MHT-CET Physics · Motion in a Plane

Equations of Motion and Motion Graphs

Under constant acceleration v = u + at, s = ut + ½at² and v² = u² + 2as; when the acceleration changes, velocity is the slope of the position–time graph, acceleration the slope of the velocity–time graph, and each area under a graph gives the change in the quantity above it.

All Motion in a Plane notes MHT-CET Physics strategy

Why this matters

19 PYQs, none HARD. Ten use the equations of motion — braking distances, free fall from a tower, a ball thrown up from a bridge, the distance in the nth second, average speed. Nine read a graph or differentiate a position: speed from x(t) and y(t), when acceleration is zero, distance and displacement from a velocity graph. Two cards.

Concept 1 of 2: The Equations of Motion

Three equations cover constant acceleration; pick the one that leaves out what you are not given. Stopping from speed v at deceleration a takes a distance v²/2a, so n times the speed needs n² times the deceleration for the same distance. A body falling from rest covers distances in the ratio 1 : 3 : 5… in successive seconds, and in the nth second covers a(2n − 1)/2. Starting at the midpoint of two speeds v₁ and v₂, the speed is √((v₁² + v₂²)/2), not their average. Over equal distances at speeds v and v/3 the average speed is the harmonic mean 2v₁v₂/(v₁ + v₂). Check whether a braking vehicle stops before the time asked about.

Definition

$v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , $v^2 = u^2 + 2as$ ; nth second: $s_n = u + \tfrac{a}{2}(2n - 1)$ .
Stopping distance $\dfrac{v^2}{2a}$ ⇒ speed × n needs deceleration × n².
Free fall for t/2 of a fall lasting t: H/4 fallen; for T/4: H/16.
Midpoint speed: $\sqrt{\dfrac{v_1^2 + v_2^2}{2}}$ (20 and 30 ⇒ 25.5 m/s).
Equal distances: $v_{\text{avg}} = \dfrac{2v_1v_2}{v_1 + v_2}$ .

Equations of motion

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

Worked example

A ball is thrown up at 10 m/s from a 15 m high bridge. When does it hit the water? (g = 10 m/s²)

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q13Moderate

Example 1 · Motion in a Plane · Kinematics — Equations of Motion, Free Fall, and Graphs

A body travelling with uniform acceleration crosses two points A and B with velocities 20 m/s and 30 m/s respectively. The speed of the body at mid point of A and B is (nearly)

Averaging speeds over equal distances

Equal DISTANCES take unequal times, so the average speed is the harmonic mean. V and V/3 give V/2, not 2V/3.

Running the equations past the stop

A vehicle braking from 15 m/s at 0.3 m/s² stops at 50 s. After that it stays put; s = ut + ½at² at 60 s would have it reversing.

Drill 9 more on the equations of motion

Concept 2 of 2: Motion Graphs and Calculus

Velocity is dx/dt and acceleration dv/dt; going the other way, integrate. On an x–t graph a flat line means rest and a straight sloping line uniform velocity; on a v–t graph the slope is acceleration and the area is displacement. Distance counts every area as positive, displacement subtracts the parts below the axis. On an a–t graph the area is the change in velocity. In two dimensions, find vₓ and v_y separately and combine: x = at², y = bt² gives speed 2t√(a² + b²).

Definition

$v = \dfrac{dx}{dt}$ , $a = \dfrac{dv}{dt}$ ; $x = at^2 - bt^3$ ⇒ a = 0 at $t = \dfrac{a}{3b}$ .
a varying: integrate twice ( $a = 6t + 5$ from rest ⇒ 18 m in 2 s).
v–t: slope = acceleration, area = displacement; distance adds |areas| (areas 6, −2, 4, −2, 4 ⇒ 10 : 18).
a–t area = Δv (triangle 8 m/s² × 10 s ⇒ 40 m/s).
2-D: $v = \sqrt{v_x^2 + v_y^2}$ .

Calculus of motion

v = \frac{dx}{dt}, \qquad a = \frac{dv}{dt}, \qquad \Delta x = \int v\,dt

Worked example

A particle's position is x = t³ − 6t² + 9t. When is it momentarily at rest?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q2Moderate

Example 2 · Motion in a Plane · Kinematics — Equations of Motion, Free Fall, and Graphs

A point moves along X-axis initially at rest. Its acceleration is

a = (6t + 5)\ \text{m/s}^2

. The distance covered in 2 s, if it starts from origin is given by

Using s = ½at² when a changes with time

With a = 6t + 5, the equations of motion do not apply. Integrate a to get v, then v to get x.

Drill 8 more on motion graphs and calculus

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The Equations of Motion
Equations of motion
$v = u + at, \qquad s = ut + \tfrac{1}{2}at^2, \qquad v^2 = u^2 + 2as$
Motion Graphs and Calculus
Calculus of motion
$v = \frac{dx}{dt}, \qquad a = \frac{dv}{dt}, \qquad \Delta x = \int v\,dt$

Watch out for (3)

Averaging speeds over equal distances
→ The Equations of Motion
Running the equations past the stop
→ The Equations of Motion
Using s = ½at² when a changes with time
→ Motion Graphs and Calculus

Test yourself on Motion in a Plane

15 past MHT-CET questions from this chapter, timed at 14 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Motion in a Plane chapter test Drill all 19 questions