Motion and Kinematics — Free Physics Review Games.

Speed is the rate at which an object covers distance, calculated as distance divided by time.

Speed is a scalar (magnitude only), while velocity is a vector that includes both magnitude and direction.

Acceleration is the rate at which velocity changes with time, measured in m/s^2.

A horizontal line on a position-time graph means position is not changing, so the object is at rest.

The SI unit of velocity is meters per second (m/s).

Acceleration = change in velocity / time = (20 - 0) / 4 = 5 m/s^2.

The slope of a velocity-time graph gives the acceleration of the object.

The acceleration due to gravity near Earth's surface is approximately 9.8 m/s^2 downward.

In projectile motion without air resistance, horizontal acceleration is zero; only gravity acts vertically.

The equation d = v0*t + 1/2*a*t^2 relates displacement to initial velocity, acceleration, and time.

At the top, v = 0. Using v = v0 - gt: 0 = 30 - 9.8t, so t = 30/9.8 = 3.06 s.

In the absence of air resistance, all objects fall at the same rate regardless of mass, as shown by Galileo.

The car covers a greater distance in the second interval (200 m vs 100 m), indicating increasing speed and therefore acceleration.

A launch angle of 45 degrees maximizes the range of a projectile on level ground in the absence of air resistance.

Displacement is a vector; 5 m east minus 3 m west = 2 m east (net displacement).

Acceleration is the rate of change of velocity, so its unit is velocity per time, \(m/s\) divided by \(s\), giving \(m/s^2\). The unit "\(m/s\)" is wrong because that is the unit for velocity, not its rate of change. Always check that units for derived quantities come from dividing the units of the quantities in their definition.

A straight line on a position-time graph has a constant slope, and since slope equals velocity, the object moves at constant velocity. "Constant acceleration" is incorrect because constant acceleration produces a curved, parabolic position-time graph, not a straight line. Remember that the slope of a position-time graph always represents velocity, whether the line is straight or curved.

Instantaneous velocity is defined as the velocity at one specific instant, mathematically the derivative of position with respect to time evaluated at that moment. "The total distance traveled divided by total time" describes average speed, not an instantaneous quantity. Distinguishing instantaneous quantities from average quantities is essential for correctly reading graphs and solving kinematics problems.

Since there is no horizontal acceleration in projectile motion, horizontal velocity stays constant, producing a horizontal straight line on a velocity-time graph. "A line with a downward slope" is incorrect because that describes the vertical velocity, which decreases due to gravity. Remember that horizontal and vertical motions in projectile motion are independent, with only vertical velocity affected by gravitational acceleration.

The area under a velocity-time graph equals the integral of velocity over time, which gives displacement. "Acceleration" is incorrect because acceleration is represented by the slope of the velocity-time graph, not the area beneath it. Knowing which graphical feature (slope vs. area) corresponds to which quantity is key for interpreting motion graphs.

Acceleration measures the rate of change of velocity, and if velocity is unchanging, there is no change to measure, so acceleration is zero. "Negative" is wrong because a negative acceleration would mean the velocity is decreasing, which contradicts the constant velocity described. A constant velocity always corresponds to zero net force and zero acceleration according to Newton's first law.

Speed only describes magnitude, how fast something moves, without any reference to direction, which makes it a scalar quantity. "Vector" is incorrect because vectors require both magnitude and direction, and speed has no directional component, unlike velocity. Distinguishing scalar quantities like speed and distance from vector quantities like velocity and displacement is fundamental throughout kinematics.

Displacement is a vector quantity defined as the straight-line change in position from start to finish, including direction. "The total path length traveled by an object" describes distance, which is a scalar that can differ from displacement if the path is not straight. Remember that displacement can be smaller than distance traveled, or even zero, if the object returns to its starting point.

A horizontally launched projectile has no initial upward or downward motion, so its initial vertical velocity component is zero. "Equal to \(9.8\ m/s\)" is incorrect because \(9.8\ m/s^2\) describes the acceleration due to gravity, not an initial velocity value. In projectile problems, always split the initial velocity into horizontal and vertical components before analyzing each independently.

A curved position-time graph means the slope, which represents velocity, is continuously changing, and a changing velocity is the definition of acceleration. "The object has constant velocity" is wrong because constant velocity produces a straight line, not a curve, on a position-time graph. Curvature on a position-time graph is a direct visual signal that acceleration is present.

Average velocity is calculated as the total displacement, a vector, divided by the total time elapsed for the trip. "Total distance divided by total time" instead defines average speed, which uses distance rather than displacement. Because displacement accounts for direction, average velocity can be zero even if an object traveled a long distance, such as in a round trip.

