Q.Two particles are projected in air with speed $v_o$ at angles $\theta_1$ and $\theta_2$ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices (Note: more than one of the given options may be correct.)

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A point object is at rest on the principal axis at a distance of 20 cm from a concave mirror of radius of curvature 30 cm. Under the action of a horizontal force, the object moves away from the mirror along the axis and reaches a distance of 40 cm from the mirror in a time of 3 s. The acceleration of the image is (A) $8\ \mathrm{cm\,s^{-2}}$ (B) $4\ \mathrm{cm\,s^{-2}}$ (C) $16\ \mathrm{cm\,s^{-2}}$ (D) $12\ \mathrm{cm\,s^{-2}}$

   ›Reveal solutionSolution

   The image moves from $60$ cm to $24$ cm (a $36$ cm displacement) starting from rest in $3$ s, so $a=\dfrac{2s}{t^2}=8~\mathrm{cm\,s^{-2}}$.

   Concave mirror: $R=30$ cm $\Rightarrow f=15$ cm. Using $\dfrac1v+\dfrac1u=\dfrac1f$ (magnitudes), $v=\dfrac{uf}{u-f}$.

   • At $u=20$ cm: $v=\dfrac{20\times15}{20-15}=60$ cm.
   • At $u=40$ cm: $v=\dfrac{40\times15}{40-15}=24$ cm.

   The object starts from rest (moved by a force from rest), so the image also starts from rest. Its displacement over the $3$ s interval:

   $s_{\text{image}}=|60-24|=36~\text{cm}.$ …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.A block slides down a rough inclined plane with a constant velocity $9.8\,\mathrm{ms}^{-1}$. The coefficient of kinetic friction between the block and the surface is $\frac{1}{\sqrt{3}}$. If the block is pushed up along the inclined plane from bottom of the plane with a velocity $9.8\,\mathrm{ms}^{-1}$, then the distance travelled by the block before coming to rest is (A) $9.80$ m (B) $2.45$ m (C) $14.75$ m (D) $4.90$ m

   ›Reveal solutionSolution

   The block slides down at constant speed, so the net force is zero — friction balances the downhill component of gravity. Using that to find the incline angle, then applying the work–energy theorem for the upward motion gives the stopping distance as $4.90$ m.

   The key insight is that constant velocity down the incline means no net acceleration — the downhill gravitational pull is exactly cancelled by kinetic friction. That lets us find the angle of the incline. Then, when the block is pushed upward with the same initial speed, friction now acts downhill (opposing the motion), so both gravity and friction work together to slow it down. The work–energy theorem gives the stopping distance directly.

   ---

   1. Find the incline angle from the downward motion. For a block sliding down at constant speed, the net force along the incline is zero:

   $mg\sin\theta = \mu_k mg\cos\theta$

   Cancel $mg$:

   $\sin\theta = \mu_k \cos\theta \quad\Rightarrow\quad \tan\theta = \mu_k$

   Given $\mu_k = \frac{1}{\sqrt{3}}$, we get:

   $\tan\theta = \frac{1}{\sqrt{3}} \quad\Rightarrow\quad \theta = 30^\circ$

   2. Set up the upward motion. When the block is pushed up with initial speed $u = 9.8\ \mathrm{m/s}$, friction acts down the incline (opposing the upward motion), and gravity also pulls it down the incline. The net decelerating force along the incline is:

   $F_{\text{net}} = mg\sin\theta + \mu_k mg\cos\theta$

   Using $\theta = 30^\circ$:

   $\sin30^\circ = \frac12,\quad \cos30^\circ = \frac{\sqrt3}{2}$

   So:

   $F_{\text{net}} = mg\left(\frac12\right) + \frac{1}{\sqrt3}\,mg\left(\frac{\sqrt3}{2}\right) = mg\left(\frac12 + \frac12\right) = mg$ …

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3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Two bodies A and B separated by some distance on the ground are projected simultaneously in opposite directions towards each other. Body B is projected with a velocity of $20\ \mathrm{ms}^{-1}$ at an angle of $45^\circ$ with the horizontal and body A is projected at an angle of $30^\circ$ with the horizontal. If the two bodies collide at a time of $\sqrt{2}\ \mathrm{s}$ from the beginning of their motion, then the initial distance between the bodies A and B is nearly (A) $24.12\ \mathrm{m}$ (B) $48.24\ \mathrm{m}$ (C) $27.32\ \mathrm{m}$ (D) $54.64\ \mathrm{m}$

   ›Reveal solutionSolution

   For two bodies projected simultaneously from the ground to collide, their initial vertical velocity components must be equal. The initial separation is the sum of the horizontal distances each body travels until collision. The initial distance between the bodies A and B is $\boxed{54.64\ \mathrm{m}}$.

