Vardaan Learning Institute

Sol: For half a revolution, distance travelled (arc length) = $\pi R$. Displacement (straight line from start to end) = Diameter = $2R$.
Ratio = $\frac{\text{Distance}}{\text{Displacement}} = \frac{\pi R}{2R} = \frac{\pi}{2}$.

Sol: Total time $t = 2 \times 60 + 20 = 140\text{ s}$.
Number of rounds = $\frac{140}{40} = 3.5$ rounds.
After 3 complete rounds, displacement is $0$. After the remaining $0.5$ round, he is at the diametrically opposite point. Displacement = $2R$.

Sol: Total Distance = $4\text{ m} + 3\text{ m} = 7\text{ m}$.
Displacement is the hypotenuse of the right-angled triangle formed: $s = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\text{ m}$.

Sol: In one cycle of $8$ steps ($8\text{ s}$), net distance covered = $5 - 3 = 2\text{ m}$.
To reach $13\text{ m}$, he needs to fall into the pit on a forward stroke. Total distance needed before final burst = $13 - 5 = 8\text{ m}$.
Time for $8\text{ m}$ net displacement = $4 \text{ cycles} \times 8\text{ s} = 32\text{ s}$.
In the next $5\text{ s}$, he takes $5$ forward steps, reaching $8+5 = 13\text{ m}$ and falls. Total time = $32 + 5 = 37\text{ s}$.

Sol: Yes, displacement can be zero if the particle returns to its starting position (e.g., completing a full circle).
No, distance travelled by a moving particle is always positive; it can never be zero or negative.

Sol: (d) Equal to or less than $1$. Displacement is the shortest path, so $|\text{Displacement}| \le \text{Distance}$. Ratio $\le 1$.

Sol: After half a revolution, the point moves horizontally by $\pi R$ and vertically up by $2R$.
Net displacement = $\sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$. Since $R=1$, displacement = $\sqrt{\pi^2 + 4}\text{ m}$.

Sol: For equal distances, $v_{avg} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{2(30)(50)}{30 + 50} = \frac{3000}{80} = 37.5\text{ km/h}$.

Sol: Average Velocity = $0$ (since net displacement is zero).
Average Speed (equal distances) = $\frac{2(10)(20)}{10+20} = \frac{400}{30} = 13.33\text{ m/s}$.

Sol: Total Distance = $S$. Total Time = $\frac{S/3}{u} + \frac{S/3}{v} + \frac{S/3}{w} = \frac{S}{3} (\frac{1}{u} + \frac{1}{v} + \frac{1}{w})$.
$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{S}{\frac{S}{3} \frac{vw + uw + uv}{uvw}} = \frac{3uvw}{vw + uw + uv}$.

Sol: Total Displacement = $v_1 t_1 + v_2 t_2$. Total Time = $t_1 + t_2$.
$v_{avg} = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2}$.

Sol: Time to market $t_1 = \frac{2.5}{5} = 0.5\text{ h} = 30\text{ min}$.
Remaining time for interval = $40 - 30 = 10\text{ min} = \frac{1}{6}\text{ h}$.
Distance back in $10\text{ min}$ = $7.5 \times \frac{1}{6} = 1.25\text{ km}$.
Net Displacement = $2.5 - 1.25 = 1.25\text{ km}$.
Average Velocity = $\frac{\text{Displacement}}{\text{Total Time (hours)}} = \frac{1.25}{40/60} = \frac{1.25}{2/3} = 1.875\text{ km/h}$.

Sol: Since time intervals are equal ($t_1 = t_2 = t$), $v_{avg} = \frac{v_1 t + v_2 t}{2t} = \frac{v_1 + v_2}{2}$.

Sol: Total distance = $20 + 16 = 36\text{ m}$. Total time = $2 + 2 = 4\text{ s}$.
Average Speed = $\frac{36}{4} = 9\text{ m/s}$.

Sol: Yes, velocity is a vector. If direction changes while magnitude (speed) remains constant, velocity varies. Example (2D): Uniform Circular Motion.
In strict 1D motion, it is NOT possible, because to change direction in a straight line, the object must stop and reverse, meaning speed cannot remain constant.

