Motion in a Straight Line

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 2.

Your speedometer says 60. Your trip averaged 55. Who is lying?

Nobody. The speedometer reports how fast you are moving at this instant. The trip average reports total distance over total time. Both numbers are honest, and CUET questions depend on knowing which one is being asked for. Kinematics is the art of describing motion precisely, without asking what causes it. The causes (forces) arrive in Laws of Motion. Here we only ask where, how fast and how quickly the speed is changing.

Setting the stage: position, path and displacement

We study motion along a straight line, called rectilinear motion. We treat the moving object as a point object, which is fair whenever the object is much smaller than the distance it travels (a train between two cities, but not a train at a platform).

To describe position you need a frame of reference: an origin, an axis and a choice of which direction is positive. After that, every quantity along the line is an algebraic number whose sign carries the direction. Choosing the positive direction first and sticking to it is the single habit that prevents most sign mistakes.

Dividing each by the time taken gives two different averages.

Worked example: the return trip. A car goes from city A to city B at 40 km h⁻¹ and returns along the same road at 60 km h⁻¹. Find its average speed and average velocity for the whole trip.

The common wrong step: averaging the two speeds to get 50 km h⁻¹. The car spends more time at the slower speed, so the true average is pulled toward 40. For equal distances the average speed is the harmonic mean, $2v_1v_2/(v_1 + v_2)$.

Instantaneous velocity and acceleration

If you shrink the time interval Δt towards zero, the average velocity settles at a single number: the instantaneous velocity.

$$v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}$$

On a position–time graph the average velocity over an interval is the slope of the chord joining two points, and the instantaneous velocity is the slope of the tangent at one point. Instantaneous speed is the magnitude of instantaneous velocity, so the two are equal at every instant. This is different from average speed, which can exceed the magnitude of average velocity.

Acceleration is how quickly velocity changes. Average acceleration is $\Delta v/\Delta t$, and the instantaneous value is

$$a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$$

Its SI unit is m s⁻². Because velocity has direction, a body can accelerate by changing its speed, its direction, or both. In one dimension the question is whether the sign of a helps or hurts the motion.

So the sign of acceleration alone never tells you whether the body is speeding up. It is the relation of a to v that matters: same sign speeds you up, opposite signs slow you down.

Pause and predict: you throw a ball straight up. At the very top, for an instant, the ball is at rest. What is its acceleration there?

Check your answer: it is still g = 9.8 m s⁻² downward. Velocity is zero, but velocity is changing from upward to downward, and that rate of change is exactly the acceleration. "Momentarily at rest" and "zero acceleration" are different statements. If a were zero at the top the ball would hover forever.

Worked example: motion from a formula. A particle moves along a line with $x = 2t^3 - 9t^2 + 12t$, where x is in metres and t in seconds. When is it momentarily at rest, and what is its acceleration then?

The particle moves forward to 5 m, stops, comes back to 4 m, stops again and then moves forward. Its displacement in the first two seconds is 4 m, but the path length is 5 + 1 = 6 m.

The common wrong step: setting a = 0 to find "when it stops". Acceleration zero means velocity is momentarily not changing, which is a different instant (here t = 1.5 s, when the velocity is at its most negative).

Reading graphs

Graphs are the fastest way to answer a kinematics question, if you remember that slopes give rates and areas give accumulated changes.

The shape of the curve then tells you the type of motion.

The area under a v–t graph is a signed area. Area above the time axis is displacement forward, and area below the axis is displacement backward. To get the distance you must add the areas without signs.

The three kinematic equations

For constant acceleration, three equations connect five quantities: initial velocity $v_0$, final velocity v, acceleration a, time t and displacement x.

$$v = v_0 + at$$

$$x = v_0 t + \dfrac{1}{2}at^2$$

$$v^2 = v_0^2 + 2ax$$

Where they come from. The first is just the definition of acceleration rewritten. For the second, notice that with constant acceleration the velocity changes uniformly, so the average velocity is exactly halfway: $\bar{v} = (v_0 + v)/2$. Then $x = \bar{v}t = 1/2\,(v_0 + v)\,t$, and putting $v = v_0 + at$ gives $x = v_0t + 1/2\,at^2$. On a v–t graph this is a trapezium, and its area is the displacement. For the third, eliminate t between the first two. If the body starts at $x_0$ instead of the origin, replace $x$ by $(x - x_0)$.

The same equations come from calculus: integrate $a = dv/dt$ to get v, then integrate $v = dx/dt$ to get x. Calculus also handles non-uniform acceleration, where a can be written as $a = v\,(dv/dx)$.

