Kinematics
Apply concepts from Kinematics to problem-solving. Focus on numerical practice and real-world applications.
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/), and t is the time (s).
Dimensional check: [] = [] + [][] = [].
Second: s = ut + , where s is the displacement (m).
Third: = + 2as.
A fourth useful relation gives displacement in the nth second: = 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/; with downward positive, g = +9.8 m/.
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: = A cos(), = A sin(). The scalar product A.B = AB cos() yields a scalar; the vector product A x B = AB sin() 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 with the horizontal (g = 9.8 m/): time of flight T = 2u sin()/g [] (s), maximum height H = ()/2g [] (m), and range R = sin(2theta)/g [] (m).
Maximum range occurs at = 45 degrees. Complementary angles ( and 90 - ) produce the same range but different maximum heights.
Throughout the flight, the horizontal component = u cos() 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 = v/r [] (rad/s).
Centripetal acceleration = /r = 2 r [] (m/) 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 = 30 degrees to the horizontal (g = 10 m/).
= 20 cos 30 = 20 x 0.866 = 17.32 m/s.
= 20 sin 30 = 20 x 0.5 = 10 m/s.
Time of flight: T = 2u sin()/g = 2 x 20 x 0.5 / 10 = 2.0 s.
Maximum height: H = ()/2g = (20)2 x (0.5)2 / (2 x 10) = 400 x 0.25 / 20 = 5.0 m.
Range: R = 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 = 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: = /r = (10)^.5 = 200 m/. Dimensional formula: [].
Angular velocity: = v/r = .5 = 20 rad/s. Dimensional formula of : [].
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
| Equation | Variables | Dimensional Check | When to Use |
|---|---|---|---|
| v = u + at | v: final velocity, u: initial velocity, a: acceleration, t: time | [] = [] + [][T] | When displacement is not needed |
| s = ut + | s: displacement, u: initial velocity, a: acceleration, t: time | [L] = [][T] + [][] | When final velocity is not needed |
| = + 2as | v: final velocity, u: initial velocity, a: acceleration, s: displacement | [L^{2}$$T^{-2}] = [L^{2}$$T^{-2}] + [][L] | When time is not needed |
| = u + a(2n-1)/2 | : displacement in nth second, n: the specific second | [L] = [] + [][T] | For displacement in a particular second |
B) Projectile Motion Formulae
| Quantity | Formula | At = 30 degrees | At = 45 degrees | At = 60 degrees |
|---|---|---|---|---|
| Time of flight T (s) | 2u sin()/g | 2u(0.5)/g = u/g | 2u(0.707)/g = 1.414u/g | 2u(0.866)/g = 1.732u/g |
| Max height H (m) | ()/2g | (0.25)/2g | (0.5)/2g | (0.75)/2g |
| Range R (m) | sin(2theta)/g | (0.866)/g | (1)/g = /g (max) | (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 Type | Slope Meaning | Area Meaning | Shape for Uniform Motion | Shape for Accelerated Motion |
|---|---|---|---|---|
| Position-time (x vs t) | Instantaneous velocity | Not used directly | Straight line (constant slope) | Parabola (increasing slope) |
| Velocity-time (v vs t) | Instantaneous acceleration | Displacement | Horizontal line (zero slope) | Straight line with non-zero slope |
| Acceleration-time (a vs t) | Rate of change of acceleration (jerk) | Change in velocity | Horizontal line | Varies with motion type |
D) Scalar vs Vector Quantities
| Quantity | Type | Magnitude Only / Direction Too | Example |
|---|---|---|---|
| Distance | Scalar | Magnitude only | Total path = 10 m |
| Displacement | Vector | Magnitude + direction | 6 m east |
| Speed | Scalar | Magnitude only | 5 m/s |
| Velocity | Vector | Magnitude + direction | 5 m/s north |
| Time | Scalar | Magnitude only | 3 s |
| Acceleration | Vector | Magnitude + direction | 9.8 m/ downward |
Study Materials
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100 Flashcards
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Frequently Asked Questions
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