Sol: $v = \frac{dx}{dt} = \frac{d}{dt}(8.5 + 2.5t^2) = 5t$
At $t=0$, $v = 5(0) = 0\text{ m/s}$.
At $t=2$, $v = 5(2) = 10\text{ m/s}$.

Sol: At $t=0$, Position $x = 25\text{ m}$.
Velocity $v = \frac{dx}{dt} = 8t - 15$. At $t=0$, $v = -15\text{ m/s}$.
Acceleration $a = \frac{dv}{dt} = 8\text{ m/s}^2$ (Constant).

Sol: $v = \int a dt = \int (3t^2 + 2t + 2) dt = t^3 + t^2 + 2t + C$.
At $t=0, v=2 \implies C = 2$. So, $v = t^3 + t^2 + 2t + 2$.
At $t=2$, $v = (2)^3 + (2)^2 + 2(2) + 2 = 8 + 4 + 4 + 2 = 18\text{ m/s}$.

Sol: $v = \frac{ds}{dt} = 3t^2 - 12t + 3$.
$a = \frac{dv}{dt} = 6t - 12$. Set $a = 0 \implies 6t = 12 \implies t = 2\text{ s}$.
Substitute $t=2$ in velocity equation: $v = 3(2)^2 - 12(2) + 3 = 12 - 24 + 3 = -9\text{ m/s}$.

Sol: By principle of homogeneity, all terms must have dimensions of velocity $[LT^{-1}]$.
$[A] = [LT^{-1}]$.
$[Bt] = [LT^{-1}] \implies [B][T] = [LT^{-1}] \implies [B] = [LT^{-2}]$.
$[Ct^2] = [LT^{-1}] \implies [C][T^2] = [LT^{-1}] \implies [C] = [LT^{-3}]$.

Sol: Use chain rule: $a = v \frac{dv}{dx}$.
$\frac{dv}{dx} = \alpha \cdot \frac{1}{2\sqrt{x}}$.
$a = (\alpha \sqrt{x}) \cdot (\frac{\alpha}{2\sqrt{x}}) = \frac{\alpha^2}{2}$.

Sol: Differentiate wrt $t$: $1 = 2ax \frac{dx}{dt} + b \frac{dx}{dt} = v(2ax + b)$.
Thus, $v = (2ax + b)^{-1}$.
Diff wrt $t$ again: $a = \frac{dv}{dt} = -1(2ax + b)^{-2} \cdot \frac{d}{dt}(2ax+b) = -(2ax+b)^{-2} \cdot (2av)$.
Substitute $(2ax+b)^{-1} = v$: $a = -v^2 \cdot (2av) = -2av^3$.

Sol: Roots of $v=0$ are $t=1, 3$. The particle turns at these times.
Dist = $\int_0^1 |v|dt + \int_1^3 |v|dt + \int_3^4 |v|dt$.
$S(t) = \int v dt = \frac{t^3}{3} - 2t^2 + 3t$.
$\Delta S_{0 \to 1} = (\frac{1}{3} - 2 + 3) - 0 = \frac{4}{3}$.
$\Delta S_{1 \to 3} = (9 - 18 + 9) - \frac{4}{3} = -\frac{4}{3} \implies \text{Dist} = \frac{4}{3}$.
$\Delta S_{3 \to 4} = (\frac{64}{3} - 32 + 12) - 0 = \frac{64}{3} - 20 = \frac{4}{3}$.
Total Distance = $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4\text{ m}$.

Sol: $a = v \frac{dv}{dx} = -kx \implies \int_{v_0}^v v dv = \int_0^x -kx dx$.
$\left[ \frac{v^2}{2} \right]_{v_0}^v = \left[ -\frac{kx^2}{2} \right]_0^x \implies \frac{v^2}{2} - \frac{v_0^2}{2} = -\frac{kx^2}{2}$.
$v^2 = v_0^2 - kx^2 \implies v = \sqrt{v_0^2 - kx^2}$.