Gravitational acceleration near Earth's surface always points downward, toward the planet's center, regardless of the object's velocity direction. "There is no acceleration during free fall" is incorrect because free fall is defined by the constant downward acceleration of \(9.8\ m/s^2\) due to gravity. This downward acceleration applies to falling objects even when they are momentarily moving upward, such as a ball thrown into the air.

Range refers specifically to the horizontal distance covered by a projectile from its launch point to where it lands. "The maximum height it reaches" describes a different quantity, the peak vertical displacement, not the horizontal range. Range depends on both the launch angle and initial speed, and it is maximized at a launch angle of 45 degrees for equal launch and landing heights.

Acceleration equals change in velocity divided by time, so \(a = \frac{25 - 5}{10} = 2\ m/s^2\). The choice "\(20\ m/s^2\)" mistakenly uses only the numerator, the velocity change, without dividing by the time interval. Always confirm units cancel correctly and remember acceleration requires dividing by elapsed time, not just subtracting velocities.

When acceleration is opposite in sign to velocity, it acts against the direction of motion, causing the object to decelerate while still moving forward. "It is speeding up in the positive direction" is wrong because speeding up requires acceleration and velocity to share the same sign, not opposite signs. A key exam skill is recognizing that the sign relationship between velocity and acceleration determines whether an object speeds up or slows down.

The displacement equals the area under the velocity-time graph, which for this triangular region is \(\frac{1}{2} \times 6\ s \times 12\ m/s = 36\ m\). The choice "\(72\ m\)" incorrectly omits the factor of \(\frac{1}{2}\) needed for a triangular area rather than a rectangular one. When velocity changes linearly, always use the triangle area formula rather than simply multiplying velocity by time.

Since horizontal and vertical motions are independent, the fall time depends only on the height, found using \(h = \frac{1}{2}gt^2\), giving \(t = \sqrt{\frac{2(45)}{9.8}} \approx 3.0\ s\). The horizontal speed of "\(20\ m/s\)" does not affect the fall time because there is no vertical component to that initial velocity. A common exam strategy is to solve the vertical motion equation independently of any horizontal velocity when finding time of flight.

Using \(d = \frac{1}{2}at^2\) with \(a = 2\ m/s^2\) and \(t = 6\ s\) gives \(d = \frac{1}{2}(2)(36) = 36\ m\). The choice "\(12\ m\)" incorrectly uses \(d = at\) instead of the correct quadratic kinematic equation for constant acceleration starting from rest. Whenever an object starts from rest, remember displacement grows with the square of time, not linearly.

Because the slope of a parabola opening upward becomes steeper over time, and slope represents velocity, the velocity must be continuously increasing. "The velocity is constant" is incorrect since a constant velocity corresponds to a straight line, not a curved parabola, on a position-time graph. This pattern reflects positive constant acceleration, where velocity grows steadily as time passes.

Acceleration is \(\frac{\Delta v}{\Delta t} = \frac{10 - 30}{4} = -5\ m/s^2\), with the negative sign indicating deceleration opposite to the direction of motion. The choice "\(5\ m/s^2\)" has the correct magnitude but omits the necessary negative sign that shows the car is slowing down. Always keep track of signs in kinematics, since a missing negative sign can turn a deceleration into an acceleration in your answer.

Average speed is total distance divided by total time, so \(\frac{300\ km}{2.5\ h} = 120\ km/h\). The choice "\(750\ km/h\)" comes from multiplying instead of dividing the given values, an inverted operation. Always double check that you are dividing distance by time, not the reverse, when computing average speed.

If motion is along a straight line without reversing direction, distance traveled equals the magnitude of displacement, making average speed equal the magnitude of average velocity. "When the object returns to its starting point" is actually the opposite case, where displacement becomes zero while distance traveled remains positive, making the two very different. A useful check is to ask whether the path ever curves or doubles back, since any deviation from a straight, one-directional path makes speed and velocity magnitudes diverge.

Maximum height is found by setting final vertical velocity to zero in \(v_y^2 = v_{y0}^2 - 2gh\), which rearranges to \(h = \frac{v_{y0}^2}{2g}\). The choice "\(h = \frac{v_{y0}}{g}\)" actually gives the time to reach maximum height, not the height itself, confusing two related but distinct quantities. Keeping track of which kinematic equation isolates which variable is essential for correctly solving projectile motion problems.

Since the two velocities are perpendicular, they combine using the Pythagorean theorem, giving \(\sqrt{4^2 + 3^2} = \sqrt{25} = 5\ m/s\). The choice "\(7\ m/s\)" incorrectly adds the magnitudes directly, which only works when vectors point in the same direction, not perpendicular directions. Perpendicular velocity components must always be combined using vector addition rather than simple arithmetic addition.

