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PhysicsME

Kinematics

Apply concepts from Kinematics to problem-solving. Focus on numerical practice and real-world applications.

3-4 Qs/year50 minPhase 1 · APPLICATION

Concept Core

Kinematics describes the motion of objects without considering the forces causing them. The fundamental distinction is between scalars (distance, speed) and vectors (displacement, velocity, acceleration). Distance is the total path length (always positive), while displacement is the shortest straight-line separation between initial and final positions (can be zero even when distance is non-zero).

For motion with constant acceleration, three equations of motion govern every calculation.
First: v = u + at, where v is the final velocity (m/s), u is the initial velocity (m/s), a is the acceleration (m/s2s^{2}), and t is the time (s).
Dimensional check: [L1  T1L^{1} \; T^{-1}] = [L1  T1L^{1} \; T^{-1}] + [L1  T2L^{1} \; T^{-2}][T1T^{1}] = [L1  T1L^{1} \; T^{-1}].
Second: s = ut + 12\frac{1}{2} at2at^{2}, where s is the displacement (m).
Third: v2v^{2} = u2u^{2} + 2as.
A fourth useful relation gives displacement in the nth second: sns_{n} = u + a(2n - 1)/2, with SI unit m. Sign convention is critical: choose one direction as positive and maintain it.
For free fall with upward positive, g = -9.8 m/s2s^{2}; with downward positive, g = +9.8 m/s2s^{2}.

Graphical analysis is a powerful tool. In a position-time graph, the slope at any point gives the instantaneous velocity. In a velocity-time graph, the slope gives acceleration, and the area under the curve gives displacement (positive area for forward motion, negative area for backward motion).

Vectors are resolved into components: AxA_{x} = A cos(θ\theta), AyA_{y} = A sin(θ\theta). The scalar product A.B = AB cos(θ\theta) yields a scalar; the vector product A x B = AB sin(θ\theta) yields a vector perpendicular to both A and B (direction by right-hand rule).

Projectile motion combines uniform horizontal motion with uniformly accelerated vertical motion.
For a projectile launched at speed u at angle θ\theta with the horizontal (g = 9.8 m/s2s^{2}): time of flight T = 2u sin(θ\theta)/g [M0  L0  T1M^{0} \; L^{0} \; T^{1}] (s), maximum height H = u2  sin2u^{2} \; sin^{2}(θ\theta)/2g [M0  L1  T0M^{0} \; L^{1} \; T^{0}] (m), and range R = u2u^{2} sin(2theta)/g [M0  L1  T0M^{0} \; L^{1} \; T^{0}] (m).
Maximum range occurs at θ\theta = 45 degrees. Complementary angles (θ\theta and 90 - θ\theta) produce the same range but different maximum heights.
Throughout the flight, the horizontal component uxu_{x} = u cos(θ\theta) remains constant; only the vertical component changes under gravity.

In uniform circular motion, an object moves at constant speed v along a circle of radius r.
The angular velocity is ω\omega = v/r [M0  L0  T1M^{0} \; L^{0} \; T^{-1}] (rad/s).
Centripetal acceleration aca_{c} = v2v^{2}/r = ω\omega2 r [M0  L1  T2M^{0} \; L^{1} \; T^{-2}] (m/s2s^{2}) always points toward the centre, changing the direction of velocity without altering its magnitude.

Solved Numerical 1: A ball is thrown with velocity u = 20 m/s at θ\theta = 30 degrees to the horizontal (g = 10 m/s2s^{2}).
uxu_{x} = 20 cos 30 = 20 x 0.866 = 17.32 m/s.
uyu_{y} = 20 sin 30 = 20 x 0.5 = 10 m/s.
Time of flight: T = 2u sin(θ\theta)/g = 2 x 20 x 0.5 / 10 = 2.0 s.
Maximum height: H = u2  sin2u^{2} \; sin^{2}(θ\theta)/2g = (20)2 x (0.5)2 / (2 x 10) = 400 x 0.25 / 20 = 5.0 m.
Range: R = u2u^{2} sin(2theta)/g = (20)2 x sin 60 / 10 = 400 x 0.866 / 10 = 34.64 m.

