Motion in a Straight Line

Every motion you have ever seen — a bus pulling away from a signal, a ball tossed straight up, a raindrop falling, even the blood moving through your own veins — can, for a short enough object over a short enough time, be reduced to a single number changing with another single number: position and time. This chapter builds the first complete grammar of motion, restricted to the simplest and most common case, an object moving along a straight line. Kinematics, the branch of mechanics used here, deliberately ignores why something moves (that is the story of the next two chapters, on force) and asks only how to describe the motion precisely: where the object is, how fast it is getting there, and whether that speed is itself changing. You will meet position and displacement, and see why displacement can be negative or even zero while an object has covered a large distance. You will sharpen the everyday, blurred notion of speed into two distinct quantities, average velocity and instantaneous velocity, the second obtained only through the idea of a limit — the same idea that underlies all of calculus. You will do the same for acceleration, discover that the area under a velocity-time graph is displacement, and use that geometric fact to derive three short equations that govern every uniformly accelerated straight-line motion: a falling stone, a braking car, a ball thrown skyward. Finally you will learn to combine two independent motions, such as two cars on a highway or a bullet fired from a moving vehicle, into a single relative velocity. By the end you should be able to read a position-time or velocity-time graph and extract velocity, acceleration and displacement from its shape alone; choose signs correctly for objects moving against gravity or against your chosen positive direction; and select, without hesitation, the right one of the three kinematic equations for a numerical problem — the single most tested skill this chapter builds toward.

Key takeaways

Displacement is the straight-line change in position (can be negative or zero); distance is the total path length covered and is always greater than or equal to the magnitude of displacement.
Average velocity equals displacement divided by the time interval; average speed equals total path length divided by the time interval; the two are equal only when the motion never reverses direction.
Instantaneous velocity is the limit of average velocity as the time interval shrinks to zero, $v = dx/dt$ — geometrically, the slope of the tangent to the position-time graph at that instant.
Instantaneous acceleration is $a = dv/dt$ , the slope of the tangent to the velocity-time graph; it can be positive, negative or zero, and its sign alone does not tell you whether the object is speeding up or slowing down.
The area under a velocity-time graph between two instants equals the displacement in that interval — this single geometric fact is the source of every kinematic equation in the chapter.
For uniform (constant) acceleration: $v = v_0 + at$ , then $x = v_0t + \tfrac12 at^2$ , and $v^2 = v_0^2 + 2ax$ , with $x$ replaced by $(x - x_0)$ whenever the motion does not start at the origin.
Free fall near Earth's surface is uniformly accelerated motion with magnitude $g \approx 9.8\ \text{m/s}^2$ directed downward; Galileo's law of odd numbers says the distances covered from rest in successive equal time intervals stand in the ratio 1 to 3 to 5 to 7 and so on.
Relative velocity of A with respect to B is $v_{AB} = v_A - v_B$ ; two objects moving toward each other close the gap at the sum of their speeds, and two moving the same way close it at the difference of their speeds.
Key idea: every straight-line motion is completely described by position as a function of time, x(t), and its first two rates of change, velocity v = dx/dt and acceleration a = dv/dt; when acceleration is constant this reduces to three simple algebraic equations connecting v0, v, a, t and x.

The core insight

Once you know how position changes with time along a line, velocity and acceleration follow purely from the slopes of that change, and for the common special case of constant acceleration this collapses into three linked equations that solve almost any straight-line motion problem.

What this chapter covers

This chapter builds straight-line kinematics from the ground up, in the same order NCERT presents it. It opens by restricting attention to point objects moving along a single line (rectilinear motion), and separating kinematics, describing motion, from dynamics, explaining its cause — dynamics waits until the chapter on laws of motion. It then constructs instantaneous velocity as a limit of average velocity, using a numerical table of shrinking time intervals and a tangent-line construction on a position-time graph. It defines acceleration the same way, as a limit of average acceleration, and shows how positive, negative and zero acceleration produce distinct, recognisable shapes on position-time and velocity-time graphs. A key geometric result, that the area under a velocity-time graph equals displacement, is then used to derive the three kinematic equations of uniformly accelerated motion, first graphically and then, exactly as NCERT does in its own worked example, by direct integration using calculus. These equations are applied throughout to free fall, to Galileo's law of odd numbers, to the stopping distance of vehicles, and to measuring human reaction time. The chapter closes by extending the description of a single object to two objects at once, through relative velocity.

