AP Physics 1 Unit 1: Kinematics — Free Review Games.

a = (v - v0)/t = (25 - 0)/5 = 5 m/s^2.

The area under a v-t graph gives the displacement of the object over that time interval.

v = v0 + gt = 0 + 10(3) = 30 m/s downward.

Displacement is a vector with both magnitude and direction, unlike speed and distance which are scalars.

With no air resistance, horizontal velocity stays constant because there is no horizontal force acting on the projectile.

Using v^2 = v0^2 - 2gh at the top (v=0): 0 = 400 - 20h, h = 20 m.

A parabolic position-time graph indicates constant acceleration, as position changes quadratically with time.

Complementary angles (30 and 60) produce the same range for the same initial speed on level ground.

Displacement = final position - initial position = -3 - 5 = -8 m.

Average velocity = (30 + 10)/2 = 20 m/s. Distance = 20 x 4 = 80 m.

Vertical: h = 1/2 gt^2, so 80 = 1/2(10)t^2, t^2 = 16, t = 4 s. Horizontal speed does not affect fall time.

The velocity change vector has magnitude \(\sqrt{10^2 + 10^2} = 14.14 \text{ m/s}\). Average acceleration = \(14.14/2 = 7.07 \text{ m/s}^2\).

Vertical component = v*sin(30) = 40 x 0.5 = 20 m/s.

v_A = dx/dt = 4t + 3, v_B = 5. Setting equal: 4t + 3 = 5, t = 0.5 s.

Using \(y = v_0 t - \frac{1}{2} g t^2\) with \(y = -45 \text{ m}\): \(-45 = 20t - 5t^2\). Rearranging: \(5t^2 - 20t - 45 = 0\), \(t^2 - 4t - 9 = 0\). Using quadratic formula: \(t = \frac{4 + \sqrt{16+36}}{2} = \frac{4 + \sqrt{52}}{2} = \text{about } 5 \text{ s}\).

Instantaneous velocity is defined as the limit of the average velocity as the time interval approaches zero, essentially the derivative of position with respect to time. The choice 'The average velocity over the entire trip' describes average velocity, which uses total displacement over total time rather than a single instant. Students should remember that instantaneous velocity can be read as the slope of a tangent line on a position-time graph at one point.

Velocity includes both magnitude and direction, making it a vector, whereas speed only describes magnitude, making it a scalar. The claim that 'Velocity is always positive while speed can be negative' is backwards since speed, being a magnitude, is never negative. Recognizing which quantities are vectors versus scalars is essential for correctly applying kinematic equations with signs.

The slope of a position-time graph equals the rate of change of position with respect to time, which is by definition velocity. 'Acceleration' is incorrect because acceleration is found from the slope of a velocity-time graph, not a position-time graph. This graphical relationship is a core AP Physics 1 skill for interpreting motion diagrams.

Acceleration is the rate of change of velocity per unit time, so its units are velocity units divided by time again, giving \(m/s^2\). The unit '\(m/s\)' is incorrect because that is the unit of velocity, not acceleration. Keeping track of units helps verify that a kinematic equation has been applied correctly.

The standard value used for free-fall acceleration near Earth's surface is approximately \(9.8\, m/s^2\), directed downward, due to Earth's gravitational pull. The value '\(6.67\, m/s^2\)' is incorrect because \(6.67 \times 10^{-11}\) is instead the gravitational constant \(G\), a different physical quantity entirely. Students should memorize \(g \approx 9.8\, m/s^2\) since it appears in nearly every projectile and free-fall problem.

Displacement measures only the net change in position as a vector, while distance sums the total length of the path traveled regardless of direction. The statement 'Displacement always equals distance for any path' is wrong because they are only equal when motion is in a single direction without reversal. Recognizing this distinction prevents errors when objects change direction during motion.

Acceleration is defined as the rate of change of velocity over time, and since velocity is a vector, acceleration must also be a vector with both magnitude and direction. The choice 'Acceleration always points opposite to the velocity vector' is false because acceleration can point in the same direction as velocity, such as when an object speeds up. Understanding acceleration as a vector is crucial for correctly analyzing both speeding-up and slowing-down motion.

Average velocity is defined as total displacement divided by the total time interval, expressed as \(\bar{v} = \frac{\Delta x}{\Delta t}\). The expression '\(\bar{v} = \frac{\Delta t}{\Delta x}\)' inverts the correct ratio and would give units of seconds per meter, not meters per second. This formula is foundational and used throughout kinematics to relate displacement, velocity, and time.

Since acceleration measures the rate of change of velocity, an object with unchanging velocity has zero acceleration by definition. The choice 'A constant nonzero value' is incorrect because any nonzero constant acceleration would cause velocity to change steadily, contradicting the premise of constant velocity. This concept underlies the distinction between uniform motion and uniformly accelerated motion.

