IMAT Physics · Formula sheet

Kinematics formulas

9 formulas and 12 common traps for IMAT Physics Kinematics, grouped by subtopic.

Displacement, Velocity, Relative Motion and Acceleration

Distance and displacement, speed and velocity

Average speed and average velocity

\text{average speed} = \frac{\text{distance}}{t} \qquad \bar{v} = \frac{\Delta s}{t}

$\bar{v}$ average velocity, a vector
$\Delta s$ displacement, from start to finish, in m
$t$ total time taken, in s

Relative velocity and Galilean transformations

Relative velocity

v_{AB} = v_A - v_B \qquad t_{\text{catch}} = \frac{\text{gap}}{v_A - v_B}

$v_A, v_B$ velocities of A and B measured by the same observer
$t_{\text{catch}}$ time for A to close a gap on B ahead of it, moving the same way

Acceleration: the rate of change of velocity

Average acceleration

a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}

$u$ initial velocity, in m/s
$v$ final velocity, in m/s
$t$ time taken, in s

Common traps

Average speed is not the average of the speeds

Average speed is total distance over total time. Going 60 km at 60 km/h and 60 km back at 30 km/h gives 40 km/h, not 45 km/h, because more time is spent at the lower speed.

Average velocity uses the straight-line displacement

For a trip with legs at right angles, the displacement comes from Pythagoras, not from adding the legs. Dividing the total path by the time gives the average speed, and IMAT lists it as a wrong option.

Galilean transformations need uniform straight-line relative motion

They hold only when one observer moves at a constant velocity in a straight line relative to the other. Accelerating, braking or rotating observers are excluded, even when the rotation is at a constant rate.

Check the units before comparing a gap with a speed

When speeds are in km/h and a gap is in metres, convert first. A gap of 400 m closed at 36 km/h (10 m/s) takes 40 s; dividing 400 by 36 gives a meaningless 11.

Zero velocity does not mean zero acceleration

At the top of its flight a ball thrown straight up is momentarily still, but gravity is still changing its velocity at

9.8\ \text{m/s}^2

downwards. Acceleration depends on how velocity is changing, not on its value.

Equations of Motion, Motion Graphs and Free Fall

Learn this subtopic in the notes

The equations of motion for constant acceleration

Equations of uniformly accelerated motion

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

$s$ displacement, in m
$u, v$ initial and final velocity, in m/s
$a$ constant acceleration, in m/s²
$t$ time, in s

Motion graphs: gradients and areas

Reading a v-t graph

a = \text{gradient} = \frac{\Delta v}{\Delta t} \qquad s = \text{area under the line}

$\Delta v / \Delta t$ rise over run of the v-t line

Free fall and vertical throws

Falling from rest and the highest point of a throw

h = \tfrac{1}{2}gt^2, \quad v = gt \qquad h_{\max} = \frac{u^2}{2g}

$g$ acceleration of free fall, about 9.8 m/s²
$h$ height fallen or risen, in m
$u$ speed of an upward throw, in m/s

Common traps

Stopping distance grows with the square of the speed

From

v^2 = u^2 + 2as

, the braking distance is

u^2 / 2a

. Double the speed and the distance is four times as long, not twice.

The equations hold only while the acceleration is constant

A car that accelerates and then cruises has two stages with different accelerations. Work out each stage on its own and carry the speed at the end of one into the next.

The same flat line means different things on different graphs

A horizontal line on an s-t graph means the body is standing still. On a v-t graph it means the body moves at a constant velocity. Read the axis labels before reading the shape.

Heavier objects do not fall faster

Without air resistance every body falls with the same acceleration

g

. A heavier body has a larger weight, but also more mass to accelerate, and the two cancel. Air resistance is what makes a feather fall slowly on Earth.

Projectile Motion and Uniform Circular Motion

Learn this subtopic in the notes

Projectile motion: horizontal and vertical parts are independent

Launch from level ground

t_{\text{flight}} = \frac{2u_y}{g} \qquad R = u_x\, t_{\text{flight}}

$u_x, u_y$ horizontal and vertical components of the launch velocity
$R$ horizontal range on level ground

Period, frequency and angular velocity in circular motion

Circular motion quantities

f = \frac{1}{T} \qquad \omega = \frac{2\pi}{T} = 2\pi f \qquad v = \omega r

$T$ period, in s
$f$ frequency, in Hz
$\omega$ angular velocity, in rad/s
$r$ radius of the circle, in m

Centripetal acceleration: always towards the centre

Centripetal acceleration

a = \frac{v^2}{r} = \omega^2 r

$v$ speed along the circle, in m/s
$r$ radius, in m
$\omega$ angular velocity, in rad/s

Common traps

At the top of its path a projectile is still moving

Only the vertical velocity is zero at the highest point. The horizontal velocity is unchanged, so the speed there equals

u\cos\theta

, not zero.

Change rpm into hertz before using 2π

Angular velocity in rad/s needs revolutions per SECOND. A value in revolutions per minute must be divided by 60 first, or the answer is 60 times too big.

Constant speed in a circle still means acceleration

Uniform circular motion has a constant speed but not a constant velocity. The acceleration

v^2/r

points towards the centre. An option giving zero acceleration for a body moving steadily round a circle is wrong.

Kinematics formulas

Displacement, Velocity, Relative Motion and Acceleration

Distance and displacement, speed and velocity

Relative velocity and Galilean transformations

Acceleration: the rate of change of velocity

Common traps

Equations of Motion, Motion Graphs and Free Fall

The equations of motion for constant acceleration

Motion graphs: gradients and areas

Free fall and vertical throws

Common traps

Projectile Motion and Uniform Circular Motion

Projectile motion: horizontal and vertical parts are independent

Period, frequency and angular velocity in circular motion

Centripetal acceleration: always towards the centre

Common traps

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