   The core idea behind solving this problem is to understand that projectile motion can be broken down into independent horizontal and vertical components. For two bodies projected towards each other to collide, they must reach the same horizontal position and the same vertical position at the exact same time.

   Since both bodies are projected from the ground, their initial vertical positions are identical. Therefore, for them to collide, their vertical displacements from the ground must be equal at the time of collision. This condition will allow us to find the unknown initial velocity of body A. The horizontal motion, being uniform (constant velocity), will then allow us to calculate the distance each body covers horizontally. The sum of these horizontal distances will be the initial separation.

   1. Determine the initial velocity of body A ($v_A$) using the condition for vertical collision.

      For the two bodies to collide, they must be at the same height at the collision time $t = \sqrt{2}\ \mathrm{s}$. The vertical displacement for a projectile launched from the ground is given by $y = (v_0 \sin \theta)t - \frac{1}{2}gt^2$.

      For body A, the vertical displacement is $y_A = (v_A \sin \theta_A)t - \frac{1}{2}gt^2$.

      For body B, the vertical displacement is $y_B = (v_B \sin \theta_B)t - \frac{1}{2}gt^2$.

      Since $y_A = y_B$ at the collision time $t$:

      $(v_A \sin \theta_A)t - \frac{1}{2}gt^2 = (v_B \sin \theta_B)t - \frac{1}{2}gt^2$

      The term $-\frac{1}{2}gt^2$ cancels out on both sides, leading to:

      $v_A \sin \theta_A = v_B \sin \theta_B$

      This means that for two projectiles launched simultaneously from the same horizontal level to collide, their initial vertical components of velocity must be equal.

      We are given $v_B = 20\ \mathrm{ms}^{-1}$, $\theta_B = 45^\circ$, and $\theta_A = 30^\circ$.

      Substitute these values:

      $v_A \sin 30^\circ = 20 \sin 45^\circ$

      $v_A \left(\frac{1}{2}\right) = 20 \left(\frac{1}{\sqrt{2}}\right)$

      $v_A = 2 \times \frac{20}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2}\ \mathrm{ms}^{-1}$

   2. Calculate the horizontal distance covered by body B ($x_B$).

      The horizontal component of velocity for body B is $v_{Bx} = v_B \cos \theta_B$.

      $v_{Bx} = 20 \cos 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}\ \mathrm{ms}^{-1}$

      The horizontal distance covered by body B until collision at $t = \sqrt{2}\ \mathrm{s}$ is: …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If a body is projected vertically from the surface of the earth with a speed of $8000~\mathrm{ms}^{-1}$, then the maximum height reached by the body is (Radius of the earth = $6400~\mathrm{km}$ and acceleration due to gravity = $10~\mathrm{ms}^{-2}$) (A) 1600 km (B) 9600 km (C) 6400 km (D) 3200 km

   ›Reveal solutionSolution

   The key idea is that for a projectile launched at near-orbital speed, the usual constant‑g formula fails because gravity weakens with height; using energy conservation (kinetic + gravitational potential) gives the true maximum height. The correct answer is (D) 3200 km.

   ---

   Concept and intuition

   If you throw a ball upward at 8000 m/s, you might be tempted to use $v^2 = u^2 - 2gh$ with $g = 10~\mathrm{m/s^2}$. That would give $h = u^2/(2g) = (8000)^2/20 = 3.2\times 10^6~\mathrm{m} = 3200~\mathrm{km}$. That’s exactly option (D). But is that correct?

   The catch: $g = 10~\mathrm{m/s^2}$ is only valid near Earth’s surface. At a height of 3200 km (half Earth’s radius), gravity is much weaker. The constant‑g formula overestimates the work gravity does, so the true height would be larger than 3200 km if we used constant g. Wait — that suggests (D) might be too small. But let’s check: 8000 m/s is actually close to the orbital speed near Earth’s surface (about 7900 m/s). So the body is almost in orbit; it will go very high indeed. Let’s compute properly.