With air resistance neglected, no horizontal force acts on a projectile, so by Newton's second law the horizontal acceleration is zero and horizontal velocity stays constant. The option 'Equal to \(g\)' is wrong because \(g\) only affects the vertical component of motion, causing the vertical velocity to change while the horizontal component remains unaffected. This separation of horizontal and vertical motion is the key strategy for solving all projectile motion problems.

Integrating acceleration over a time interval yields the change in velocity, so the area under an acceleration-time graph equals \(\Delta v\). 'The displacement' is incorrect because displacement instead comes from the area under a velocity-time graph, one level removed from acceleration. Recognizing this pattern of areas under graphs representing the next higher-order quantity is a valuable graphical analysis skill in kinematics.

Gravity continues to act on the ball throughout its flight, including at the peak, so the acceleration remains approximately \(-9.8\, m/s^2\) even though the velocity is momentarily zero. The choice 'It is zero because the velocity is zero' incorrectly assumes acceleration depends on the instantaneous value of velocity rather than on the constant gravitational force. A common exam pitfall is conflating zero velocity with zero acceleration, so students must remember these are independent quantities.

Uniformly accelerated motion means the acceleration vector does not change, allowing the kinematic equations for constant acceleration to be applied directly. 'Velocity remains constant throughout the motion' describes zero acceleration, not uniform acceleration, so it is incorrect. This constant-acceleration condition is the assumption underlying every standard kinematic equation used on the AP exam.

Using \(x = \frac{1}{2}at^2\) with \(a = 4\, m/s^2\) and \(t = 6\, s\) gives \(x = 0.5(4)(36) = 72\, m\). The value '\(24\, m\)' results from forgetting to square the time or omitting the factor of one-half correctly, an easy algebra slip. This equation is one of the core kinematic formulas for constant acceleration starting from rest and should be memorized precisely.

Because the runner returns to the exact starting point, the net displacement is zero, so average velocity, which depends on displacement, must also be zero. The choice '\(5\, m/s\)' is actually the average speed calculated from total distance divided by time, not the average velocity. This problem highlights that average speed and average velocity can differ dramatically for closed-loop paths.

The vertical velocity component is \(v_y = 50(0.6) = 30\, m/s\), and time of flight for a level launch and landing is \(t = \frac{2v_y}{g} = \frac{60}{10} = 6\, s\). The choice '\(3\, s\)' represents only the time to reach maximum height, forgetting to double it for the full up-and-down trip. Time of flight problems require separating the vertical component from the total speed before applying the symmetric flight-time formula.

Using \(v^2 = v_0^2 - 2as\) with \(v = 0\), \(v_0 = 20\, m/s\), and \(a = 5\, m/s^2\) gives \(s = \frac{400}{10} = 40\, m\). The value '\(20\, m\)' would result from incorrectly dividing \(v_0\) by \(a\) rather than applying the correct kinematic relationship. This equation is especially useful when time is not given and only velocities and acceleration are known.

A straight line on a position-time graph indicates a constant slope, and a negative slope means the object moves with constant velocity in the negative direction. 'Increasing speed over time' would instead require a curved, steepening graph rather than a straight line. Straight lines on position-time graphs always signal zero acceleration, regardless of the sign of the slope.

Total displacement is \(30 + 40 = 70\, m\) and total time is \(10\, s\), so average velocity is \(\frac{70}{10} = 7\, m/s\) north. The choice '\(5\, m/s\) north' incorrectly averages the two individual velocities of \(6\, m/s\) and \(8\, m/s\) using a simple arithmetic mean rather than total displacement over total time. Average velocity must always be computed from the whole trip's net displacement and total elapsed time, not by averaging segment velocities.

Acceleration equals the slope of a velocity-time graph, calculated as \(\frac{10-2}{4} = 2\, m/s^2\). The value '\(8\, m/s^2\)' mistakenly uses only the change in velocity without dividing by the time interval. Reading acceleration as the slope of a velocity-time graph is a standard AP Physics 1 graphical skill.

Since horizontal velocity is unaffected by gravity, the range is simply \(x = v_x t = 12 \times 2 = 24\, m\). The value '\(12\, m\)' incorrectly forgets to multiply by the full flight time, treating the calculation as if \(t = 1\, s\). Horizontal range problems always rely on the constant horizontal velocity combined with the independently determined fall time.

Using \(h = \frac{1}{2}gt^2\), solving for \(t\) gives \(t = \sqrt{\frac{2(45)}{10}} = \sqrt{9} = 3\, s\). The choice '\(4.5\, s\)' comes from dividing height by gravity without taking a square root, a common algebraic error. Free-fall time-to-fall problems always require solving the quadratic relationship between height, gravity, and time correctly.

