Explanation:
Displacement: x = at² – bt³
Velocity: v = dx/dt = 2at – 3bt²
Acceleration: a_acc = dv/dt = 2a – 6bt
Setting acceleration to zero: 2a – 6bt = 0 ⇒ t = 2a / 6b = a / (3b).

Explanation:
– Distance covered in half a revolution = half of circumference = πR.
– Displacement = shortest straight line distance (diameter) = 2R.
– Ratio = Distance / Displacement = πR / 2R = π : 2.

Explanation:
Let the total distance be 2d.
– Time taken for the first half: t₁ = d / v₁.
– Time taken for the second half: t₂ = d / v₂.
– Average speed = Total Distance / Total Time = 2d / (d/v₁ + d/v₂) = 2v₁v₂ / (v₁ + v₂) (Harmonic mean).

Explanation:
From the third equation of motion: v² = u² – 2as.
For stopping, final velocity v = 0 ⇒ s = u² / 2a.
Since retardation a is constant: s ∝ u².
If the initial speed is tripled (3v), the stopping distance becomes 3² = 9 times the original distance, which is **9d**.

Explanation:
– Distance travelled from top in time t is: h = 1/2 g t².
– For total time T: H = 1/2 g T².
– Distance travelled from top in time T/2 is: h’ = 1/2 g (T/2)² = H / 4.
– Height from the ground at T/2 = H – h’ = H – H/4 = 3H / 4.

Explanation:
The displacement in the n-th second is: S_nth = u + a/2 (2n – 1).
Since u = 0: S_nth ∝ (2n – 1).
– For n = 1: S₁ ∝ 1.
– For n = 2: S₂ ∝ 3.
– For n = 3: S₃ ∝ 5.
The ratio is **1 : 3 : 5** (Galileo’s Law of Odd Numbers).

Explanation:
By definition, acceleration is: a = dv/dt ⇒ dv = a dt.
Integrating both sides:
Δv = ∫ dv = ∫ a dt.
The integral of a dt is the area under the a-t curve, which represents the **change in velocity**.

Explanation:
The area under the velocity-time (v-t) graph represents the displacement.
Since the graph is a triangle:
Displacement = Area of triangle = 1/2 × base × height.
Here, base = t₀ and height = v₀.
Displacement = 1/2 v₀ t₀.

Explanation:
The horizontal range of a projectile is: R = (u² sin 2θ) / g.
– For complementary angle (90° – θ):
R’ = [u² sin 2(90° – θ)] / g = [u² sin(180° – 2θ)] / g = (u² sin 2θ) / g.
Since R = R’, the horizontal ranges are equal, and their ratio is **1 : 1**.

Explanation:
Horizontal Range: R = (2 u² sinθ cosθ) / g.
Maximum Height: H = (u² sin²θ) / 2g.
Given R = 4H:
(2 u² sinθ cosθ) / g = 4 × [(u² sin²θ) / 2g].
Simplifying: cosθ = sinθ ⇒ tanθ = 1 ⇒ θ = 45°.

Explanation:
– Time of flight: T = 2u sinθ / g ⇒ T² = 4u² sin²θ / g².
– Maximum Height: H = u² sin²θ / 2g ⇒ u² sin²θ = 2gH.
– Substituting the height equation into the squared time of flight equation:
T² = 4(2gH) / g² = 8H / g ⇒ gT² = 8H.

Explanation:
The standard equation of trajectory of a projectile is: y = x tanθ – (g x²) / (2 u² cos²θ).
Comparing this with the given equation y = √3 x – (g x²) / 2:
tanθ = √3 ⇒ θ = 60°.

Explanation:
In projectile motion, there is no acceleration in the horizontal direction (a_x = 0).
– Consequently, the horizontal component of velocity remains constant throughout the flight: v_x = u cosθ.
– At the highest point of its trajectory, the vertical component of velocity becomes zero (v_y = 0).
– Hence, the net velocity at the highest point is only the horizontal component: **u cosθ**.

Explanation:
Analyzing the motion along the vertical direction:
– Initial vertical velocity: u_y = 0.
– Vertical displacement: y = h.
– Using second equation of motion: y = u_y t + 1/2 g t².
– h = 1/2 g t² ⇒ t = √(2h / g).
(The time of flight is independent of the horizontal projection velocity u).