The physical picture

Position, displacement and distance

To describe motion along a line, first fix an origin O and a positive direction, by convention usually to the right. The position of the object is then a single signed number x, measured from O. If the object moves from position $x_1$ at time $t_1$ to position $x_2$ at time $t_2$ , its displacement is $\Delta x = x_2 - x_1$ . Displacement is signed: it is positive if the object ends up further along the positive direction than it started, negative if it ends up further along the negative direction, and exactly zero if the object returns to its starting point, no matter how far it wandered in between.

Distance, by contrast, is the total length of the path actually travelled, and it can never be negative and never decreases. A drunkard's walk that goes five steps forward and three steps back, repeated again and again, covers a distance of eight steps in every cycle even though the net displacement after one cycle is only two steps forward. This is the first and most persistent source of numerical errors in this chapter: distance and displacement agree in size only when the motion is entirely one-directional; the moment an object reverses, distance overtakes the magnitude of displacement.

Average velocity and average speed

Average velocity over the interval from $t_1$ to $t_2$ is the displacement divided by the elapsed time,

\bar{v} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}.

where: $x_1, x_2$ are positions (metres) at times $t_1, t_2$ (seconds), and $\bar{v}$ is the average velocity (metres per second), which inherits the sign of the displacement. Average speed over the same interval is the total path length divided by the same elapsed time, and it is always a positive quantity. For any motion that reverses direction, average speed is strictly greater than the magnitude of average velocity; for a motion that never reverses, the two coincide exactly. This distinction is exactly the one NCERT stresses with the example of a person walking to the market and back: the round trip has zero net displacement, hence zero average velocity, but a very real, non-zero average speed.

Instantaneous velocity — velocity at a single instant

Average velocity over a long interval hides how the motion actually behaved inside that interval. To capture the velocity at one exact instant $t$ , shrink the interval $\Delta t$ around $t$ down toward zero and see what the average velocity approaches. This limiting process defines the instantaneous velocity:

v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}.

where: $v$ is the instantaneous velocity at the chosen instant, and $dx/dt$ is the derivative of position with respect to time in the language of calculus — the same limiting ratio, made precise. Physically, this is the rate of change of position at that instant, and graphically it is the slope of the tangent line to the position-time curve at that point. NCERT illustrates this numerically for the motion $x = 0.08\,t^3$ : computing $\Delta x/\Delta t$ centred at $t = 4.0\ \text{s}$ for successively smaller $\Delta t$ values of $2.0\ \text{s}$ , $1.0\ \text{s}$ , $0.5\ \text{s}$ , $0.1\ \text{s}$ and $0.01\ \text{s}$ produces a sequence of average velocities that converges to $3.84\ \text{m/s}$ , which is exactly $dx/dt$ evaluated at $t = 4.0\ \text{s}$ . Instantaneous speed is simply the magnitude of instantaneous velocity, and — unlike the average quantities — the two are always equal at any single instant, because over an infinitesimally short interval the path cannot double back on itself.

Figure

A position time graph with a curved trajectory and two nearby points whose connecting line becomes a tangent as the time gap shrinks toward zero

Acceleration — average and instantaneous

Velocity itself usually changes as motion proceeds, and the rate of that change is acceleration. Over an interval from $t_1$ to $t_2$ , the average acceleration is

\bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t},

where: $v_1, v_2$ are the instantaneous velocities at $t_1, t_2$ , and $\bar{a}$ is in metres per second squared. Taking the same limit as before, as $\Delta t \to 0$ , gives the instantaneous acceleration,

a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt},

the slope of the tangent to the velocity-time graph at that instant. Because velocity has both magnitude and direction, acceleration can arise from a change in speed, a change in direction, or both, and it can be positive, negative or exactly zero regardless of which way the object happens to be moving at that instant — a point the exam traps below return to. On a position-time graph, positive acceleration curves the graph upward (increasingly steep), negative acceleration curves it downward (decreasingly steep, or curving back), and zero acceleration means the graph is a straight line.

