CLASS 11 · CHAPTER 2 · MECHANICS

Motion in a Straight Line

Before forces, before vectors in two dimensions, there's this: a position, a clock, and the two derivatives that describe everything else. Every later chapter quietly assumes you've internalised this one.

01 · See it

Watch it happen

Initial velocity v₀: 8 m/sAcceleration a: -2 m/s²

Blue shaded area = displacement while moving one way; orange = displacement after it reverses. They subtract, not add.

0.0 m

Position

8.0 m/s

Velocity

0.0 m

Displacement

0.0 m

Distance

The top graph is position, the bottom is velocity, same clock for both. Watch the shaded area accumulate under the velocity graph, that area, not the curve’s height, is the displacement. Set the acceleration to work against the initial velocity and watch the object reverse: the position graph turns around, the velocity graph crosses zero, and the shaded area splits into two colours that partially cancel.

02 · Derive it

Where the formula comes from

Start from the definitions themselves. Acceleration is the rate of change of velocity, so for constant $a$ :

a = \dfrac{dv}{dt} \;\;\Rightarrow\;\; \int_{v_0}^{v} dv = \int_0^t a\,dt \;\;\Rightarrow\;\; \textcolor{#e08a1e}{v = v_0 + at}

Velocity is the rate of change of position, so integrating this result once more:

v = \dfrac{dx}{dt} \;\;\Rightarrow\;\; \int_{x_0}^{x} dx = \int_0^t (v_0+at)\,dt \;\;\Rightarrow\;\; \textcolor{#e08a1e}{x = x_0 + v_0t + \tfrac{1}{2}at^2}

For the third equation, eliminate time instead of integrating over it directly, using the chain rule $a = v\frac{dv}{dx}$ :

\int_{v_0}^{v} v\,dv = \int_{x_0}^{x} a\,dx \;\;\Rightarrow\;\; \textcolor{#e08a1e}{v^2 = v_0^2 + 2a(x-x_0)}

All three connect the same five quantities, $x_0,x,v_0,v,a$ and $t$ , and hold only while the acceleration truly stays constant.

03 · Break it

Where the shortcut stops working

“Zero velocity means zero acceleration” feels obvious and is simply false. Throw a ball straight up: at its highest point its velocity is exactly zero for an instant, but gravity never stopped acting on it. The acceleration there is still $-g$ , completely unchanged, it’s precisely what curves the ball back downward a moment later.

A related trap: the sign of acceleration does not tell you whether something is speeding up or slowing down, only the direction it points in. That effect depends on how the sign of $a$ compares to the sign of $v$ at that instant. The same $a = -g$ slows a rising ball down and speeds a falling one up, identical acceleration, opposite effect, because the velocity’s sign flipped in between.

And distance is not displacement. Average speed uses total path length; average velocity uses net displacement. They agree only when the motion never reverses, which is exactly the special case most textbook problems quietly hand you, and exactly the case real motion (a walk to a shop and back, a ball bouncing) usually isn’t.

04 · Master it

Apply it under exam conditions

Q1. A car moving at 25 m/s brakes to a stop over 100 m. Find the deceleration and the time it takes to stop.

0 = (25)^2 + 2a(100) \;\Rightarrow\; a = \textcolor{#e08a1e}{-3.125\text{ m/s}^2}

0 = 25 + (-3.125)t \;\Rightarrow\; t = \textcolor{#e08a1e}{8\text{ s}}

Q2. A ball is thrown straight up at 15 m/s from the ground. Find the maximum height and the total time before it lands again. (g = 10 m/s²)

H = \dfrac{v_0^2}{2g} = \dfrac{225}{20} = \textcolor{#e08a1e}{11.25\text{ m}}

T = \dfrac{2v_0}{g} = \dfrac{30}{10} = \textcolor{#e08a1e}{3\text{ s}}

05 · FAQs

Quick answers

What's actually different between speed and velocity?+

Velocity carries a sign (direction along the line); speed is just its size. A velocity of −5 m/s and +5 m/s are different velocities but the same speed, 5 m/s.

Why is the area under a velocity-time graph equal to displacement?+

Displacement is the integral of velocity over time, and a definite integral is exactly the signed area under the curve. It's not an analogy, it's the same calculation viewed geometrically.

Can an object have zero velocity but nonzero acceleration?+

Yes, constantly. A ball thrown straight up has exactly zero velocity for an instant at its highest point, while gravity keeps accelerating it downward the entire time, including at that instant.

Does negative acceleration always mean an object is slowing down?+

No. It means the acceleration points in the negative direction. Whether that speeds the object up or slows it down depends entirely on which way it's already moving, the sign alone never tells you.

Isn't distance the same thing as displacement?+

Only if the object never changes direction. Distance is the total path length covered (always adds up); displacement is the net change in position (can partially or fully cancel). Walk 5 m forward and 3 m back and you've covered 8 m of distance but only 2 m of displacement.

Related concepts

Physics doesn’t stay inside chapter boundaries. Neither should you.

Mechanics

Motion in a Plane

Drag angle and speed, watch the parabola respond, and see why 45° isn't always optimal.

Mechanics

Laws of Motion

Push a block and watch friction switch from static to kinetic the instant it slips.