A straight line with a constant, non-zero slope on a velocity-time graph means velocity changes at a steady rate, which is the definition of constant acceleration. "A horizontal straight line" is incorrect because a horizontal line has zero slope, indicating zero acceleration, not a constant non-zero value. Recognizing that slope on a velocity-time graph equals acceleration helps quickly interpret motion graphs on exams.

Because position depends on time squared under constant acceleration (\(x = \frac{1}{2}at^2\)), the graph curves upward increasingly steeply, forming an upward-opening parabola. "A straight diagonal line" is incorrect because that shape only occurs when velocity, not acceleration, is constant. Recognizing the algebraic relationship \(x \propto t^2\) helps predict the parabolic shape whenever acceleration is constant and nonzero.

At maximum height the vertical velocity component momentarily equals zero, but the horizontal component remains unchanged throughout the flight, so the total velocity equals the horizontal component alone. "Zero" is incorrect because that would only be true for a projectile launched straight upward with no horizontal component. This is a classic point of confusion, so remember that only the vertical velocity, not the total velocity, is zero at the peak of an angled trajectory.

Air resistance creates a drag force opposing motion, which reduces the net downward force and therefore the object's acceleration compared to the idealized value of \(g\) in a vacuum. "It increases the object's acceleration beyond \(g\)" is incorrect because drag force always opposes motion and thus can only decrease, not increase, the net acceleration. Real-world falling objects eventually reach terminal velocity, where drag force balances gravity and acceleration becomes zero.

A negative slope means acceleration is negative, and since the velocity stays positive, the object continues moving forward but its speed is decreasing over time. "The object is moving backward and speeding up" is incorrect because the velocity values are stated to remain positive, meaning the object never reverses direction. Always check both the sign of velocity and the sign of the slope together to correctly describe whether an object is speeding up or slowing down.

Converting requires multiplying by \(\frac{1000\ m}{1\ km}\) and dividing by \(\frac{3600\ s}{1\ h}\), so \(90 \times \frac{1000}{3600} = 25\ m/s\). The choice "\(32.4\ m/s\)" results from multiplying instead of dividing by \(3.6\), an inverted conversion factor. Remembering that dividing by \(3.6\) converts \(km/h\) to \(m/s\) is a useful shortcut for kinematics problems.

With zero acceleration, displacement is simply velocity multiplied by time, giving \(8\ m/s \times 12\ s = 96\ m\). The choice "\(1.5\ m\)" incorrectly divides velocity by time instead of multiplying, reversing the correct operation. For constant velocity motion, always use \(d = vt\) rather than any equation involving acceleration.

Slope on a velocity-time graph represents acceleration, so a steeper slope directly indicates a greater magnitude of acceleration for Object A. "Object A is moving faster than Object B at all times" is incorrect because slope reveals the rate of change of velocity, not the actual velocity values at any given moment. Steepness of a velocity-time graph should always be interpreted as acceleration, never confused with the object's actual speed.

For uniform deceleration to rest, displacement equals average velocity times time, and average velocity is \(\frac{v_0 + 0}{2}\), so \(50 = \frac{v_0}{2} \times 5\), giving \(v_0 = 20\ m/s\). The choice "\(10\ m/s\)" mistakenly treats the final velocity as if it equaled the average velocity rather than solving the full averaging equation. When an object decelerates to rest, using the average velocity form of the displacement equation is often faster than solving for acceleration first.

First find fall time from \(h = \frac{1}{2}gt^2\), giving \(t = \sqrt{\frac{2(20)}{9.8}} \approx 2.02\ s\), then multiply by horizontal velocity to get range \(\approx 15 \times 2.02 \approx 30.3\ m\). The choice "\(20\ m\)" incorrectly reuses the height value instead of computing the actual horizontal distance traveled during the fall. Solving projectile range problems always requires two steps: finding time of flight vertically, then applying it to the constant horizontal velocity.

Since the cars close the distance together, their combined closing speed is \(60 + 40 = 100\ km/h\), so time to meet is \(\frac{300}{100} = 3\ h\). The choice "\(5\ h\)" incorrectly uses only the slower car's speed of \(40\ km/h\) rather than the combined closing speed of both cars. When objects move toward each other, always add their speeds to find the rate at which the gap between them closes.

Using \(h = \frac{1}{2}gt^2\) with \(t = 2.5\ s\), the depth is \(h = \frac{1}{2}(9.8)(6.25) \approx 30.6\ m\). The choice "\(24.5\ m\)" incorrectly uses \(t = 2\ s\) instead of the given \(2.5\ s\), showing the importance of using the exact time value provided. Free-fall depth problems require squaring the full time before multiplying by half of \(g\), since the relationship between depth and time is quadratic, not linear.