   ---

   Step‑by‑step solution

   1. Recognize that gravity is not constant The acceleration due to gravity varies with distance $r$ from Earth’s centre as

   $g(r) = \frac{GM}{r^2}$

   where $GM = g R^2$ with $R = 6400~\mathrm{km} = 6.4\times 10^6~\mathrm{m}$ and $g = 10~\mathrm{m/s^2}$.

   So $GM = 10 \times (6.4\times 10^6)^2 = 4.096\times 10^{14}~\mathrm{m^3/s^2}$.

   2. Use conservation of mechanical energy At launch (surface, $r = R$), speed $u = 8000~\mathrm{m/s}$. At maximum height, speed = 0, distance from centre = $r_{\max}$. Energy conservation:

   $\frac{1}{2} m u^2 - \frac{GMm}{R} = 0 - \frac{GMm}{r_{\max}}$

   Cancel $m$:

   $\frac{1}{2} u^2 - \frac{GM}{R} = -\frac{GM}{r_{\max}}$

   3. Solve for $r_{\max}$ Rearranging:

   $\frac{GM}{r_{\max}} = \frac{GM}{R} - \frac{1}{2} u^2$

   $\frac{1}{r_{\max}} = \frac{1}{R} - \frac{u^2}{2GM}$

   Plug numbers:

   $\frac{1}{R} = \frac{1}{6.4\times 10^6} \approx 1.5625\times 10^{-7}~\mathrm{m^{-1}}$

   $\frac{u^2}{2GM} = \frac{(8000)^2}{2 \times 4.096\times 10^{14}} = \frac{6.4\times 10^7}{8.192\times 10^{14}} \approx 7.8125\times 10^{-8}~\mathrm{m^{-1}}$

   So

   $\frac{1}{r_{\max}} = 1.5625\times 10^{-7} - 7.8125\times 10^{-8} = 7.8125\times 10^{-8}~\mathrm{m^{-1}}$

   Hence

   $r_{\max} = \frac{1}{7.8125\times 10^{-8}} \approx 1.28\times 10^7~\mathrm{m} = 12800~\mathrm{km}$

   4. Find height above surface Height $h = r_{\max} - R = 12800 - 6400 = 6400~\mathrm{km}$. …

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5. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.A body is falling freely under gravity from certain height from the ground. If the sum of the displacements of the body in the second and third seconds of its motion is 32% of the height from which it is falling, then the speed with which the body hits the ground is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $25~\mathrm{ms}^{-1}$ (B) $50~\mathrm{ms}^{-1}$ (C) $100~\mathrm{ms}^{-1}$ (D) $75~\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The key is to express the sum of displacements in the 2nd and 3rd seconds in terms of the total height $H$, then solve for $H$ and finally the impact speed. The answer is $50~\mathrm{ms}^{-1}$.

   When a body falls freely from rest under gravity, the distance travelled in the $n$th second is given by $s_n = u + \frac{g}{2}(2n-1)$, but here $u=0$, so $s_n = \frac{g}{2}(2n-1)$. This formula comes directly from subtracting the distance covered in $(n-1)$ seconds from that in $n$ seconds. The problem gives a relation between the sum of two such displacements and the total height — that lets us find the height, and from it the final velocity.

   1. Displacements in the 2nd and 3rd seconds

      For $g = 10~\mathrm{ms}^{-2}$:

      $s_2 = \frac{10}{2}(2 \times 2 - 1) = 5 \times 3 = 15~\mathrm{m}$

      $s_3 = \frac{10}{2}(2 \times 3 - 1) = 5 \times 5 = 25~\mathrm{m}$

      Their sum: $s_2 + s_3 = 15 + 25 = 40~\mathrm{m}$.

   2. Relating to the total height

      Let the total height from which the body falls be $H$ metres. The problem states:

      $40 = \frac{32}{100} H$

      So $H = \frac{40 \times 100}{32} = \frac{4000}{32} = 125~\mathrm{m}$.

   3. Speed on hitting the ground

      For free fall from rest, $v^2 = 2gH$.

      $v^2 = 2 \times 10 \times 125 = 2500$

      $v = \sqrt{2500} = 50~\mathrm{ms}^{-1}$. …

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6. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.A body of mass 50 g is projected vertically upwards from the ground. If the kinetic energy of the body at a height of 32 m is 100% more than its potential energy, then the height at which the potential energy of the body becomes 60% of its kinetic energy is (A) 24 m (B) 72 m (C) 18 m (D) 36 m

   ›Reveal solutionSolution

   Use conservation of mechanical energy and the given ratio of kinetic to potential energy at 32 m to find the total energy, then find the height where PE = 0.6 × KE. The answer is 36 m.

