JEE Mains Physics · Formula sheet

Motion in a Straight Line formulas

13 formulas, 1 reference table and 31 common traps for JEE Mains Physics Motion in a Straight Line, grouped by subtopic.

Full notes with worked examples

Average Speed, Average Velocity and Relative Velocity

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Average speed over a journey in legs

Average speed

\bar v = \frac{\text{total distance}}{\text{total time}} \qquad \text{equal distances: } \bar v = \frac{2v_1v_2}{v_1 + v_2}

Relative velocity in one dimension

Relative velocity

v_{BA} = v_B - v_A \qquad t_{cross} = \frac{L_1 + L_2}{v_{rel}}

Common traps

Averaging two speeds over equal distances

Equal distances take unequal times, so the plain mean is wrong. At 3 km/h and 5 km/h over equal distances the average is 2·3·5/8 = 3.75 km/h, not 4 km/h.

Average velocity on a round trip is zero

If the body comes back to its start, the displacement is zero, so the average velocity is zero whatever the speeds. The average speed is not zero.

Equal times inside an equal-distance leg

When half the distance is split into two equal TIMES, average those two speeds plainly first. Only then take the harmonic mean with the first half.

Forgetting the train's own length

A train has cleared a tunnel only when its last coach leaves, so the distance is tunnel length plus train length. Using the tunnel alone gives too short a time.

Subtracting speeds for trains approaching each other

With one direction taken positive, the other train's velocity is negative. v_B − v_A then has the size of the SUM of the speeds.

Leaving speeds in km/h

Lengths are in metres and times in seconds, so convert first. 108 km/h is 30 m/s; 18 km/h is 5 m/s.

Equations of Uniformly Accelerated Motion

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Choosing the right equation of motion

Equations of motion

v = u + at \qquad s = ut + \tfrac{1}{2}at^{2} \qquad v^{2} = u^{2} + 2as \qquad s = \frac{u + v}{2}\,t

Braking and stopping distance

Stopping distance

s = \frac{u^{2}}{2a} \qquad \frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^{2}

Distance in the nth second

s_n = u + \frac{a}{2}(2n - 1)

Common traps

Mid-distance speed is not the mean speed

The middle of a train passes the post when half the LENGTH has gone by, so its speed is √((u² + v²)/2). The mean (u + v)/2 is the speed at half the TIME.

The ratio of times is upside down

Speeding up at a₁ and braking at a₂ over the same speed change gives t₁/t₂ = a₂/a₁. The larger acceleration takes the shorter time.

Stopping distance is not proportional to speed

Halving the speed quarters the stopping distance. Scaling the distance by the speed ratio alone is the usual wrong option.

Loses one third, or keeps one third?

"Loses one third of its velocity" leaves 2u/3. "Velocity becomes one third" leaves u/3. The two give very different answers.

"In the nth second" is not "in n seconds"

In the 5th second means between t = 4 s and t = 5 s. In 5 seconds means from t = 0 to t = 5 s. Read the stem twice.

Seconds two apart differ by 2a

The nth and (n + 2)th seconds differ by 2a, not a. Dividing the difference by 1 doubles the acceleration.

Motion Under Gravity

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Free fall from rest

Falling from rest

h = \tfrac{1}{2}gt^{2} \qquad v = gt \qquad v = \sqrt{2gh}

Thrown up or down: one equation for the whole flight

Whole flight, up positive

s = ut - \tfrac{1}{2}gt^{2} \qquad H = \frac{u^{2}}{2g} \qquad t_{drop} = \sqrt{t_1t_2}

Two bodies, or drops falling at regular intervals

Later body's position

y_2 = u_2(t - \tau) - \tfrac{1}{2}g(t - \tau)^{2}

Common traps

A rebound changes the sign of the velocity

Down at 10 m/s and up at 5 m/s is a change of 15 m/s, not 5 m/s. Speeds add when the direction reverses.

Using 9.8 when the stem gives 10

Use the g the question states. The options are usually built on it, and the other value lands near a wrong option.

Dropped from a moving body is not dropped from rest

A stone let go from a balloon rising at 8 m/s first rises at 8 m/s. Taking u = 0 gives a shorter fall and the wrong answer.

Zero velocity is not zero acceleration

At the top the body is momentarily at rest, but gravity still acts. If a were zero there, the body would stay at the top.

Splitting the flight when one equation will do

Going up, then down, as two stages is slower and invites sign errors. With a = −g throughout, s = ut − ½gt² handles the whole flight in one line.

Counting drops instead of intervals

If the first drop lands as the sixth begins, five intervals fit in one fall time, not six.

Giving both bodies the same clock

A body launched later has been moving for t − τ, not t. Using t for both makes them meet at the wrong moment.