Using \(x = v_0 t + \frac{1}{2}at^2 = 5(3) + 0.5(2)(9) = 15 + 9 = 24\, m\). The value '\(15\, m\)' only accounts for the constant-velocity term and ignores the additional displacement contributed by acceleration. This equation is essential whenever both an initial velocity and a nonzero acceleration are present.

Using \(v = v_0 + at = 10 + (10)(3) = 40\, m/s\) downward, since both the initial velocity and gravity act in the same downward direction. The choice '\(30\, m/s\) downward' comes from omitting the initial \(10\, m/s\) contribution and using only the gravitational term. When an object is thrown downward, initial velocity and gravitational acceleration add together rather than opposing each other.

Using \(v = v_0 + at = 2 + (1.5)(10) = 2 + 15 = 17\, m/s\). The value '\(15\, m/s\)' comes from computing only the change in velocity from acceleration while forgetting to add the initial velocity of \(2\, m/s\). This simple linear equation is one of the most frequently used kinematic formulas on the AP exam.

Taking the derivative gives \(v(t) = 6t - 2\), and substituting \(t = 2\) yields \(v = 12 - 2 = 10\, m/s\). The choice '\(6\, m/s\)' comes from only using the coefficient of the \(t^2\) term without completing the differentiation and substitution correctly. Differentiating a position function is the calculus-based approach to finding instantaneous velocity, an important skill for AP Physics 1 with Calculus-adjacent reasoning.

By the symmetry of projectile motion under constant gravitational acceleration, the speed at a given height is the same going up and coming down, though the vertical velocity component points downward on descent instead of upward. The choice 'Speed on the way down is always greater' is wrong because horizontal velocity is unchanged and vertical speed at equal heights is identical in magnitude on both sides of the trajectory. This up-down symmetry is a powerful shortcut for solving projectile motion problems without redoing full calculations.

Relative velocity is found by subtracting velocities as vectors, so \(v_{A/B} = 25 - 15 = 10\, m/s\) east. The choice '\(40\, m/s\) east' incorrectly adds the two velocities instead of subtracting them, which would only be appropriate if the cars moved in opposite directions. Relative velocity problems require careful attention to the direction of each vector before combining them.

An increasingly steep curve on a velocity-time graph means the slope, which represents acceleration, is growing larger over time. 'Acceleration is constant' is incorrect because constant acceleration would produce a straight line rather than a curve of changing steepness. Recognizing curvature on a velocity-time graph as a sign of non-constant acceleration is essential for advanced graph interpretation.

Using \(h = \frac{v_y^2}{2g} = \frac{900}{20} = 45\, m\), derived from setting final vertical velocity to zero at the peak. The choice '\(90\, m\)' comes from forgetting to divide by 2 in the denominator, doubling the correct answer. Maximum height depends only on the vertical velocity component, independent of any horizontal motion.

When acceleration is opposite in sign to velocity, the object is decelerating, meaning its speed decreases over time even though it is still moving in the positive direction. 'It is increasing' would only be true if acceleration and velocity shared the same sign, reinforcing each other. Determining whether an object speeds up or slows down requires comparing the signs of velocity and acceleration, not just their magnitudes.

Using \(h = \frac{1}{2}gt^2\), solving for \(t\) gives \(t = \sqrt{\frac{2(1.25)}{10}} = \sqrt{0.25} = 0.5\, s\). The value '\(1\, s\)' overestimates the fall time by neglecting to properly take the square root of the computed ratio. The time to fall in projectile motion depends only on the vertical height and gravity, never on the horizontal velocity.

Setting the positions equal, \(4t = 30 - 2t\), gives \(6t = 30\), so \(t = 5\, s\), since their combined closing speed of \(6\, m/s\) must cover the initial \(30\, m\) gap. The choice '\(10\, s\)' incorrectly uses only Object A's speed to cover the full 30 m gap, ignoring Object B's contribution to closing the distance. Meeting-point problems require combining the relative closing speed of both objects rather than analyzing only one object's motion.

With \(v_x = 24\, m/s\) and \(v_y = 18\, m/s\), solving \(-20 = 18t - 5t^2\) gives \(t \approx 4.49\, s\), so the horizontal range is \(x = v_x t \approx 24 \times 4.49 \approx 108\, m\). The choice 'About \(75\, m\)' underestimates the time of flight by ignoring the extra fall time contributed by the platform's height above the ground. Launching from an elevated platform requires solving a full quadratic in time rather than using the simple symmetric flight-time formula.