Sol: $v = \frac{dx}{dt} = 6t^2 - 6t + 4$.
At $t=1, v_1 = 6(1) - 6(1) + 4 = 4\text{ m/s}$.
At $t=3, v_3 = 6(9) - 6(3) + 4 = 54 - 18 + 4 = 40\text{ m/s}$.
$a_{avg} = \frac{\Delta v}{\Delta t} = \frac{40 - 4}{3 - 1} = \frac{36}{2} = 18\text{ m/s}^2$.

Sol: Length $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3)(35^2) = 1.5 \times 1225 = 1837.5\text{ m}$.
Velocity $v = u + at = 0 + 3(35) = 105\text{ m/s}$.

Sol: Initial speed $u = 54 \times \frac{5}{18} = 15\text{ m/s}$.
Distance during reaction time: $s_1 = 15 \times 0.20 = 3\text{ m}$.
Braking distance ($v=0$): $v^2 - u^2 = 2as \implies 0 - 15^2 = 2(-6)s_2 \implies s_2 = \frac{225}{12} = 18.75\text{ m}$.
Total distance = $3 + 18.75 = 21.75\text{ m}$.

Sol: $u = 200\text{ m/s}$, $v = 0$, $s = 4\text{ cm} = 0.04\text{ m}$.
$v^2 - u^2 = 2as \implies 0 - (200)^2 = 2 \cdot a \cdot 0.04$
$a = -\frac{40000}{0.08} = -5 \times 10^5\text{ m/s}^2$.

Sol: $x = \frac{1}{2}a(10)^2 = 50a$.
Distance in first $20\text{s}$ is $S_{20} = \frac{1}{2}a(20)^2 = 200a$.
Distance in next $10\text{s} = y = S_{20} - x = 200a - 50a = 150a$.
Thus, $y = 3(50a) \implies y = 3x$.

Sol: Let time of accel be $t_1$, decel be $t_2$. $t_1 + t_2 = t$.
$v_{max} = \alpha t_1 \implies t_1 = \frac{v_{max}}{\alpha}$.
$0 = v_{max} - \beta t_2 \implies t_2 = \frac{v_{max}}{\beta}$.
$\frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta} = t \implies v_{max}(\frac{\alpha+\beta}{\alpha\beta}) = t \implies v_{max} = \frac{\alpha\beta t}{\alpha+\beta}$.

Sol: Distance in $n^{\text{th}}$ second = Dist in $n$ sec - Dist in $(n-1)$ sec.
$S_n = (un + \frac{1}{2}an^2) - [u(n-1) + \frac{1}{2}a(n-1)^2]$
$S_n = un + \frac{1}{2}an^2 - un + u - \frac{1}{2}a(n^2 - 2n + 1)$
$S_n = u + \frac{1}{2}an^2 - \frac{1}{2}an^2 + an - \frac{1}{2}a = u + a(n - \frac{1}{2}) = u + \frac{a}{2}(2n - 1)$.

Sol: Using $S_n = u + \frac{a}{2}(2n - 1)$:
$12 = u + \frac{a}{2}(3) \quad \text{...(i)}$
$20 = u + \frac{a}{2}(7) \quad \text{...(ii)}$
Subtracting (i) from (ii): $8 = 2a \implies a = 4\text{ m/s}^2$.
Substitute $a=4$ in (i): $12 = u + \frac{4}{2}(3) \implies 12 = u + 6 \implies u = 6\text{ m/s}$.

Sol: From $v_f^2 - v_i^2 = 2as \implies 0 - v^2 = -2ad \implies d = \frac{v^2}{2a}$.
If speed is $nv$, new distance $d' = \frac{(nv)^2}{2a} = n^2 (\frac{v^2}{2a}) = n^2 d$.

Sol: For full train: $v^2 - u^2 = 2aL \implies aL = \frac{v^2 - u^2}{2}$.
For midpoint, distance is $L/2$. Let velocity be $v_m$.
$v_m^2 - u^2 = 2a(L/2) = aL$. Substitute $aL$:
$v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2} \implies v_m = \sqrt{\frac{u^2 + v^2}{2}}$.

Sol: Since $d \propto v^2$ (from constant deceleration).
Speed is doubled ($40 \to 80$). Therefore, distance increases by $2^2 = 4$ times.
New distance = $4 \times 2 = 8\text{ m}$.