Motion Graphs: Slopes and Areas

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Areas under velocity–time and acceleration–time graphs

Areas

\Delta x = \int v\,dt \qquad \Delta v = \int a\,dt

Velocity–position graphs: a = v dv/dx

Acceleration from velocity and position

a = v\frac{dv}{dx} = \frac{1}{2}\frac{d(v^{2})}{dx}

Matching graph shapes to the motion

Motion	Position–time	Velocity–time	Acceleration–time
At rest	Horizontal line	On the time axis (v = 0)	On the time axis (a = 0)
Constant velocity	Straight sloping line	Horizontal line	On the time axis (a = 0)
Speeding up from rest, constant a	Parabola opening upward, $x \propto t^{2}$	Straight line through the origin	Horizontal line above the axis
Slowing to rest, constant a	Curve bending over, flat where it stops	Straight line falling to zero	Horizontal line below the axis
Thrown up and caught (up positive)	Downward-opening parabola	Straight line from +u to −u, zero at the top	Horizontal line at −g The velocity line crosses zero at the top, but the acceleration line never moves.
Falling from rest with drag kv (down positive)	Curve that straightens into a line of slope mg/k	Rises from 0 and levels off at mg/k	Starts at g and falls towards zero
Constant velocity, then reversed at equal speed	Rising line, then falling line	Positive constant, then negative constant	Zero, with a spike at the reversal

Each column is the slope of the column to its left.

Common traps

Area gives displacement, not distance

The statement "area under a v–t graph is the distance" is false as soon as the graph dips below the axis. Distance needs every area counted as positive.

Average velocity comes from areas on a v–t graph

Average velocity is total displacement over total time, so on a v–t graph it comes from the areas, not from a slope. When the areas above and below the axis are equal, the average velocity is zero even though the body has moved.

Average velocity is a chord, not a tangent

Between two times, join the two points on the x–t graph and take that slope. The tangent at one instant gives the instantaneous velocity instead.

Velocity zero does not make the acceleration zero

Where the v–t line crosses the axis the body is at rest for an instant, but the slope there, the acceleration, is unchanged.

The slope of v–x is not the acceleration

dv/dx has units of 1/s, not m/s². Multiply by v first. A straight v–x line does not mean constant acceleration.

Halving the slope of v² against x

v² = u² + 2ax, so the slope is 2a. Reading the slope as a doubles the answer.

Variable Acceleration: Differentiate and Integrate

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Position as a function of time: differentiate

Differentiate

v = \frac{dx}{dt} \qquad a = \frac{dv}{dt} = \frac{d^{2}x}{dt^{2}}

Velocity as a function of position: a = v dv/dx

Chain rule

a = v\frac{dv}{dx} \qquad v = k\sqrt{x} \Rightarrow a = \frac{k^{2}}{2}

Acceleration or force as a function of time: integrate

Integrate with the starting values

v(t) = u + \int_0^{t} a\,dt \qquad x(t) = x_0 + \int_0^{t} v\,dt

Common traps

A turning point is v = 0, not a = 0

The body reverses where its velocity changes sign. Where a = 0 the velocity is largest or smallest, which is a different instant.

Distance across a turning point

If the body turns inside the interval, x(end) − x(start) is only the displacement. Split at the turning point and add the sizes.

dv/dx is not the acceleration

"Velocity increases at 5 m/s per metre" is dv/dx. The acceleration is v times that, so it depends on where the body is.

Inverting the derivative but not the function

From t(x), v is 1/(dt/dx), not dt/dx. Check the units: dt/dx is in seconds per metre.

Dropping the starting value

An integral fixes the change, not the value. Leaving out the initial velocity or position shifts every answer that follows.

Differentiating when you should integrate

If v(t) is given and a displacement is asked, integrate v. Differentiating gives the acceleration, which is often among the options.

Motion in a Straight Line formulas

Average Speed, Average Velocity and Relative Velocity

Average speed over a journey in legs

Relative velocity in one dimension

Common traps

Equations of Uniformly Accelerated Motion

Choosing the right equation of motion

Braking and stopping distance

Common traps

Motion Under Gravity

Free fall from rest

Thrown up or down: one equation for the whole flight

Two bodies, or drops falling at regular intervals

Common traps

Motion Graphs: Slopes and Areas

Areas under velocity–time and acceleration–time graphs

Velocity–position graphs: a = v dv/dx

Matching graph shapes to the motion

Common traps

Variable Acceleration: Differentiate and Integrate

Position as a function of time: differentiate

Velocity as a function of position: a = v dv/dx

Acceleration or force as a function of time: integrate

Common traps

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