Motion in a Straight Line

1. Observer, Object & Kinematics

• Motion of a body depends on both the object and the observer.
• e.g. a girl sitting in a moving car : her position is different w.r.t. the boy driving beside her (B₁) and w.r.t. a boy standing on the road (B₂).
• Kinematics describes motion using : position (r→), velocity (v→), acceleration (a→) and time t.

2. Position Vector

• Requirement : an origin and an axis system.

• r→_A (also written r→_A/O) = position of A(x₁, y₁, z₁) w.r.t. the origin O.
• Similarly for B(x₂, y₂, z₂) : r→_B = x₂î + y₂ĵ + z₂k̂.

3. Position of B w.r.t. A

• Triangle law (figure above) : r→_A + r→_B/A − r→_B = 0

• r→_B/A does not depend on the choice of origin (C, D = any two origins / observers).

4. Displacement & Distance Travelled

• Displacement (S→) : a vector — change in position ; straight-line distance and direction from the initial to the final point.
• Distance travelled (l) : total length of the path covered between two positions ; a scalar.

• Displacement and distance travelled do not depend on the choice of origin.
• Distance travelled can't be negative and can't decrease with time.
• Minimum possible distance between two points = |S→| (straight line). e.g. 40 m along x, then 30 m along y : S→ = 40î + 30ĵ, |S→| = 50 m, l = 70 m.

5. Displacement & Distance on a Circle

• Car on a circle of radius R = 10 m (centre O at the origin), starting at A(−10, 0), moving clockwise A → B → C → D.

l = arc length = Rθ (θ in radian = angle turned at the centre). e.g. A → B : S→ = (0î + 10ĵ) − (−10î + 0ĵ) = 10î + 10ĵ.

General point P (angle φ at the centre from OA) :

P = (−R cosφ, R sinφ) ; S→ = (−R cosφ î + R sinφ ĵ) − (−R î + 0ĵ)

|S→| = √( [R(1 − cosφ)]² + (R sinφ)² )

6. Instantaneous Speed & Velocity

• Instantaneous speed (v) : rate of change of distance travelled.

For a small element of path, distance travelled dl = |displacement| = |dr→| (dr→ = small change in position vector).

• Instantaneous velocity (v→) : rate of change of position.

• Velocity = (speed)(direction) ; |velocity| = speed.
• Instantaneous velocity is always tangential to the path.

• e.g. constant speed 10 m/s on a circle (as above) : speed = 10 m/s at A, B, C, but v→_A = 10ĵ, v→_B = 10î, v→_C = −10ĵ (m/s) ⇒ velocity changes.

Units of speed & velocity :

• x km/h = x(5/18) m/s ; e.g. 36 km/h = 36 × 5/18 = 10 m/s. Also 160 cm/s = 1.6 m/s, 767 cm/s = 7.67 m/s.

7. Average Speed & Average Velocity

• Magnitude of instantaneous velocity = instantaneous speed.
• |Avg. velocity| ≤ Avg. speed (since |S→| ≤ l).

8. Uniform Velocity Motion

• Uniform = same throughout : v→ = constant.

v→ = dr→/dt ⇒ dr→ = v→ dt ⇒ ∫_ri^r_f dr→ = v→ ∫₀^t dt ⇒ r→_f − r→_i = v→t

• v→ constant ⇒ v constant (speed constant) and v̂ constant (straight-line path).
• Velocity constant ⇒ speed constant ; but speed constant ⇏ velocity constant (e.g. motion on a circle with constant speed).
• Straight-line path in one direction ⇒ l = |S→| = vt ; avg. velocity = v→, avg. speed = v.
• Speed v along a given direction : v→ = v v̂. e.g. 60 m/s along 20î − 10ĵ + 20k̂ (magnitude 30) ⇒ v→ = 60(20î − 10ĵ + 20k̂)/30 = 40î − 20ĵ + 40k̂ ; in 6 s : S→ = 240î − 120ĵ + 240k̂, |S→| = vt = 360 m.

9. Average Speed – Motion in Parts

• Find each part's time from distance = speed × time ; then avg. speed = total distance / total time.

Half distance : t₁ = l/2V, t₂ = l/22V ⇒ avg. speed = ll/2V + l/4V = 4V3

Half time : l₁ = V(t/2), l₂ = 2V(t/2) ⇒ avg. speed = l₁ + l₂t = 3V2

• Half-distance and half-time cases give different answers — always use total distance ÷ total time.

