Looking for classes? Ksquare Career Institute, Bengaluru →
Motion in a Straight Line is the first real physics chapter of Class 11, and most of mechanics is built on it. These notes cover what NEET actually tests: distance versus displacement, reading position–time and velocity–time graphs, the equations of motion, free fall, and the sign mistakes that quietly cost marks.
The one rule for this chapter
Choose a positive direction before you start. Displacement, velocity and acceleration then each get a sign that agrees with it — and you keep that sign all the way through the calculation.
1. Distance and displacement
To describe motion you need a frame of reference: an origin, a positive direction, and a clock.
- Distance is the total length of path covered. It is a scalar and is never negative.
- Displacement is the change in position, $\Delta x = x_2 - x_1$. It is a vector, so it can be positive, negative or zero.
- The magnitude of displacement is always less than or equal to distance: $|\Delta x| \le \text{distance}$. They are equal only when the body moves in one direction without turning back.
Worked example
A student walks 5 m east and then 3 m west.
Distance $= 5 + 3 = 8\ \text{m}$. Taking east as positive, displacement $= +5 - 3 = +2\ \text{m}$, that is 2 m east.
2. Speed and velocity
- Average velocity $= \dfrac{\text{displacement}}{\text{time}} = \dfrac{\Delta x}{\Delta t}$
- Average speed $= \dfrac{\text{total distance}}{\text{total time}}$
- Instantaneous velocity $v = \dfrac{dx}{dt}$, the slope of the position–time graph at that instant. Its magnitude is the instantaneous speed.
Average speed is always greater than or equal to the magnitude of average velocity, because distance is never smaller than displacement.
Worked example: the round trip
A car goes from A to B at 30 km/h and returns from B to A at 60 km/h. Find the average speed and average velocity for the whole trip.
Let AB $= d$. Time out $= d/30$, time back $= d/60$, total time $= d/20$.
Average speed $= \dfrac{2d}{d/20} = 40\ \text{km/h}$ — not 45 km/h. For equal distances, average speed $= \dfrac{2v_1v_2}{v_1+v_2}$.
Average velocity $= 0$, because the car ends where it started.
3. Acceleration
Acceleration is the rate of change of velocity, $a = \dfrac{dv}{dt}$. In this chapter it is usually constant.
- A body speeds up when velocity and acceleration have the same sign.
- A body slows down when they have opposite signs.
- So negative acceleration does not always mean slowing down. A ball falling downward, with downward taken as negative, has negative velocity and negative acceleration — and it is speeding up.
4. Reading the graphs
Graph questions appear almost every year. Learn what the slope and the area mean on each graph.
| Graph | Slope gives | Area under it gives |
|---|---|---|
| Position–time ($x$–$t$) | Velocity | Nothing physical |
| Velocity–time ($v$–$t$) | Acceleration | Displacement (area below the time axis counts as negative) |
| Acceleration–time ($a$–$t$) | Rate of change of acceleration | Change in velocity |
- Uniform velocity gives a straight, sloped $x$–$t$ line and a flat $v$–$t$ line.
- Uniform acceleration gives a parabola on the $x$–$t$ graph and a straight sloped line on the $v$–$t$ graph.
- An $x$–$t$ graph can never show two positions at the same time, and a real one never has a vertical section, which would mean infinite speed.
- For distance from a $v$–$t$ graph, add all areas as positive. For displacement, subtract the areas below the time axis.
5. Equations of motion
These hold only for constant acceleration. Here $u$ is initial velocity, $v$ final velocity, $a$ acceleration, $t$ time and $s$ displacement.
$$v = u + at$$ $$s = ut + \tfrac{1}{2}at^2$$ $$v^2 = u^2 + 2as$$ $$s = \tfrac{1}{2}(u + v)\,t$$Displacement in the $n^{\text{th}}$ second:
$$s_n = u + \tfrac{a}{2}(2n - 1)$$Choosing the equation is simple: pick the one that contains the three values you know and the one you want, and leaves out the quantity you neither know nor need.