Integrating acceleration gives \(v(t) = 3t^2\), and substituting \(t = 3\) yields \(v = 3(9) = 27\, m/s\). The value '\(18\, m/s\)' results from simply plugging \(t = 3\) into the acceleration function itself rather than integrating it to obtain velocity. When acceleration varies with time, velocity must be found through integration rather than the constant-acceleration kinematic equations.

The range formula \(R = \frac{v^2 \sin(2\theta)}{g}\) is maximized when \(\sin(2\theta) = 1\), which occurs at \(\theta = 45^\circ\). The choice '\(90^\circ\)' actually produces zero horizontal range because the projectile travels straight up and down with no horizontal component. Understanding the range formula's dependence on \(\sin(2\theta)\) helps explain why complementary angles like \(30^\circ\) and \(60^\circ\) yield equal but smaller ranges.

Since the boat's velocity and the current's velocity are perpendicular, the resultant speed is found using the Pythagorean theorem: \(\sqrt{4^2 + 3^2} = \sqrt{25} = 5\, m/s\). The choice '\(7\, m/s\)' incorrectly adds the two perpendicular speeds directly instead of combining them as vector components. Two-dimensional relative velocity problems with perpendicular components always require vector addition using the Pythagorean theorem, not simple arithmetic sums.

During acceleration, \(x_1 = \frac{1}{2}(2)(5^2) = 25\, m\) and the velocity reached is \(v = 2(5) = 10\, m/s\); during constant velocity, \(x_2 = 10(10) = 100\, m\), giving a total of \(125\, m\). The choice '\(100\, m\)' only accounts for the constant-velocity phase and neglects the distance covered during the initial acceleration phase. Multi-stage motion problems require analyzing each phase separately before summing the distances.

Differentiating gives \(v(t) = 3t^2 - 12t + 9\), which factors as \(3(t-1)(t-3) = 0\), so the particle is at rest at \(t = 1\, s\) and \(t = 3\, s\). The choice '\(t = 2\, s\) only' incorrectly assumes the velocity function has a single root rather than correctly factoring the quadratic into two distinct solutions. Finding when a particle is momentarily at rest requires setting the derivative of position equal to zero and solving the resulting equation completely.

Ball 1's velocity is \(v_1 = 10t\), and Ball 2's velocity is \(v_2 = 15 + 10(t-1) = 10t + 5\), so the difference \(v_2 - v_1 = 5\, m/s\) remains constant for all time, meaning they never have equal velocities. The choice 'Yes, at \(t = 1.5\, s\)' incorrectly assumes the velocity difference changes over time, when in fact both balls accelerate identically under gravity, keeping their velocity gap fixed. Comparing motions with a time delay requires writing each object's velocity as a function of the same reference time before checking whether they can ever be equal.

During reaction time the car travels \(20(0.75) = 15\, m\) at constant speed, and during braking \(v^2 = v_0^2 - 2as\) gives \(0 = 400 - 10s\), so \(s = 40\, m\); the total stopping distance is \(15 + 40 = 55\, m\). The choice '\(40\, m\)' ignores the distance traveled during the reaction time before braking even begins. Realistic stopping-distance problems always require adding a constant-velocity reaction phase to the deceleration phase.

The fall time is \(t = \sqrt{\frac{2(20)}{10}} = 2\, s\), and horizontal displacement is \(x = v_x t + \frac{1}{2}a_x t^2 = 10(2) + 0.5(2)(4) = 20 + 4 = 24\, m\). The choice '\(20\, m\)' ignores the additional displacement contributed by the horizontal acceleration from the wind. When a horizontal acceleration is present in projectile motion, the horizontal displacement must include both the constant-velocity term and the acceleration term.

The first segment contributes an area of \(6 \times 4 = 24\, m\), and the second segment contributes \(-3 \times 4 = -12\, m\), giving a net displacement of \(24 - 12 = 12\, m\). The choice '\(24\, m\)' only accounts for the positive segment and ignores the negative contribution from the reversed motion. When calculating displacement from a velocity-time graph, areas below the time axis must be subtracted rather than added.

Relative velocities in the same direction add together, so the ball's velocity relative to the ground is \(8 + 5 = 13\, m/s\) upward. The choice '\(3\, m/s\) upward' incorrectly subtracts the platform's velocity from the ball's velocity, which would only apply if the platform were moving downward. Properly combining reference frame velocities is essential whenever motion is described relative to a moving object rather than a stationary ground frame.

Total distance is \(10 + 6 + 4 = 20\, m\), while net displacement is \(10 - 6 + 4 = 8\, m\) east, giving a ratio of \(\frac{20}{8} = 2.5\). The choice '\(1.0\)' would only be correct if the object never reversed direction, making distance and displacement magnitudes equal. This problem illustrates that direction reversals always cause total distance to exceed the magnitude of net displacement.