Chapter 3: Kinematics

NEB Physics, Grade XI. Motion along a line and in a plane, described without asking what causes it.

3.1 Instantaneous velocity and acceleration

Displacement, average velocity

Displacement Δx = x₂ − x₁ is a vector: it depends only on the initial and final positions. The average velocity over a time interval Δt is

It hides detail: a car may speed up and slow down within the interval. To capture motion at one instant, shrink the interval.

Instantaneous velocity

The velocity at an instant is the limit of the average velocity as Δt → 0:

Graphically, it is the slope of the tangent to the position–time graph at that instant. The slope of the chord (secant) joining two points gives the average velocity; as the second point slides towards P, the chord turns into the tangent.

Speed is the magnitude of instantaneous velocity. For motion in a straight line without reversal, average speed equals the magnitude of average velocity; otherwise, average speed = total distance / total time is larger.

Acceleration

Average acceleration is a_av = Δv/Δt. Instantaneous acceleration is

It is the slope of the velocity–time graph. Unit: m s⁻²; dimensions [LT⁻²]. If a and v have the same sign the body speeds up; if opposite, it slows down (retardation). Constant acceleration gives a straight-line v–t graph.

3.2 Relative velocity

Velocity is always measured relative to some observer (frame). The velocity of A relative to B is

In one dimension: two bodies moving in the same direction at speeds v_A and v_B have relative speed |v_A − v_B|; in opposite directions, v_A + v_B. Note v_BA = −v_AB.

River–boat problems

Let v_r = river speed (relative to ground), v_b = boat speed in still water, width = d. The boat's ground velocity is v_bg = v_b + v_r.

(a) Shortest time: point the boat perpendicular to the bank. t = d/v_b; downstream drift = v_rt.
(b) Shortest path (zero drift): aim upstream at angle θ to the perpendicular with sin θ = v_r/v_b (possible only if v_b > v_r). Effective speed = √(v_b² − v_r²), so t = d/√(v_b² − v_r²).

Rain and the moving man

Rain falls vertically at v_R; a man walks at v_M. Rain relative to the man: v_RM = v_R − v_M, of magnitude √(v_R² + v_M²), inclined to the vertical at tan θ = v_M/v_R, slanting toward him. He must tilt his umbrella forward by θ.

3.3 Equations of motion (graphical treatment)

For uniform acceleration along a straight line, let the initial velocity be u, final velocity v after time t, acceleration a, displacement s.

1. Velocity–time relation

Slope of the line = acceleration: a = (v − u)/t, so

2. Position–time relation

The shaded area is a rectangle (u × t) plus a triangle (½ × t × (v − u) = ½ × t × at):

3. Velocity–displacement relation

Area of the trapezoid: s = ½(u + v)t. Substitute t = (v − u)/a:

Distance in the nth second

Difference of displacements in n and (n − 1) seconds:

Use these only when a is constant. Choose a positive direction and give every vector its sign.

3.4 Motion of a freely falling body

A body moving under gravity alone (air resistance neglected) is in free fall, whether it is dropped, thrown up or thrown down. All bodies fall with the same acceleration g ≈ 9.8 m s⁻² (about 10 m s⁻² for rough work) directed towards Earth's centre. Galileo's observation is consistent with a feather and a hammer landing together in a vacuum.

Body dropped from height h (u = 0, downward positive)

Body thrown vertically upward with speed u (upward positive)

At the top, v = 0. Time to rise t_up = u/g; maximum height H = u²/2g; total time of flight (back to start) T = 2u/g. The motion is symmetric: it passes any height with equal speed going up and down, and it returns with speed u. At the top the velocity is zero but the acceleration is still g.

3.5 Projectile motion and its applications

A projectile is a body thrown into the air that then moves under gravity alone. The key idea: horizontal and vertical motions are independent. Horizontally, a_x = 0 (constant velocity); vertically, a_y = −g.

A. Projectile fired at angle θ (from ground level)

Components: u_x = u cos θ, u_y = u sin θ. At time t:

Equation of the trajectory

Eliminate t using t = x/(u cos θ):

This has the form y = bx − cx², a parabola.

Time of flight, maximum height, range

Time of flight: the projectile returns to y = 0: T = 2u sin θ/g.
Maximum height: at the top v_y = 0, so t = u sin θ/g and H = u² sin²θ/2g.
Horizontal range: R = (u cos θ)T = u² sin 2θ/g.

Consequences. (i) R is maximum, R_max = u²/g, when θ = 45°. (ii) Angles θ and (90° − θ) give the same range, since sin 2θ = sin(180° − 2θ). (iii) R = 4H/tan θ, so at θ = 45° R = 4H. (iv) Speed at any time = √(v_x² + v_y²), direction tan φ = v_y/v_x. (v) On level ground, landing speed = u, at angle θ below the horizontal.

B. Horizontal projection from a height h

Trajectory: y = gx²/2u² measured downward from the launch point. The time of fall does not depend on u: a bullet fired horizontally and a bullet dropped from the same height hit level ground together.

Applications

Sports (a football kick, a javelin, a cricket shot, a basketball throw), water from a hose or fountain, bomb release from an aircraft, and artillery. In reality air resistance shortens range and height, and makes the path asymmetric.