10. Minimum Time Problems

• Constant speed ⇒ l = vt ⇒ minimum time ⇔ minimum path length.

Ant on a cubical room (side l), corner to opposite corner, speed V (can't fly) :

• If it could fly : along the body diagonal √3 l ⇒ t = √3 l/V (not allowed).
• Floor face diagonal + vertical edge : t = (√2 l + l)/V.
• Shortest : unfold the floor and the wall into one plane ⇒ straight line across a 2l × l rectangle : √((2l)² + l²) = √5 l (it crosses the floor–wall edge at its midpoint).

11. Least-Time Path : Cylinder & Two Speeds

• Body moving on surfaces (cannot fly) : unfold the surfaces into one plane ; the shortest path is the straight line joining the two points in the unfolded figure.

• Hollow cylinder (radius R, height h), A on the bottom rim to B on the top rim, diametrically opposite : half the curved surface opens into a rectangle πR × h.

Two regions with different speeds :

• Straight line = shortest length but more path at low speed ; going around = more length at high speed. The least-time path bends at the boundary like light (Snell's law, n = c/v) :

i, r → angles with the normal to the boundary in regions 1 and 2 ; v₁, v₂ → speeds there. (Or write t(x) and put dt/dx = 0.)

12. Acceleration

• Instantaneous acceleration : rate of change of velocity ; unit m/s².

• In 3-D a vector needs components, e.g. v→ = (12î − 14ĵ + 6k̂) m/s.
• On a straight line only two directions are possible ⇒ the sign alone gives the direction : v→ = 12î ⇔ v = +12 m/s ; v→ = −12î ⇔ v = −12 m/s.

13. Sign of Displacement, Velocity & Acceleration

• The sign of s, v, a only gives direction (not size).
• First decide which direction is +ve (e.g. → +ve, or ↑ +ve).
• Quantity along the chosen +ve direction ⇒ +ve sign ; opposite to it ⇒ −ve sign. e.g. (→ +ve) a = 4 m/s² to the right ⇒ a = +4 m/s² ; to the left ⇒ a = −4 m/s².
• Displacement S is measured from initial to final position ; it does not depend on the path (distance l does).

Vector form : S = +20 m ⇔ S→ = 20î m ; S = −30 m (↑ +ve) ⇔ S→ = −30ĵ m.

14. Speeding Up & Slowing Down

• Negative acceleration does NOT mean speed decreases. Speeding up or slowing down depends on the signs of v and a together.

(|v| = speed.) e.g. (→ +ve) v = −10 m/s, a = −2 m/s² ⇒ after 1 s v = −12 m/s : speed rises to 12 m/s although a is negative. v = +10, a = +2 ⇒ +12 m/s : also speeding up.

• v = +ve, a = −ve ⇒ slowing down ; v = −ve, a = +ve ⇒ slowing down.

15. Sign of Acceleration due to Gravity

• Near the earth's surface g is constant.
• It is the force mg that acts on the body ; in free fall only mg acts ⇒ by F = ma :

• Taking ↑ as +ve : a = −g both while going up and while coming down.
• Going up : v = +ve, a = −ve ⇒ speed down. Coming down : v = −ve, a = −ve ⇒ speed up.

16. Equations of Uniformly Accelerated Motion

• Valid only when a→ is uniform (constant in magnitude and direction).

Derivation :

a = dvdt ⇒ ∫_u^v dv = a∫₀^t dt ⇒ v − u = at ⇒ v = u + at

v = drdt = u + at ⇒ ∫_ri^r_f dr = u∫₀^t dt + a∫₀^t t dt ⇒ S = r_f − r_i = ut + ½at²

1-D motion (s, u, v, a with signs) :

S_n^th = S_n − S_n−1 = [un + ½an²] − [u(n − 1) + ½a(n − 1)²] = u + a2(2n − 1)

• Put s, u, v, a with their signs : along the chosen +ve direction +, opposite −.
• These equations give displacement, not distance. Distance = |S| only if the motion is uni-directional (no turning back).

17. Standard Results – Uniform Acceleration

• Start from rest, a = 2 m/s² for 10 s (one direction) : S = ½ × 2 × 10² = 100 m.
• Distance in the last k seconds = S_{in t} − S_{in (t − k)} ; e.g. last 2 s of the above : 100 − 64 = 36 m.
• Constant force ⇒ a = F/m = constant ⇒ use these equations. From rest, 18 m in the first 3 s : 18 = ½a(3)² ⇒ a = 4 m/s².
• Same force, mass doubled ⇒ a′ = a/2 = 2 m/s² ⇒ distance in 3 s = ½ × 2 × 3² = 9 m (half).