Explanation:
– Since the trains are moving in opposite directions, their relative velocity is the sum of their individual speeds: v_rel = v₁ + v₂ = 20 + 30 = 50 m/s.
– The total distance to be covered to cross each other is the sum of their lengths: d_rel = L₁ + L₂ = 120 + 80 = 200 m.
– Time taken: t = d_rel / v_rel = 200 / 50 = 4 s.

Explanation:
To cross along the shortest path (directly perpendicular to the bank), the swimmer’s upstream velocity component must cancel the river’s flow velocity:
– If the swimmer makes an angle α with the line perpendicular to flow:
v sinα = u ⇒ sinα = u / v ⇒ α = sin⁻¹(u/v).
– The total angle with the downstream direction of river flow is: θ = 90° + α = 90° + sin⁻¹(u/v).

Explanation:
The time taken to cross the river depends only on the velocity component perpendicular to the river banks:
t = d / (v cosθ).
– For time t to be minimum, cosθ must be maximum, which is 1 (meaning θ = 0°).
– Thus, the swimmer must head directly perpendicular to the flow. The minimum time is: t_min = d / v.

Explanation:
The umbrella must be held in the direction of the velocity of rain relative to the man (v_rm):
– Velocity of rain: v⃗_r = -4ĵ.
– Velocity of man: v⃗_m = 3î.
– Relative velocity: v⃗_rm = v⃗_r – v⃗_m = -3î – 4ĵ.
– Angle with the vertical θ is given by:
tanθ = |v_x / v_y| = 3 / 4 ⇒ θ = tan⁻¹(3/4).

Explanation:
In uniform circular motion, speed is constant but the direction of velocity changes continuously.
– The acceleration responsible for changing the direction of velocity is called centripetal acceleration.
– Its magnitude is a_c = v² / r and it is always directed **radially inward, towards the center** of the circle.

Explanation:
Acceleration: a = dv/dt ⇒ v = ∫ a dt.
Integrating the given acceleration expression:
v = ∫ (3t² + 2t + 2) dt = t³ + t² + 2t + C.
– At t = 0, v = 2 m/s ⇒ 2 = C.
– Thus, the velocity equation is: v = t³ + t² + 2t + 2.
– At t = 2 s: v = 2³ + 2² + 2(2) + 2 = 8 + 4 + 4 + 2 = 18 m/s.

Explanation:
– First displacement: d₁ = v₁ × t₁.
– Second displacement: d₂ = v₂ × t₂.
– Average velocity = Total Displacement / Total Time = (d₁ + d₂) / (t₁ + t₂) = (v₁t₁ + v₂t₂) / (t₁ + t₂).

Explanation:
Given: t = α x² + β x.
Differentiating with respect to x:
dt/dx = 2αx + β ⇒ v = dx/dt = 1 / (2αx + β).
Differentiating v with respect to t (using chain rule):
a = dv/dt = v (dv/dx) = v × d/dx [ (2αx + β)⁻¹ ].
a = v × [ – (2αx + β)⁻² × 2α ] = -2α v³.
Thus, retardation (negative acceleration) is **2α v³**.

Explanation:
Since there is no horizontal acceleration, the horizontal component of velocity remains unchanged:
v_x = u_x.
– The vertical motion experiences a constant acceleration of -g. Using the first equation of motion:
v_y = u_y – gt.
– Therefore, the velocity vector at any time t is: v⃗ = u_x î + (u_y – gt) ĵ.

Explanation:
The slope of the tangent to the curve on a displacement-time (x-t) graph is mathematically represented by the derivative dx/dt.
– By definition, dx/dt at any given instant is the **instantaneous velocity** of the particle.

Explanation:
Both freely falling bodies experience the same acceleration due to gravity g directed vertically downwards.
– Acceleration of the first body: a₁ = -g ĵ.
– Acceleration of the second body: a₂ = -g ĵ.
– Relative acceleration: a_rel = a₁ – a₂ = (-g ĵ) – (-g ĵ) = 0.
Therefore, the relative acceleration is always **zero**.