Kinematics (1‑D & 2‑D) – KVPY Study Notes
KVPY – Kishore Vaigyanik Protsahan Yojana
Subject – Physics
Topic – Kinematics (1D & 2D)
Aim – Understand the language of motion, derive the basic equations, and solve typical KVPY‑style problems.
1. 1‑Dimensional Kinematics
Definition – Motion along a straight line.
Key Variables
- (x) – Position (m)
- (x_0) – Initial position (m)
- (v) – Instantaneous velocity (m s⁻¹)
- (v_0) – Initial velocity (m s⁻¹)
- (a) – Acceleration (m s⁻²)
- (t) – Time (s)
1.1 Uniform Motion (Constant (v), (a = 0))
[ x = x_0 + v,t ]
1.2 Uniformly Accelerated Motion (Constant (a))
[ \begin{aligned} v &= v_0 + a t \tag{1}\ x &= x_0 + v_0 t + \frac{1}{2} a t^2 \tag{2}\ v^2 &= v_0^2 + 2 a (x - x_0) \tag{3} \end{aligned} ]
(2) – “Time–displacement” equation
(3) – Useful when time is not given
1.3 Free‑Fall (Near Earth)
- (a = g \approx 9.81,\text{m s}^{-2}) (downward is negative)
1.4 Relative Motion (1‑D)
If two observers move with velocities (v_A) and (v_B) along the same line,
the velocity of (B) relative to (A) is
[ v_{B/A} = v_B - v_A ]
2. 2‑Dimensional Kinematics
Definition – Motion in a plane; studied using vector components.
2.1 Vector Decomposition
Any vector (\vec{r}) or (\vec{v}) can be written in Cartesian form:
[ \vec{r} = x,\hat{i} + y,\hat{j}, \qquad \vec{v} = v_x,\hat{i} + v_y,\hat{j} ]
- (\hat{i}) – unit vector along (+x)
- (\hat{j}) – unit vector along (+y)
2.2 Projectile Motion (Parabolic Trajectory)
Assumptions:
- No air resistance.
- Acceleration only in (y)-direction: (a_x = 0), (a_y = -g).
Initial conditions – launch at speed (u) making angle (\theta) with (+x):
[ u_x = u\cos\theta,\qquad u_y = u\sin\theta ]
Equations
[ \begin{aligned} x(t) &= u_x t = u\cos\theta, t \tag{4}\ y(t) &= u_y t - \frac{1}{2} g t^2 = u\sin\theta, t - \frac{1}{2} g t^2 \tag{5} \end{aligned} ]
Range (horizontal distance when (y=0) again):
[ R = \frac{u^2 \sin 2\theta}{g} ]
Maximum height:
[ H = \frac{u_y^2}{2g} = \frac{u^2 \sin^2\theta}{2g} ]
Time of flight:
[ T = \frac{2 u_y}{g} = \frac{2u\sin\theta}{g} ]
2.3 Relative Motion (2‑D)
If (A) moves with velocity (\vec{v}_A) and (B) with (\vec{v}_B),
[ \vec{v}_{B/A} = \vec{v}_B - \vec{v}_A ]
Component‑wise:
[ v_{B/A,x} = v_{B,x} - v_{A,x}, \quad v_{B/A,y} = v_{B,y} - v_{A,y} ]
3. Worked Example (KVPY‑Style)
Problem
A stone is thrown horizontally from the top of a 45‑m cliff with speed (15,\text{m s}^{-1}).
- (a) Time of flight.
- (b) Horizontal distance from the base of the cliff.
- (c) Speed just before impact.
- (d) Launch angle that gives the same range if the stone is thrown from ground level at the same speed.
Solution
-
Horizontal launch → (u_y = 0), (u_x = 15,\text{m s}^{-1})
-
Time of flight – Solve (y(t)=0) using (5):
[
0 = 0\cdot t - \frac{1}{2} g t^2 + 45
\implies t = \sqrt{ \frac{2\times45}{g} } \approx \sqrt{ \frac{90}{9.81} } \approx 3.03,\text{s}
]
- Horizontal range
[ R = u_x t = 15 \times 3.03 \approx 45.5,\text{m} ]
- Speed before impact
[ \begin{aligned} v_x &= u_x = 15,\text{