Velocity-time graphs and the area-displacement connection

For motion at a genuinely constant velocity $u$ over a duration $T$ , the velocity-time graph is a horizontal line at height $u$ , and the rectangle it forms with the time axis has area $u \times T$ , which is exactly the displacement covered. This is not a coincidence restricted to constant velocity: for any velocity-time graph, however curved, the area between the curve and the time axis over an interval equals the net displacement in that interval — a result that in general needs calculus (the definition $v = dx/dt$ integrated back up) but that you can see directly for straight-line segments, which is exactly the case of uniform acceleration this chapter focuses on.

Figure

A velocity time graph that is a straight line rising from an initial velocity to a final velocity, with the area beneath it split into a lower rectangle and an upper triangle representing the displacement

Relative velocity in one dimension

Everything so far describes a single object against a fixed origin. When two objects A and B both move along the same line, it is often the change between them, not either one's motion alone, that matters — will a car overtake, will two trains collide, at what speed does a bullet strike a fleeing target. The relative velocity of A with respect to B is defined as

v_{AB} = v_A - v_B,

where: $v_A$ and $v_B$ are the (signed) velocities of A and B in the fixed frame, and $v_{AB}$ is the velocity A appears to have to an observer riding along with B. If A and B move in the same direction, $v_{AB}$ is the difference of their speeds; if they move in opposite directions, the signs make $v_{AB}$ come out as the sum of their speeds, because approaching objects close the gap faster than either moves alone. A car doing 25 m/s that is overtaking a car doing 15 m/s in the same lane, same direction, gains on it at only $25 - 15 = 10\ \text{m/s}$ ; two cars approaching each other at 25 m/s and 15 m/s close the gap between them at $25 + 15 = 40\ \text{m/s}$ .

Derivation AKinematic equations from the area under a velocity-time graph

The goal of this derivation is to obtain all three standard equations of uniformly accelerated motion — the relation between final and initial velocity, the relation giving displacement in terms of time, and the relation connecting velocity directly to displacement without any explicit time — using nothing but the geometry of a straight-line velocity-time graph and the fact, established above, that the area under such a graph equals displacement. No calculus is used in this version; it is the purely graphical route NCERT takes before introducing the calculus method as an alternative.

Step 1. Consider an object with initial velocity $v_0$ at time $t = 0$ , moving with constant acceleration $a$ , so that its velocity at a later time $t$ is $v$ . Because the acceleration is constant, the velocity-time graph of this motion is a straight line joining the point $(0, v_0)$ to the point $(t, v)$ . The slope of that line is, by definition, the acceleration, $a = (v - v_0)/t$ , which rearranges immediately to the first kinematic equation, $v = v_0 + at$ .

Step 2. The displacement $x$ in this interval is the area under that straight line, and this area splits naturally into two pieces: a rectangle of height $v_0$ and base $t$ , contributing area $v_0 t$ , sitting below a triangle of base $t$ and height $(v - v_0)$ , contributing area $\tfrac12 (v - v_0)\, t$ . Adding the two pieces gives the total displacement, $x = v_0 t + \tfrac12 (v - v_0)\, t$ .

Step 3. Substitute the Step 1 result, $v - v_0 = at$ , into the triangle term from Step 2. This turns $\tfrac12 (v - v_0)\, t$ into $\tfrac12 a t \cdot t = \tfrac12 a t^2$ , giving the second kinematic equation, $x = v_0 t + \tfrac12 a t^2$ .

Step 4. To eliminate time altogether, first write the displacement using average velocity: since velocity grows uniformly from $v_0$ to $v$ , the average is simply the arithmetic mean, $\bar{v} = (v + v_0)/2$ , so $x = \bar{v}\, t = \tfrac12 (v + v_0)\, t$ . Now substitute $t = (v - v_0)/a$ from Step 1 into this expression: $x = \tfrac12 (v + v_0) \cdot \dfrac{v - v_0}{a} = \dfrac{v^2 - v_0^2}{2a}$ , which rearranges to the third kinematic equation, $v^2 = v_0^2 + 2ax$ . If the motion does not begin at the origin but at some initial position $x_0$ , every occurrence of $x$ above is replaced by $(x - x_0)$ , giving the general forms $x = x_0 + v_0 t + \tfrac12 a t^2$ and $v^2 = v_0^2 + 2a(x - x_0)$ .