Speed at the mid-point :

• Speeds u at A and v at B, constant acceleration : A → B : v² = u² + 2al ; A → mid-point C : v_C² = u² + 2a(l/2).

e.g. 10 m/s and 20 m/s ⇒ √250 = 5√10 m/s.

Reaction time :

• Reaction time : time interval between the moment the driver sees the need to brake and the moment he actually applies the brakes.
• During the reaction time the brakes are not applied ⇒ the car keeps its constant speed ; then it retards uniformly.

18. Changing Acceleration – Motion in Parts

• If the acceleration changes (in magnitude or direction) during the motion, break the motion into parts with constant acceleration ; the final velocity of one part is the initial velocity u of the next part.

Similarly : a train starting from rest with constant a ; its last two cars (each length l) pass a passenger in 6 s and 4 s ⇒ l = 6v₁ + 18a, 2l = 10v₁ + 50a ⇒ v₁ = 7a = at₁ ⇒ the passenger is 7 s late.

19. Free Fall from Rest

• Released from rest (↑ +ve, a = −g, g = 10 m/s²) — distances fallen in successive seconds :

0 → 1 s : −x₁ = ½(−10)(1)² ⇒ x₁ = 5 m. 1 → 2 s (u′ = −10 m/s) : −x₂ = −10(1) + ½(−10)(1)² ⇒ x₂ = 15 m (or x₁ + x₂ = ½(10)(2)² = 20 m).

20. Motion under Gravity – Ground to Ground

• Ball projected up with speed u from A, highest point B (v = 0), returns to the ground at C. Take ↑ +ve ⇒ a = −g throughout.

• Time of flight T : time for which the ball remains in air. A → C : S = 0 ⇒ 0 = uT − ½gT².
• Time of ascent (A → B) : 0 = u − gt₁. Time of descent : t₂ = T − t₁.
• Maximum height (A → B) : 0 = u² − 2gH.
• Speed on hitting the ground (A → C, S = 0) : v² = u² ⇒ v = ±u ; v = u − gT = −u ⇒ speed = u (downward).

Only for ground-to-ground motion under mg only.

• Displacement A → C = 0 ; A → B = +H ; B → C = −H. Distance has no formula : split the path where the ball turns ⇒ distance = 2H.
• e.g. u = 50 m/s, g = 10 m/s² : H = 125 m, T = 10 s, t₁ = t₂ = 5 s, v = −50 m/s, distance = 250 m.
• With g = 10 m/s² the speed changes by 10 m/s every second : 50, 40, 30, 20, 10, 0 going up and 10, 20, …, 50 coming down — at the same height the speed is the same while going up and coming down. Distance in a second : S = (u + v2)t, e.g. 1st second (50 + 40)/2 × 1 = 45 m.
• Passing a height h : S = +h in ut − ½gt² = h gives two times (going up and coming down). e.g. u = 50 m/s, h = 80 m : t² − 10t + 16 = 0 ⇒ t = 2 s (v = +30 m/s) and t = 8 s (v = −30 m/s) ⇒ speed 30 m/s both times.

21. Motion under Gravity – From a Tower

• Ball released (u = 0) from height H (↑ +ve, A top → B ground) : −H = 0 − ½gT².

v = 0 + (−g)√(2H/g) = −√(2gH) ⇒ speed √(2gH).

• Displacement is from the initial point (top of the tower) to the ground ⇒ S = −H in every case ; only the sign of u changes.

• Projected down at 20 m/s : −160 = −20t − 5t² ⇒ t² + 4t − 32 = 0 ⇒ t = 4 s ; v = −20 − 10(4) = −60 m/s ⇒ speed 60 m/s ; max height 160 m ; distance 160 m.
• Of the two roots of the quadratic, reject the negative time.

Same speed (60 m/s) on reaching the ground whether thrown up or down at 20 m/s.

Thrown up : no need to split into "up" and "down" parts — a = −g is constant for the whole flight, so one equation from top to ground works.

22. Vertical Motion with Air Resistance

• Air resistance force F_R is always opposite to velocity ⇒ it gives a constant retardation a (opposite to v).
• A → C (up and back) : s = 0, but the acceleration is not constant (it changes when the direction of motion changes) ⇒ s = ut + ½at² cannot be used from A to C directly. Break into A → B (up) and B → C (down).

e.g. u = 60 m/s, a = 2 m/s² : up (a₁ = −12 m/s²) : 0 = 60 − 12t₁ ⇒ t₁ = 5 s ; H = 60(5) − ½(12)(5)² = 150 m.