Sol: At max height, $v = 0$. $u = 20\text{ m/s}, g = 10\text{ m/s}^2$ (downward).
$h = \frac{u^2}{2g} = \frac{20^2}{2(10)} = \frac{400}{20} = 20\text{ m}$ above the thrower.
Total height from ground = $25 + 20 = 45\text{ m}$.

Sol: Using $S = ut + \frac{1}{2}at^2$. Taking upward as positive: $S = -25\text{ m}$ (since it lands below start), $u = 20$, $a = -10$.
$-25 = 20t - 5t^2 \implies 5t^2 - 20t - 25 = 0 \implies t^2 - 4t - 5 = 0$.
Factoring: $(t - 5)(t + 1) = 0$. Time must be positive, so $t = 5\text{ s}$.

Sol: Velocity on hitting ground $v = \sqrt{2gh}$. Momentum $P = m\sqrt{2gh} \propto \sqrt{h}$.
If $h' = 2h$, $P' = \sqrt{2}P \approx 1.414 P$.
Change = $1.414P - P = 0.414P$. Percentage change = $41.4\%$.

Sol: Let interval be $t$. Distances fallen from rest in times $t, 2t, 3t \dots$ are:
$y_1 = \frac{1}{2}gt^2$
$y_2 = \frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)$
$y_3 = \frac{1}{2}g(3t)^2 = 9(\frac{1}{2}gt^2)$
Distances in successive intervals: $d_1 = y_1 = 1(\frac{1}{2}gt^2)$. $d_2 = y_2 - y_1 = 3(\frac{1}{2}gt^2)$. $d_3 = y_3 - y_2 = 5(\frac{1}{2}gt^2)$. Ratio = $1:3:5\dots$

Sol: From $S = ut + \frac{1}{2}at^2$, since $u=0$, $h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}}$.
Hence, $t \propto \sqrt{h}$. Ratio $\frac{t_1}{t_2} = \sqrt{\frac{h_1}{h_2}}$.

Sol: Time for 1st drop to hit ground: $t = \sqrt{2h/g} = \sqrt{10/10} = 1\text{s}$.
3 drops mean 2 time intervals in $1\text{s}$. So each drop falls every $0.5\text{s}$.
The 2nd drop has been falling for $0.5\text{s}$. Its distance fallen = $\frac{1}{2}(10)(0.5)^2 = 1.25\text{ m}$.
Height above ground = $5 - 1.25 = 3.75\text{ m}$.

Sol: Packet has initial velocity $u = 9.8\text{ m/s}$ (upwards), $S = -39.2\text{ m}$.
$-39.2 = 9.8t - 4.9t^2 \implies 4.9t^2 - 9.8t - 39.2 = 0 \implies t^2 - 2t - 8 = 0$
$(t-4)(t+2) = 0 \implies t = 4\text{ s}$.
Velocity $v = u - gt = 9.8 - 9.8(4) = 9.8 - 39.2 = -29.4\text{ m/s}$ (downwards).

Sol: Projection $u = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = 20\text{ m/s}$.
Total flight time $T = 2u/g = 4\text{s}$. With 4 balls juggling continuously, one is thrown every $\frac{4}{4} = 1\text{s}$.
If ball 4 is leaving ($t=0$), ball 3 is at $t=1\text{s}$, ball 2 at $t=2\text{s}$, ball 1 at $t=3\text{s}$.
$h(t) = ut - \frac{1}{2}gt^2 = 20t - 5t^2$.
$h(1) = 20(1) - 5 = 15\text{m}$.
$h(2) = 20(2) - 20 = 20\text{m}$ (max height).
$h(3) = 20(3) - 45 = 15\text{m}$.

Sol: Dist in $n^{\text{th}}$ sec: $S_n = u + \frac{g}{2}(2n - 1) \implies 45 = 0 + 5(2n - 1)$.
$9 = 2n - 1 \implies 2n = 10 \implies n = 5\text{s}$ (Total time of flight).
Total height $h = \frac{1}{2}gt^2 = 5(5^2) = 125\text{ m}$.

Sol: No. At the highest point, velocity is momentarily zero, but gravity continues to pull the object down. Acceleration is strictly $g$ ($9.8\text{ m/s}^2$) downwards throughout the flight.