Go deeper · Derivation B · 2 minKinematic equations by direct integration

Derivation BKinematic equations by direct integration

The goal of this derivation is to obtain the same three kinematic equations by treating $a = dv/dt$ and $v = dx/dt$ as genuine differential relations and integrating them, exactly the calculus method NCERT presents as its own worked example. The advantage over the graphical route above is that this method still works, in principle, even when acceleration is not constant, so long as it is known as a function of time or position; here it is applied to the constant-acceleration case to recover the familiar equations.

Step 1. Start from the definition of acceleration, $a = dv/dt$ , with $a$ constant. Separate the variables to write $dv = a\, dt$ , then integrate both sides between the appropriate limits, velocity from $v_0$ to $v$ as time runs from $0$ to $t$ : $\displaystyle\int_{v_0}^{v} dv = a \int_0^t dt$ . The left side evaluates to $v - v_0$ and the right side to $at$ , giving $v - v_0 = at$ , that is, $v = v_0 + at$ .

Step 2. Now use $v = dx/dt$ with $v$ known from Step 1, so $dx = (v_0 + at)\, dt$ . Integrate position from $x_0$ to $x$ as time runs from $0$ to $t$ : $\displaystyle\int_{x_0}^{x} dx = \int_0^t (v_0 + at)\, dt$ . The left side gives $x - x_0$ ; the right side, term by term, gives $v_0 t$ from the constant piece and $\tfrac12 a t^2$ from the linear-in- $t$ piece. This yields $x - x_0 = v_0 t + \tfrac12 a t^2$ , the second kinematic equation in its general form.

Step 3. For the third relation, use the chain rule to rewrite acceleration in terms of $v$ and $x$ instead of $t$ : $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ , so $a\, dx = v\, dv$ . Integrating both sides, position from $x_0$ to $x$ and velocity from $v_0$ to $v$ : $\displaystyle\int_{x_0}^{x} a\, dx = \int_{v_0}^{v} v\, dv$ . Since $a$ is constant this gives $a(x - x_0) = \tfrac12(v^2 - v_0^2)$ , which rearranges to $v^2 = v_0^2 + 2a(x - x_0)$ .

Step 4. All three results match the purely graphical derivation exactly, as they must, since both start from the same physical definitions. The calculus route is the more powerful of the two: nothing in Steps 1 to 3 actually required $a$ to be constant except the final integration step, so the same method, redone with $a(t)$ or $a(x)$ instead of a fixed number, extends directly to non-uniform acceleration — a generalisation this chapter deliberately sets aside but that later mechanics chapters will need.

Go deeper · Derivation C · 1 minGalileo's law of odd numbers for a body falling from rest

Derivation CGalileo's law of odd numbers for a body falling from rest

The goal of this derivation is to prove Galileo's own result: for a body released from rest and falling freely under constant acceleration $g$ , the distances it covers in successive, equal time intervals $\tau$ stand in the ratio $1 : 3 : 5 : 7 : \ldots$ , the odd numbers in order. This is a direct consequence of the second kinematic equation applied at the boundaries of each interval and subtracted, and it is historically the first quantitative signature that free fall is uniformly accelerated motion rather than motion at constant speed.

Step 1. Since the object starts from rest, $v_0 = 0$ , and the position fallen after any time $t$ , measured from the release point, is $y(t) = \tfrac12 g t^2$ (taking downward distance as positive). After $n$ complete intervals of duration $\tau$ , the total distance fallen is $y_n = \tfrac12 g (n\tau)^2 = \tfrac12 g \tau^2\, n^2$ .

Step 2. The distance covered specifically during the $n$ -th interval, that is, between time $(n-1)\tau$ and time $n\tau$ , is the difference of two successive totals, $d_n = y_n - y_{n-1} = \tfrac12 g \tau^2 \left[ n^2 - (n-1)^2 \right]$ . Expanding the bracket, $n^2 - (n-1)^2 = n^2 - (n^2 - 2n + 1) = 2n - 1$ .