Down (u = 0, a₂ = −8 m/s², s = −150 m) : v² = 2(8)(150) = 2400 ⇒ v = 20√6 ≈ 49 m/s ; t₂ = √(75/2) = 5√6/2 ≈ 6.1 s ; T = t₁ + t₂ ≈ 11.1 s.

• So with air resistance : t₂ > t₁ and the ball returns with speed less than u.

23. Release of an Object from a Moving Platform

• When an object is released (not thrown) from a moving platform (train, balloon, aeroplane), its initial velocity = velocity of the platform at the instant of release.
• After release, the object and the platform may have different accelerations, according to a = forcemass. A released object moves under gravity only ⇒ a = g downward.
• e.g. balloon rising with v : the released ball first moves up with v, then falls.

Balloon rising with constant v = 20 m/s ; stone dropped at t = 5 s (H = 20 × 5 = 100 m) : −100 = 20t − 5t² ⇒ t² − 4t − 20 = 0 ⇒ t = 2 + 2√6 ≈ 6.9 s.

24. Variable Acceleration – Tools

• When v and a vary, use differentiation and integration.

• If a and v are opposite (retardation of magnitude a) : a = −dvdt = −vdvdx.

• Definite integration : the initial values give the lower limits, the final (unknown) values the upper limits, e.g. ∫_xi^x dx , ∫_ti^t dt.
• ∫y dx is not possible without knowing y in terms of x (∫x dx = x²/2 + C, but ∫y dx ≠ xy or y²/2). Similarly ∫a dt, ∫v dt, ∫a dv can be done only when the relation between the two variables is known.
• So choose the form to match the given data : a(t) ⇒ a = dv/dt ; a(x) or v(x) ⇒ a = v dv/dx ; v(t) ⇒ v = dx/dt. (With a = 6t + 10, ∫v dv = ∫(6t + 10)dx fails — the right side mixes t and x.)

25. Calculus Results for Variable Motion

• Differentiation : ddx(xⁿ) = nxⁿ⁻¹ ; with time use the chain rule : d(x²)dt = 2xdxdt.

To remove the log at the end : ln x = y ⇒ x = eʸ.

26. Position Given as a Function of Time

• Differentiate : v = dx/dt, a = dv/dt. "Initial" means t = 0.

• v = 0 : t² − 5t + 6 = 0 ⇒ t = 2, 3 s. a = 0 : t = 5/2 s.
• Displacement in 5 s = x_f − x_i = 65 − 10 = 55 m. (Distance in 5 s : it turns back at t = 2 and 3 s ⇒ 28 + 1 + 28 = 57 m.)

27. Distance vs Displacement (Turning Points)

• Displacement is easy : s = x_f − x_i. There is no direct formula for distance.
• In 1-D motion a particle can change its direction only when its velocity becomes zero for an instant. (In 2-D it can turn without v becoming zero.)
• Method : solve v = 0, check that v changes sign there, then add |Δx| of each part.

28. Acceleration Given as a Function of Time

• Use a = dv/dt, then v = dx/dt, with limits from the initial conditions.

a = 6t + 10 ; at t = 0 : x = 11 m, v = 4 m/s

∫₄^v dv = ∫₀^t (6t + 10) dt ⇒ v − 4 = 3t² + 10t

∫₁₁^x dx = ∫₀^t (3t² + 10t + 4) dt ⇒ x − 11 = t³ + 5t² + 4t

At t = 1 s : a = 16 m/s², v = 17 m/s, x = 21 m.

29. Velocity Given as a Function of Position

• Acceleration : a = v dvdx (or differentiate w.r.t. t using the chain rule and dx/dt = v).

v = 6√x ⇒ a = (6√x) · ddx(6√x) = (6√x)(62√x) = 18 m/s²

or a = 6 · 12√x · dxdt = 3√x · 6√x = 18 m/s²

• Position with time (x = 0 at t = 0) : separate variables, v = dx/dt :

∫₀^x dx√x = ∫₀^t 6 dt ⇒ 2√x = 6t ⇒ x = 9t²  (∫x^−1/2dx = 2√x)

30. Acceleration Given as a Function of Position

• a = f(x) : use a = vdvdx ⇒ ∫v dv = ∫f(x) dx.
• Wrong ways : ∫dv = ∫f(x) dt or ∫f(x) dxv = ∫dv — each side must contain only one variable.
• Once v(x) is known : v = dxdt ⇒ ∫dxv(x) = ∫dt gives x(t).