Summary

• v = dx/dt, a = dv/dt; slope of x–t is velocity, slope of v–t is acceleration, area under v–t is displacement.
• v_AB = v_A − v_B. River: shortest time is straight across; shortest path is aimed upstream with sin θ = v_r/v_b.
• Constant a: v = u + at, s = ut + ½at², v² = u² + 2as.
• Free fall: a = g downward always; H = u²/2g, T = 2u/g.
• Projectile: independent x and y motion; R = u² sin 2θ/g, maximum at 45°.

Practice questions

Take g = 9.8 m s⁻² unless stated. Try each question first, then open the answer.

A. Multiple choice

1. The slope of a position–time graph at an instant gives: (a) acceleration (b) instantaneous velocity (c) displacement (d) average speed
   Answer

   (b)
2. The area under an acceleration–time graph represents: (a) displacement (b) velocity (c) change in velocity (d) jerk
   Answer

   (c)
3. At the highest point of a ball thrown vertically upward: (a) v = 0, a = 0 (b) v = 0, a = g downward (c) v = max, a = 0 (d) v = 0, a = g upward
   Answer

   (b). Gravity keeps acting.
4. For maximum range of a projectile on level ground the angle of projection is: (a) 30° (b) 45° (c) 60° (d) 90°
   Answer

   (b)
5. Two projectiles with equal speeds are thrown at 30° and 60°. Their ranges are: (a) in ratio 1:2 (b) equal (c) in ratio 2:1 (d) in ratio 1:3
   Answer

   (b), since sin 60° = sin 120°.
6. A boat must cross a river in the least time. It should head: (a) upstream (b) downstream (c) perpendicular to the bank (d) at 45° upstream
   Answer

   (c)

B. Conceptual (short answer)

7. Can a body have zero velocity and non-zero acceleration? Give an example.
   Answer

   Yes. A ball at the top of its flight has v = 0 but a = g.
8. Why is the path of a projectile a parabola?
   Answer

   The horizontal position grows linearly with time and the vertical position quadratically; eliminating t gives y = bx − cx².
9. A bullet is fired horizontally and another is dropped from the same height at the same instant. Which reaches the ground first?
   Answer

   Both together. Vertical motion is independent of horizontal velocity.
10. Distinguish between average speed and the magnitude of average velocity.
    Answer

    Speed uses total path length; velocity uses net displacement. They are equal only for straight-line motion without reversal.

C. Numerical problems

11. A particle has x = 3t² + 2t + 1 (SI). Find its velocity and acceleration at t = 2 s.
    Answer

    v = 6t + 2 = 14 m s⁻¹; a = 6 m s⁻².
12. Car A moves east at 60 km h⁻¹ and car B west at 40 km h⁻¹. Find the velocity of A relative to B, and of A relative to B if B also moved east at 40 km h⁻¹.
    Answer

    100 km h⁻¹ east; then 20 km h⁻¹ east.
13. A swimmer whose speed in still water is 4 m s⁻¹ crosses a 120 m wide river flowing at 3 m s⁻¹. Find the minimum crossing time and the drift, then the time to land directly opposite.
    Answer

    Minimum time = 120/4 = 30 s; drift = 90 m. Directly opposite: effective speed √(16 − 9) = √7 ≈ 2.65 m s⁻¹, so t ≈ 45.3 s.
14. A train at 72 km h⁻¹ applies brakes and stops in 50 m. Find the retardation and stopping time.
    Answer

    u = 20 m s⁻¹; 0 = 400 + 2a(50) gives a = −4 m s⁻²; t = 5 s.
15. A body starts from rest with a = 2 m s⁻². Find its displacement in the 5th second.
    Answer

    s₅ = 0 + ½(2)(9) = 9 m.
16. A stone falls from a tower and reaches the ground in 3 s. Find the height and the impact speed.
    Answer

    h = ½(9.8)(9) = 44.1 m; v = 29.4 m s⁻¹.
17. A ball is thrown upward at 29.4 m s⁻¹. Find the maximum height, the time to return, and its velocity after 4 s.
    Answer

    H = 864.36/19.6 = 44.1 m; T = 6 s; v = 29.4 − 39.2 = −9.8 m s⁻¹ (9.8 m s⁻¹ downward).
18. A projectile is launched at 40 m s⁻¹ at 45°. Find its time of flight, maximum height and range.
    Answer

    T = 2(40)(0.707)/9.8 ≈ 5.77 s; H = 1600(0.5)/19.6 ≈ 40.8 m; R = 1600/9.8 ≈ 163.3 m.
19. A footballer wants a range of 100 m at 45°. What is the required launch speed?
    Answer

    u = √(Rg) = √980 ≈ 31.3 m s⁻¹.
20. An aeroplane flying horizontally at 100 m s⁻¹ at height 490 m releases a packet. How far ahead (horizontally) does it land?
    Answer

    t = √(2 × 490/9.8) = 10 s; R = 100 × 10 = 1000 m.

D. Long answer / derivation

20. Derive the three equations of motion for uniform acceleration using the velocity–time graph. (See section 3.3.)
21. Derive expressions for the time of flight, maximum height and range of a projectile fired at angle θ, and show that the trajectory is a parabola. (See section 3.5.)
22. A river-boat problem: explain with a vector diagram how a boatman should steer to cross in the shortest path. (See Fig. 3.2b.)