Step 3. Substituting this back gives $d_n = \tfrac12 g \tau^2 (2n - 1)$ . As $n$ runs through the positive integers $1, 2, 3, 4, \ldots$ , the factor $(2n-1)$ runs through exactly the odd numbers $1, 3, 5, 7, \ldots$ , and since the prefactor $\tfrac12 g \tau^2$ is the same fixed number for every interval, the successive distances $d_1, d_2, d_3, d_4, \ldots$ are in the ratio $1 : 3 : 5 : 7 : \ldots$ , which is Galileo's law.

Step 4. As a numerical check, take $\tau = 1\ \text{s}$ and $g = 9.8\ \text{m/s}^2$ : the distances fallen in the first, second, third and fourth seconds work out to $4.9\ \text{m}$ , $14.7\ \text{m}$ , $24.5\ \text{m}$ and $34.3\ \text{m}$ . Dividing each by $4.9\ \text{m}$ gives exactly $1, 3, 5, 7$ , confirming the ratio and matching the table NCERT builds for this same result.

Worked examples

Board level — velocity from a given position function

The position of an object moving along the x-axis is given by $x = a + bt^2$ , where $a = 8.5\ \text{m}$ , $b = 2.5\ \text{m/s}^2$ and $t$ is in seconds. Find its velocity at $t = 0\ \text{s}$ and $t = 2.0\ \text{s}$ , and its average velocity between $t = 2.0\ \text{s}$ and $t = 4.0\ \text{s}$ .

Step 1. Differentiate the position function: $v = \dfrac{dx}{dt} = \dfrac{d}{dt}(a + bt^2) = 2bt = 5.0\,t\ \text{m/s}$ , using $b = 2.5\ \text{m/s}^2$ .

Step 2. At $t = 0\ \text{s}$ , $v = 5.0 \times 0 = 0\ \text{m/s}$ . At $t = 2.0\ \text{s}$ , $v = 5.0 \times 2.0 = 10\ \text{m/s}$ .

Step 3. For the average velocity between $t = 2.0\ \text{s}$ and $t = 4.0\ \text{s}$ , first find both positions: $x(2.0) = 8.5 + 2.5(2.0)^2 = 8.5 + 10 = 18.5\ \text{m}$ , and $x(4.0) = 8.5 + 2.5(4.0)^2 = 8.5 + 40 = 48.5\ \text{m}$ . The average velocity is $\bar{v} = \dfrac{48.5 - 18.5}{4.0 - 2.0} = \dfrac{30}{2.0} = 15\ \text{m/s}$ . Answer: $v(0\ \text{s}) = 0\ \text{m/s}$ , $v(2.0\ \text{s}) = 10\ \text{m/s}$ , and the average velocity from $t = 2.0\ \text{s}$ to $4.0\ \text{s}$ is $15\ \text{m/s}$ .

Board level — a ball thrown vertically upward from a building

A ball is thrown vertically upward with a velocity of $20\ \text{m/s}$ from the top of a building $25.0\ \text{m}$ above the ground. Take $g = 10\ \text{m/s}^2$ and neglect air resistance. Find (a) how high above the launch point the ball rises, and (b) the total time before it hits the ground.

Step 1. Take upward as the positive direction, so $v_0 = +20\ \text{m/s}$ and $a = -g = -10\ \text{m/s}^2$ . At the highest point the velocity is momentarily zero, $v = 0$ . Using $v^2 = v_0^2 + 2ay$ with $y$ measured from the launch point, $0 = (20)^2 + 2(-10)y$ , which gives $y = 400/20 = 20\ \text{m}$ . The ball rises $20\ \text{m}$ above the point it was thrown from, that is, $25.0 + 20 = 45.0\ \text{m}$ above the ground.

Step 2. For the time to reach the ground, work with position measured from the ground, so the launch point is at $x_0 = 25\ \text{m}$ and the ground is at $x = 0$ . Using $x = x_0 + v_0 t + \tfrac12 a t^2$ : $0 = 25 + 20t + \tfrac12(-10)t^2 = 25 + 20t - 5t^2$ .

Step 3. Rearranging gives $5t^2 - 20t - 25 = 0$ , or dividing by 5, $t^2 - 4t - 5 = 0$ . This factors as $(t - 5)(t + 1) = 0$ , so $t = 5\ \text{s}$ or $t = -1\ \text{s}$ ; the negative root is unphysical and is discarded. Answer: the ball rises $20\ \text{m}$ above its launch point (to $45\ \text{m}$ above the ground), and it reaches the ground $5\ \text{s}$ after being thrown. Notice this single equation handles the entire trip, up and down, without needing to split it into two separate stages.