6. Free fall and vertical throws
Near the Earth's surface, ignoring air resistance, every body accelerates downward at $g = 9.8\ \text{m s}^{-2}$. NEET questions often use $g = 10\ \text{m s}^{-2}$ to keep the numbers clean.
For a body thrown straight up with speed $u$:
| Quantity | Result |
|---|---|
| Time to reach the top | $t = \dfrac{u}{g}$ |
| Maximum height | $H = \dfrac{u^2}{2g}$ |
| Total time of flight | $T = \dfrac{2u}{g}$ |
| Speed on returning to the launch point | $u$, the same as it was thrown with |
- At the highest point, velocity is zero but acceleration is still $g$ downward. It is never zero.
- Galileo's law of odd numbers: a body falling from rest covers distances in the ratio $1 : 3 : 5 : 7 : \ldots$ in successive equal intervals of time.
Worked example: thrown up from a tower
A ball is thrown upward at 10 m/s from the top of a 15 m tower. When does it hit the ground, and how fast? Take $g = 10\ \text{m s}^{-2}$.
Take up as positive. Then $u = +10$, $a = -10$, and the ground is 15 m below the start, so $s = -15$.
$-15 = 10t - 5t^2 \;\Rightarrow\; t^2 - 2t - 3 = 0 \;\Rightarrow\; (t-3)(t+1) = 0$, so $t = 3\ \text{s}$.
$v^2 = u^2 + 2as = 100 + 2(-10)(-15) = 400$, so $v = -20\ \text{m/s}$: 20 m/s, moving downward.
7. Stopping distance and reaction time
A vehicle moving at $u$ that brakes with deceleration $a$ stops in a distance
$$d_s = \frac{u^2}{2a}$$Stopping distance grows with the square of speed, so doubling the speed makes it four times longer. The driver's reaction time adds a further stretch covered at full speed before the brakes act.
Worked example
A car moves at 72 km/h, and its brakes give a deceleration of $5\ \text{m s}^{-2}$. The driver reacts in 0.5 s. Find the total stopping distance.
$72\ \text{km/h} = 72 \times \tfrac{5}{18} = 20\ \text{m/s}$.
Reaction distance $= 20 \times 0.5 = 10\ \text{m}$. Braking distance $= \dfrac{20^2}{2 \times 5} = 40\ \text{m}$.
Total $= 50\ \text{m}$.
8. Relative velocity in one dimension
The velocity of A as seen from B is
$$v_{AB} = v_A - v_B$$- Bodies moving in the same direction: relative speed is the difference of their speeds.
- Bodies moving in opposite directions: relative speed is the sum of their speeds.
Worked example
Two trains run towards each other at 54 km/h and 90 km/h. Their relative speed is $54 + 90 = 144\ \text{km/h} = 40\ \text{m/s}$. If they are 2 km apart, they meet after $\dfrac{2000}{40} = 50\ \text{s}$.
Mistakes that cost marks
- Taking $g$ as $+10$ after choosing upward as positive. It must be $-10$.
- Writing acceleration as zero at the highest point of a throw.
- Using the equations of motion when acceleration is not constant. Use $v = \dfrac{dx}{dt}$ and $a = \dfrac{dv}{dt}$ instead.
- Adding the area below the time axis as positive when the question asks for displacement.
- Mixing km/h and m/s. Multiply by $\tfrac{5}{18}$ to convert km/h to m/s.
- Assuming negative acceleration always means the body is slowing down.
- Averaging the two speeds of a round trip instead of using total distance over total time.
60-second recap
- Displacement is a signed change in position; distance is total path length.
- Slope of $x$–$t$ is velocity; slope of $v$–$t$ is acceleration; area under $v$–$t$ is displacement.
- $v = u + at$, $\;s = ut + \tfrac{1}{2}at^2$, $\;v^2 = u^2 + 2as$ — for constant acceleration only.
- Vertical throw: $H = \dfrac{u^2}{2g}$, $\;T = \dfrac{2u}{g}$, and acceleration is $g$ even at the top.
- Stopping distance $\propto u^2$.
- Relative velocity: $v_{AB} = v_A - v_B$.