Velocity is the derivative of position, \(v(t) = \frac{dx}{dt} = 6t - 4\), and evaluating at \(t = 2\) gives \(v = 6(2) - 4 = 8\ m/s\). The choice "\(6\ m/s\)" comes from only using the coefficient of the \(t^2\) term without completing the differentiation and substitution correctly. Taking the derivative of a position function with respect to time is the general method for finding instantaneous velocity from any polynomial motion equation.

Acceleration is the derivative of velocity, \(a(t) = \frac{dv}{dt} = 8t - 2\), so at \(t = 3\), \(a = 8(3) - 2 = 22\ m/s^2\). The choice "\(34\ m/s^2\)" incorrectly substitutes \(t = 3\) directly into \(v(t)\) instead of first differentiating to obtain the acceleration function. Whenever velocity is given as a nonlinear function of time, acceleration must be found by differentiating, not by simply evaluating the velocity expression.

The correct setup uses only the vertical velocity component, \(20\sin30^\circ\), in the standard vertical displacement equation \(y = v_{y0}t - \frac{1}{2}gt^2\), since the landing height differs from the launch height by \(5\ m\). The choice "\(5 = (20\cos30^\circ)t\)" incorrectly uses the horizontal velocity component and a horizontal-only equation, which cannot describe vertical displacement. Setting the correct target displacement equal to the full vertical kinematic equation, including both velocity and acceleration terms, is essential when launch and landing heights differ.

During acceleration the car reaches \(v = 3 \times 4 = 12\ m/s\) and covers \(d_1 = \frac{1}{2}(3)(4^2) = 24\ m\), then during deceleration it takes \(t_2 = \frac{12}{2} = 6\ s\) to stop, covering \(d_2 = \frac{0^2 - 12^2}{2(-2)} = 36\ m\), giving a total of \(24 + 36 = 60\ m\). The choice "\(24\ m\)" only accounts for the acceleration phase and neglects the additional distance covered while decelerating to a stop. Multi-phase motion problems require breaking the trip into separate segments, solving each with the correct kinematic equations, and then summing the results.

Since the swimmer's velocity relative to the water is directed straight across, only that perpendicular component, \(2\ m/s\), determines the crossing time, giving \(t = \frac{80}{2} = 40\ s\). The choice "\(53.3\ s\)" incorrectly divides the width by a combined speed that mixes in the current, which actually only pushes the swimmer downstream and does not affect crossing time. In river-crossing problems, the perpendicular and parallel velocity components must be treated independently, with only the perpendicular component determining time to cross.

Using \(v^2 = 2ad\) for each object, \(v_A = \sqrt{2(2a)d} = \sqrt{2}\sqrt{2ad}\) and \(v_B = \sqrt{2ad}\), so the ratio \(v_A/v_B = \sqrt{2}\). The choice "Object A's final velocity is twice Object B's" incorrectly assumes a linear relationship between acceleration and final velocity rather than the actual square-root relationship that arises from the kinematic equation. When comparing final velocities for equal displacements, remember that velocity scales with the square root of acceleration, not directly with acceleration itself.

By symmetry, the ball takes half the total time, \(3\ s\), to reach maximum height, so using \(h = \frac{1}{2}gt^2\) gives \(h = \frac{1}{2}(9.8)(3^2) \approx 44.1\ m\). The choice "\(29.4\ m\)" incorrectly skips halving the total flight time before applying the fall-distance equation. For symmetric vertical throws, always divide the total time in half before finding maximum height.

Range depends on \(\sin(2\theta)\), and since \(\sin(60^\circ) = \sin(120^\circ)\), the two complementary angles relative to \(45\) degrees produce identical ranges on level ground. The choice "The \(60\)-degree projectile has a greater range" incorrectly assumes higher launch angles always increase range, ignoring the symmetric nature of the range formula around \(45\) degrees. Any two launch angles that sum to \(90\) degrees will produce equal ranges on level ground, a useful pattern for quickly comparing projectile trajectories.

Using \(d = \frac{1}{2}at^2\), the car takes $t_{car} = \sqrt{\frac{2(81)}{2}} = 9\ s$ and the motorcycle takes $t_{moto} = \sqrt{\frac{2(81)}{4.5}} = 6\ s$, so the motorcycle arrives \(9 - 6 = 3\ s\) sooner. The choice "\(4.5\ s\)" mistakenly uses the acceleration value itself as if it were the time difference, rather than solving the quadratic time equation for each vehicle separately. Comparing times for two differently accelerating objects covering the same displacement requires solving each motion equation independently before subtracting the results.