Competitive level — relative velocity and a catch-up problem

Car A, travelling at $72\ \text{km/h}$ on a straight highway, is $100\ \text{m}$ behind car B, which travels at a constant $54\ \text{km/h}$ in the same direction. At this instant the driver of A begins accelerating at $1\ \text{m/s}^2$ . Using relative velocity, find how long A takes to catch up with B, and A's speed at that instant.

Step 1. Convert to SI units: $72\ \text{km/h} = 20\ \text{m/s}$ and $54\ \text{km/h} = 15\ \text{m/s}$ . Switch to a reference frame moving with car B, in which B is at rest and only A's motion needs tracking. The initial relative velocity of A with respect to B is $u_{AB} = v_A - v_B = 20 - 15 = 5\ \text{m/s}$ , and the relative acceleration is $a_{AB} = a_A - a_B = 1 - 0 = 1\ \text{m/s}^2$ , since B moves at constant speed.

Step 2. In B's frame, A must cover the initial $100\ \text{m}$ gap. Apply the second kinematic equation to this relative motion: $s_{AB} = u_{AB}\,t + \tfrac12 a_{AB}\,t^2$ , giving $100 = 5t + \tfrac12(1)t^2$ , or $t^2 + 10t - 200 = 0$ .

Step 3. Solve the quadratic: $t = \dfrac{-10 \pm \sqrt{100 + 800}}{2} = \dfrac{-10 \pm 30}{2}$ , giving $t = 10\ \text{s}$ or $t = -20\ \text{s}$ ; the negative root is discarded. A's actual speed at that instant, back in the ground frame, is $v_A = 20 + 1 \times 10 = 30\ \text{m/s}$ . As a check, the relative velocity at $t = 10\ \text{s}$ is $u_{AB} + a_{AB} t = 5 + 10 = 15\ \text{m/s}$ , and indeed $v_A - v_B = 30 - 15 = 15\ \text{m/s}$ , consistent. Answer: car A catches up with car B after $10\ \text{s}$ , by which time A is travelling at $30\ \text{m/s}$ (108 km/h). Timing note: under 90 seconds. Switching to B's reference frame turns a two-body problem into a one-body kinematics problem, which is the single biggest time-saver here. The trap is the sign of the relative acceleration — it is $a_A - a_B$ , never $a_A + a_B$ , and it vanishes entirely if both cars accelerate identically.

Common traps

Treating distance and displacement as interchangeable in a numerical, especially for motion that goes forward and then reverses; distance only equals the magnitude of displacement when the path is one-directional.
Confusing average speed with the magnitude of average velocity; they differ whenever the motion reverses direction, and average speed is never smaller.
Reading the slope of an x-t graph as acceleration (it is velocity), or looking for meaning in the area under an x-t graph (it has no simple physical meaning here; it is the area under the v-t graph that equals displacement).
Assuming the sign of acceleration by itself tells you whether an object is speeding up or slowing down; it only does so once you also know the direction of the velocity and your chosen positive axis.
Believing that zero instantaneous velocity always means zero acceleration at that same instant.
Applying $v^2 = v_0^2 + 2ax$ , or either of the other two kinematic equations, to a phase of motion where the acceleration is not actually constant throughout.
Forgetting to reset the reference position $x_0$ and initial velocity $v_0$ when a fresh phase of motion begins, for example immediately after a bounce or a sudden change in the direction of motion.
Writing relative velocity the wrong way round; the velocity of A with respect to B is $v_A - v_B$ , not $v_B - v_A$ , and getting this backwards silently flips the sign of every following step.

Common trapA particle momentarily at rest is not a particle with zero acceleration

The single most common conceptual slip in this chapter is to assume that if an object's velocity is zero at some instant, its acceleration must be zero too. It is not. A ball thrown straight up has velocity exactly zero for an instant at the top of its flight, yet the acceleration due to gravity, $g$ , acts on it continuously, before, during and after that instant — nothing about gravity switches off just because the ball happens to be momentarily still. The same idea appears whenever a car is stationary at a traffic light with the accelerator already pressed: velocity is zero, but acceleration is not. Acceleration describes how velocity is changing, not what velocity currently equals, and the two can be completely independent at any single instant.