Choosing the right equation. Write down what is given, what is wanted, and which of the five quantities is irrelevant. Then pick the equation that leaves that quantity out.

These equations hold only for constant acceleration. The definitions $v = dx/dt$ and $a = dv/dt$ hold always. Quantities are signed, so decide the positive direction first. A deceleration is simply a negative a when v is positive.

Worked example: a braking car. A car moving at 72 km h⁻¹ brakes with a uniform retardation of 5 m s⁻². Find the stopping distance and the stopping time. What if the speed is doubled?

The common wrong step: substituting 72 into the equation without converting to m s⁻¹, or concluding that doubling the speed doubles the stopping distance. Since $x = v_0^2/(2 \times \text{magnitude of } a)$, doubling the speed quadruples the stopping distance.

Beyond the textbook. For the exam: stopping distance is $v_0^2/(2 \times \text{magnitude of } a)$. In reality: a driver also needs time to see the hazard and press the brake, and during that reaction time the car keeps moving at full speed. So the total distance is $v_0 t_r + v_0^2/(2a)$, where a is the braking magnitude. With $t_r = 1$ s, the car above needs 20 + 40 = 60 m at 20 m s⁻¹ but 40 + 160 = 200 m at 40 m s⁻¹. The braking term grows with the square of speed and the reaction term only linearly, which is why speed limits matter so much in the rain, when the braking deceleration is smaller.

Free fall and Galileo's odd numbers

Drop a stone and, if air resistance is negligible, it moves with constant acceleration g ≈ 9.8 m s⁻² downward. This is free fall. It is just the kinematic equations with a = −g (upward positive) and no new physics.

Time up equals time down only when the ball lands at its launch level. Launch from a cliff or building and the fall is longer than the climb.

Worked example: a throw from a building. A ball is thrown vertically up at 20 m s⁻¹ from the top of a 25 m building. Taking g = 10 m s⁻², find the time to reach the ground and the speed of impact.

Cross-check by stages: the ball climbs 20 m in 2 s, so it is 45 m above the ground at the top; falling 45 m takes $\sqrt{2 \times 45/10} = 3$ s. Total 5 s. The common wrong step: using the "up and down" formula $2v_0/g = 4$ s, which applies only if the ball lands back at the roof.

Pause and predict: a stone is dropped from rest. How far does it fall in the first, second and third second?

Check your answer: distance after n seconds is $1/2\,gn^2$, so the distances in successive seconds are $1/2\,g\,(1,\ 4 - 1,\ 9 - 4) = 1/2\,g\,(1, 3, 5)$. With g = 10 that is 5 m, 15 m and 25 m. This is Galileo's law of odd numbers: in equal successive time intervals a body falling from rest covers distances in the ratio 1 : 3 : 5 : 7 : … The distances covered since the start grow as 1 : 4 : 9 : 16, which is the ratio of squares. Do not mix the two.

Worked example: the last second. A stone dropped from the top of a tower covers 45 m in the last second of its fall. Find the height of the tower (g = 10 m s⁻²).

The common wrong step: treating 45 m as $1/2\,gt^2$ from the start, which gives t = 3 s and a height of 45 m. The 45 m is only the last second's share of a longer fall. Check: in 4 s the stone falls 80 m, and 125 − 80 = 45 m.

Worked example: reaction time. A friend drops a ruler, and you catch it after it has fallen 19.6 cm. Estimate your reaction time.

The common wrong step: leaving d in centimetres. Convert to metres first.

Beyond the textbook. For the exam: free fall ignores air resistance, so every object has acceleration g. In reality: air resistance grows with speed, and a falling body stops accelerating when drag balances weight, at its terminal velocity. A large raindrop falling a kilometre would reach about 140 m s⁻¹ in free fall ($\sqrt{2gh}$), but real raindrops arrive at under 10 m s⁻¹. In 1971 the Apollo 15 astronaut David Scott dropped a hammer and a feather together on the Moon, where there is no air, and they landed together. Also, g is not exactly 9.8 everywhere: it is about 9.78 m s⁻² at the equator and about 9.83 m s⁻² at the poles. Why it matters: "all bodies fall together" is a statement about vacuum, not about parachutes.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Units and Measurements taught the unit conversions and significant figures that every numerical here depends on.
• Next: Motion in a Plane splits a 2-D motion into two independent straight-line motions, so projectile motion is this chapter used twice (horizontally with a = 0, vertically with a = −g).
• Later in Class XI: Laws of Motion explains why the acceleration has the value it does, and Gravitation explains where g comes from.