31. Acceleration Given as a Function of Velocity

• a = f(v) : a = dvdt ⇒ ∫dvf(v) = ∫dt gives v(t) ; then v = dxdt gives x(t).

a = 20 − v ; at t = 0 : x = 0, v₀ = 10 m/s

∫₁₀^v dv20 − v = ∫₀^t dt ⇒ −ln20 − v10 = t ⇒ 20 − v = 10e^−t

∫₀^x dx = ∫₀^t (20 − 10e^−t) dt = 20t + 10(e^−t − e⁰)

32. Air Resistance Proportional to Speed

• If the air resistance depends on speed, the acceleration is not constant ⇒ the equations of uniform acceleration cannot be used ; write a(v) and integrate with a = dvdt.

33. Gravity Varying with Height

• If g changes linearly with height y : write g = my + c and find m, c from two known values ; then use a = vdvdy (a depends on position).

34. Slope of a Graph

θ = angle of the tangent with the +ve x-axis. Slope of a curve at a point P = slope of the tangent at P.

• Slope = d(quantity on y-axis) / d(quantity on x-axis) : v–t graph ⇒ dvdt ; a–v graph ⇒ dadv.

• θ acute ⇒ m +ve ; θ obtuse ⇒ m = tanθ = −tanφ (φ = acute angle with the x-axis) ⇒ m −ve. Parallel lines have the same slope.
• More vertical the line ⇒ larger |m|.

Other quantities on the axes :

35. Maxima, Minima & Concavity

• At a (local) maximum or minimum the tangent is horizontal : slope dydx = 0.

• Concavity = rate of change of the slope dydx, i.e. ddx(dydx) = d²ydx².

36. Shape of an Increasing Graph

• q increasing with t ⇒ dqdt +ve. Whether the curve bends up or down is fixed by d²qdt² :

37. Area under a Curve

(area enclosed between the curve and the x-axis, from x = a to x = b)

∫₁⁴ (4x + 3) dx = [2x² + 3x]₁⁴ = 44 − 5 = 39 ;  ∫₀³ 4x² dx = [4x³3]₀³ = 36

38. Equation of a Straight Line

m = slope = tanθ ; c = intercept on the y-axis

• c = 0 ⇒ line through the origin (y = mx) ; c +ve ⇒ cuts the y-axis above the origin ; c −ve ⇒ below.

39. Standard Graphs – Rules

40. Position–Time (x–t) Graph

• Curve rising ⇒ v +ve ; falling ⇒ v −ve ; peak or valley (horizontal tangent) ⇒ v = 0.
• a = dvdt = d²xdt² = rate of change of the slope ⇒ shape of the curve gives the sign of a :

• Average velocity = x_f − x_itime (read the two end values from the graph).

41. Velocity–Time (v–t) Graph

• The most useful graph : gives both a and displacement.

• Area above the t-axis is +ve, below is −ve.
• Distance = |A₁| + |A₂| + … (add magnitudes of all areas).
• At a sharp corner the slope is not defined ⇒ a not defined at that instant.
• Speed = |velocity| ⇒ speed–time graph = v–t graph with the part below the axis flipped above ; area under speed–time = distance.

42. Acceleration–Time (a–t) Graph

43. Graphs against Position

• a–x area : do not write ∫dvdtdx ; use a = vdvdx.

a–x (from rest) : area up to x = 30 = ½ × 30 × 10 = 150 = ½v² ⇒ v = 10√3 m/s.

v–x line through origin : v = 2x ⇒ at x = 5 : v = 10 m/s, a = vdvdx = 10 × 2 = 20 m/s².

Tangent at 37° to the graph at the point : v–x with v = 60 m/s ⇒ a = 60 × 34 = 45 m/s² ; v²–x ⇒ a = 3/42 = 38 m/s² ; v²–x³ at x = 10 ⇒ a = 3x²2 × 34 = 9x²8 = 112.5 m/s².

v²–x straight line from (0, 0) to (4, 36) : a = 36/42 = 4.5 m/s².

44. Possible & Impossible Graphs

• A quantity has only one value at one instant, and time only moves forward ⇒ a graph against t can never give two values at the same t.
• Distance never decreases ⇒ distance–time slope can never be −ve.
• v +ve ⇒ x keeps increasing ; the particle can come back to a smaller x only if v becomes −ve.

45. Formula Summary – Kinematics

46. Formula Summary – Equations, Gravity & Graphs