You should be able to…

Distinguish distance from displacement, and average speed from average velocity, in any motion, including one that reverses direction partway through.
Compute instantaneous velocity as $dx/dt$ from a given position function $x(t)$ , and identify it graphically as the slope of the tangent to an x-t graph at a point.
Compute instantaneous acceleration as $dv/dt$ , and correctly interpret positive, negative and zero acceleration on both x-t and v-t graphs.
Use the area under a v-t graph to find displacement, for both constant-velocity and uniformly accelerated motion.
Derive the three kinematic equations of uniformly accelerated motion, both graphically, using the area under a v-t line, and by calculus, through direct integration.
Apply $v = v_0 + at$ , $x = v_0t + \tfrac12 at^2$ and $v^2 = v_0^2 + 2ax$ with correct signs, including the general form that replaces $x$ with $(x - x_0)$ .
Solve free-fall problems, whether an object is dropped, thrown upward or thrown downward, with $a = \pm g$ depending on your chosen positive direction.
State and apply Galileo's law of odd numbers for the distances covered in successive equal time intervals of free fall.
Set up and solve stopping-distance and reaction-time problems using the kinematic equations.
Find and correctly sign the relative velocity of one object with respect to another moving along the same straight line, including catch-up and approach problems.

Ask the tutor: if you are stuck on which kinematic equation fits a problem, or want to walk through a tricky sign-convention setup step by step, ask the tutor to work it through with you.

Why is average speed sometimes different from the magnitude of average velocity?

Average velocity depends only on the net displacement, the straight-line difference between start and end position, divided by the time taken. Average speed depends on the total path length actually covered, divided by the same time. These agree only when the object never reverses direction. A person who walks 2.5 km to the market and walks the same 2.5 km straight back has zero net displacement, so an average velocity of zero, but has covered 5 km of actual path, giving a clearly non-zero average speed. Always identify whether a problem is asking about the net change in position or the total ground covered before choosing which one to compute.

How can acceleration be negative while the object is still speeding up?

The sign of acceleration only has meaning relative to the positive direction you chose for your axis; it says nothing on its own about speeding up or slowing down. If an object moves in the negative direction and its acceleration is also negative, both velocity and acceleration point the same way, so the object's speed is increasing even though the acceleration is labelled negative. Conversely, an object moving in the positive direction with negative acceleration is slowing down. The rule that actually matters is simple: if acceleration and velocity have the same sign, the object speeds up; if they have opposite signs, it slows down.

Why does the area under a velocity-time graph give displacement, and not the area under a position-time graph?

Displacement is, by its very definition, the accumulated effect of velocity over time; formally, integrating $v = dx/dt$ back up with respect to time reconstructs $x$ , and that integration is exactly what "area under the v-t curve" means. The area under an x-t graph, by contrast, would have units of position multiplied by time, metre-seconds, which corresponds to no standard physical quantity in this chapter, so it is not meaningful here. Always check which axis is velocity before treating an area as a displacement.

Do I actually need calculus to solve problems in this chapter?

Not for constant-acceleration motion, which is everything the three kinematic equations cover; those equations can be derived purely graphically, using areas under a straight-line v-t graph, exactly as shown in the first derivation above, with no calculus at all. Calculus becomes necessary once acceleration itself changes with time, a case this chapter deliberately excludes but which later mechanics chapters will need. Even so, understanding velocity and acceleration as $dx/dt$ and $dv/dt$ gives a much sharper sense of what "instantaneous" really means than the average-value definitions alone.

Why do some problems use g equals 9.8 and others use g equals 10?

Both values appear in this very chapter. The measured standard value near Earth's surface is $9.8\ \text{m/s}^2$ , and it is the value to use whenever a question does not specify otherwise or when precision matters. Many numerical problems, including several in this chapter, deliberately round it to $10\ \text{m/s}^2$ to make the arithmetic land on clean numbers, since the difference changes final answers by only about two per cent. Always use whichever value the question explicitly states; if none is given, $9.8\ \text{m/s}^2$ is the safer default.