Solved Numerical 2: From a v-t graph, v increases from 0 to 20 m/s in 4 s (linearly), then remains constant at 20 m/s for 6 s. Displacement in first 4 s = area of triangle = 12\frac{1}{2} x 4 s x 20 m/s = 40 m. Displacement in next 6 s = area of rectangle = 6 s x 20 m/s = 120 m. Total displacement = 40 m + 120 m = 160 m.

Solved Numerical 3: A particle moves in a circle of radius r = 0.5 m with constant speed v = 10 m/s.
Centripetal acceleration: aca_{c} = v2v^{2}/r = (10)^20\frac{2}{0}.5 = 200 m/s2s^{2}. Dimensional formula: [M0  L1  T2M^{0} \; L^{1} \; T^{-2}].
Angular velocity: ω\omega = v/r = 100\frac{10}{0}.5 = 20 rad/s. Dimensional formula of ω\omega: [M0  L0  T1M^{0} \; L^{0} \; T^{-1}].

The key testable concept is projectile motion formulae (T, H, R) with the complementary angle property for range and the graphical method of calculating displacement from a velocity-time graph.

Key Testable Concept

The key testable concept is projectile motion formulae (T, H, R) with the complementary angle property for range and the graphical method of calculating displacement from a velocity-time graph.

Comparison Tables

A) Equations of Motion

EquationVariablesDimensional CheckWhen to Use
v = u + atv: final velocity, u: initial velocity, a: acceleration, t: time[LT1LT^{-1}] = [LT1LT^{-1}] + [LT2LT^{-2}][T]When displacement is not needed
s = ut + 12\frac{1}{2} at2at^{2}s: displacement, u: initial velocity, a: acceleration, t: time[L] = [LT1LT^{-1}][T] + [LT2LT^{-2}][T2T^{2}]When final velocity is not needed
v2v^{2} = u2u^{2} + 2asv: final velocity, u: initial velocity, a: acceleration, s: displacement[L^{2}$$T^{-2}] = [L^{2}$$T^{-2}] + [LT2LT^{-2}][L]When time is not needed
sns_{n} = u + a(2n-1)/2sns_{n}: displacement in nth second, n: the specific second[L] = [LT1LT^{-1}] + [LT2LT^{-2}][T]For displacement in a particular second

B) Projectile Motion Formulae

QuantityFormulaAt θ\theta = 30 degreesAt θ\theta = 45 degreesAt θ\theta = 60 degrees
Time of flight T (s)2u sin(θ\theta)/g2u(0.5)/g = u/g2u(0.707)/g = 1.414u/g2u(0.866)/g = 1.732u/g
Max height H (m)u2  sin2u^{2} \; sin^{2}(θ\theta)/2gu2u^{2}(0.25)/2gu2u^{2}(0.5)/2gu2u^{2}(0.75)/2g
Range R (m)u2u^{2} sin(2theta)/gu2u^{2}(0.866)/gu2u^{2}(1)/g = u2u^{2}/g (max)u2u^{2}(0.866)/g

Note: R at 30 degrees = R at 60 degrees (complementary angles give same range), but H at 60 degrees = 3 x H at 30 degrees.

C) Graph Interpretation

Graph TypeSlope MeaningArea MeaningShape for Uniform MotionShape for Accelerated Motion
Position-time (x vs t)Instantaneous velocityNot used directlyStraight line (constant slope)Parabola (increasing slope)
Velocity-time (v vs t)Instantaneous accelerationDisplacementHorizontal line (zero slope)Straight line with non-zero slope
Acceleration-time (a vs t)Rate of change of acceleration (jerk)Change in velocityHorizontal lineVaries with motion type

D) Scalar vs Vector Quantities

QuantityTypeMagnitude Only / Direction TooExample
DistanceScalarMagnitude onlyTotal path = 10 m
DisplacementVectorMagnitude + direction6 m east
SpeedScalarMagnitude only5 m/s
VelocityVectorMagnitude + direction5 m/s north
TimeScalarMagnitude only3 s
AccelerationVectorMagnitude + direction9.8 m/s2s^{2} downward

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