effects

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Rectilinear Motion

Practice Rectilinear Motion in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rectilinear Motion, 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.

All questions

Question 1

The vehicle travels $15 \ \text{m}$ farther than in test 1 because the delayed braking reduces the effective initial speed for the kinematic equations by half
The vehicle travels $30 \ \text{m}$ farther than in test 1, but the deceleration in test 2 is greater because the vehicle has more kinetic energy to dissipate in the same distance
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 (correct answer)
The vehicle travels $45 \ \text{m}$ farther than in test 1 because the 1 s delay adds both extra distance and effectively increases the braking distance through momentum effects

Explanation: 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.

Question 2

A particle moves along a straight track. At time $t = 0$ , the particle has velocity $v_0 = +12 \ \text{m/s}$ and acceleration $a = -4 \ \text{m/s}^2$ . A second particle starts from the same position at $t = 0$ with velocity $v_0 = 0$ and acceleration $a = +2 \ \text{m/s}^2$ .

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?

$t = 2 \ \text{s}$ ; displacement of first particle $= +20 \ \text{m}$
$t = 2 \ \text{s}$ ; displacement of first particle $= +16 \ \text{m}$ (correct answer)
$t = 4 \ \text{s}$ ; displacement of first particle $= +16 \ \text{m}$
$t = 4 \ \text{s}$ ; displacement of first particle $= +20 \ \text{m}$

Explanation: 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.

Question 3

Two trains, A and B, are on parallel tracks. Train A passes a station at $t = 0$ with speed $20 \ \text{m/s}$ and constant deceleration $1 \ \text{m/s}^2$ . At the same instant, Train B is $100 \ \text{m}$ ahead of Train A (in the direction of travel) and moving at $10 \ \text{m/s}$ with constant acceleration $0.5 \ \text{m/s}^2$ .

Does Train A ever catch Train B, and if so, at what time and position (measured from the station)?

Train A catches Train B at $t = 10 \ \text{s}$ at a position $150 \ \text{m}$ from the station
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 (correct answer)
Train A catches Train B at $t = 20 \ \text{s}$ at a position $200 \ \text{m}$ from the station
Train A catches Train B at $t = 10 \ \text{s}$ at a position $175 \ \text{m}$ from the station

Explanation: 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.

Question 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?

The acceleration phase covers twice the distance of the deceleration phase; the deceleration phase accounts for $\frac{1}{3}$ of the total distance (correct answer)
The deceleration phase covers twice the distance of the acceleration phase; the deceleration phase accounts for $\frac{2}{3}$ of the total distance
Both phases cover equal distances because the magnitude of the net velocity change is the same in each phase; the deceleration phase accounts for $\frac{1}{2}$ of the total distance
The acceleration phase covers four times the distance of the deceleration phase; the deceleration phase accounts for $\frac{1}{4}$ of the total distance

Explanation: 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.

Question 5

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 $t^*$ ?

$t^* = 8 \ \text{s}$ , because the particle must travel twice the leftmost-point distance from the origin, and the travel time scales with the square root of distance
$t^* = 2 \ \text{s}$ , because the particle returns to $x_0$ immediately after reaching the leftmost point since the acceleration equals the magnitude of the initial velocity
$t^* = 3 \ \text{s}$ , because the return journey is shorter in time due to the acceleration acting in the positive direction throughout
$t^* = 4 \ \text{s}$ , because by symmetry of constant-acceleration motion, the time to return from the leftmost point equals the time to reach it from $x_0$ (correct answer)

Explanation: 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.