Sol: First phase ($5\text{s}$): $v_1 = at = 19.6 \times 5 = 98\text{ m/s}$. Height $h_1 = \frac{1}{2}at^2 = 0.5 \times 19.6 \times 25 = 245\text{ m}$.
Second phase (free fall from $v=98$ to $v=0$): $h_2 = \frac{v_1^2}{2g} = \frac{98^2}{2(19.6)} = 245\text{ m}$. (Assuming $g=9.8$, $2g=19.6$, wait, $98^2/19.6 = \frac{9604}{19.6} = 490\text{ m}$).
Total Max Height = $245 + 490 = 735\text{ m}$.

Sol: $V_A = +54\text{ km/h} = +15\text{ m/s}$ (North). $V_B = -90\text{ km/h} = -25\text{ m/s}$ (South).
$V_{BA} = V_B - V_A = -25 - 15 = -40\text{ m/s}$.
Magnitude is $40\text{ m/s}$ in South direction.

Sol: $V_A = 15\text{ m/s}$ (North). Monkey runs south wrt train: $V_{MA} = -18\text{ km/h} = -5\text{ m/s}$.
$V_{MA} = V_M - V_A \implies -5 = V_M - 15 \implies V_M = 10\text{ m/s}$ (North).

Sol: $V_P = 30\text{ km/h} = 25/3\text{ m/s}$. $V_T = 192\text{ km/h} = 160/3\text{ m/s}$.
Speed of bullet wrt ground $V_b = 150 + V_P = 150 + 25/3 = 475/3\text{ m/s}$.
Hit speed (Relative velocity of bullet wrt thief) = $V_b - V_T = 475/3 - 160/3 = 315/3 = 105\text{ m/s}$.

Sol: Total relative distance = $50 + 50 = 100\text{ m}$.
Relative speed = $10 - (-15) = 25\text{ m/s}$.
Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{100}{25} = 4\text{ s}$.

Sol: Total distance = $120\text{ m}$. They are in opposite directions.
Relative speed = $10 + 5 = 15\text{ m/s}$.
Time = $\frac{120}{15} = 8\text{ s}$.

Sol: Initial relative velocity $u_{rel} = 30 - 20 = 10\text{ m/s}$.
Relative acceleration $a_{rel} = -2 - 0 = -2\text{ m/s}^2$.
Distance required to reduce relative velocity to zero: $S_{rel} = \frac{u_{rel}^2}{2|a_{rel}|} = \frac{100}{2(2)} = 25\text{ m}$.
Since required distance $25\text{m}$ < Initial gap $240\text{m}$, they will NOT collide.
Minimum distance = $240 - 25 = 215\text{ m}$.

Sol: Relative distance $S_{rel} = 1000\text{ m}$. Time $t = 100\text{ s}$.
Required relative speed $v_{rel} = \frac{1000}{100} = 10\text{ m/s}$.
$v_{rel} = V_{scooter} - V_{bus} \implies 10 = V_{scooter} - 10 \implies V_{scooter} = 20\text{ m/s}$.

Sol: Area under v-t graph represents Displacement (with sign) or Distance (absolute area).
Slope of x-t graph represents instantaneous Velocity.

Sol: A parabolic x-t graph implies position varies quadratically with time ($x \propto t^2$). This indicates the object moves with Uniform (Constant) Acceleration.

Sol:

Sol: Distance = Area of Triangle in v-t graph.
Area = $\frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10\text{ s} \times 20\text{ m/s} = 100\text{ m}$.

Sol: Slope of x-t graph is zero (horizontal line implies position is not changing).
Therefore, Velocity = $0$. The object is at rest.

Sol: Yes, negative slope implies the particle has a negative velocity (moving towards the origin or in opposite direction).
No, a vertical line would imply the particle is in multiple positions at the exact same instant, equating to infinite velocity, which is physically impossible.

Sol: Base is $t=0$ to $t=4\text{s}$, so diameter is $4$, giving a radius $r=2$. The peak velocity (radius on v-axis) is $2\text{ m/s}$.
Displacement = Area of Semi-circle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2)^2 = \frac{1}{2} \pi (4) = 2\pi\text{ m}$.