   The key idea is that for a body moving under gravity alone (no air resistance), mechanical energy — the sum of kinetic energy (KE) and potential energy (PE) — stays constant. The problem gives you a relationship between KE and PE at one height, which lets you pin down the total energy. Then you can use that total to find the height where a different ratio holds.

   Let’s work through it step by step.

   1. Set up the variables and convert units.

      Mass $m = 50 \text{ g} = 0.05 \text{ kg}$. But notice: in energy calculations, mass will cancel out because both KE and PE are proportional to $m$. So we can work symbolically without plugging in the number until the very end — or just keep $m$ as a symbol.

   2. Interpret the condition at height $h_1 = 32 \text{ m}$.

      The problem says: “the kinetic energy at 32 m is 100% more than its potential energy.”

      “100% more than” means KE is twice the PE. If PE at that height is $mgh_1$, then

   $KE_1 = 2 \times PE_1 = 2 m g h_1.$

   3. Find the total mechanical energy $E$. At any point, $E = KE + PE$. So at $h_1 = 32 \text{ m}$:

   $E = KE_1 + PE_1 = 2 m g h_1 + m g h_1 = 3 m g h_1.$

   Substituting $h_1 = 32 \text{ m}$ and $g = 10 \text{ m/s}^2$ (standard for such problems):

   $E = 3 m \times 10 \times 32 = 960 m \text{ J}.$

   Tip

   The mass $m$ is in kg, but since it cancels later, you can treat $E$ as $960m$ and proceed.

   4. Now consider the second condition: find height $h_2$ where PE is 60% of KE. That is: $PE_2 = 0.6 \times KE_2.$ …

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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.A body P is projected at an angle of $30^\circ$ with the horizontal and another body Q is projected at angle of $30^\circ$ with the vertical. If the ratio of the horizontal ranges of the bodies P and Q is $1:2$, then the ratio of the maximum heights reached by the bodies P and Q is (A) $1:4$ (B) $1:6$ (C) $2:3$ (D) $1:1$

   ›Reveal solutionSolution

   The key is to interpret the projection angles correctly: P is at $30^\circ$ to the horizontal, Q at $30^\circ$ to the vertical (so $60^\circ$ to the horizontal). Using the range ratio $1:2$ to find the speed ratio, then computing the height ratio gives $1:6$, so option (B).

   Concept & Intuition

   Both projectiles follow parabolic paths under gravity. The horizontal range depends on the initial speed and the sine of twice the launch angle; the maximum height depends on the square of the vertical component of velocity. The trick here is that Q’s angle is given relative to the vertical, not the horizontal — a common source of error. Once we convert Q’s angle to the standard horizontal measure, we can write the range ratio, solve for the speed ratio, and then find the height ratio.

   ---

   1. Define the angles correctly

      • For P: angle with horizontal $\theta_P = 30^\circ$.
      • For Q: angle with vertical is $30^\circ$, so angle with horizontal $\theta_Q = 90^\circ - 30^\circ = 60^\circ$.

   2. Write the horizontal range formulas

      For a projectile launched with speed $u$ at angle $\theta$ to the horizontal,

   $R = \frac{u^2 \sin 2\theta}{g}.$

   Hence:

   $R_P = \frac{u_P^2 \sin 60^\circ}{g} = \frac{u_P^2 \cdot \frac{\sqrt{3}}{2}}{g}, \qquad R_Q = \frac{u_Q^2 \sin 120^\circ}{g} = \frac{u_Q^2 \cdot \frac{\sqrt{3}}{2}}{g}.$

   (Note: $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$.)

   3. Use the given range ratio

   $\frac{R_P}{R_Q} = \frac{1}{2} \quad\Rightarrow\quad \frac{\frac{u_P^2 \sqrt{3}}{2g}}{\frac{u_Q^2 \sqrt{3}}{2g}} = \frac{u_P^2}{u_Q^2} = \frac{1}{2}.$

   So

   $\frac{u_P}{u_Q} = \frac{1}{\sqrt{2}}.$

   4. Write the maximum height formulas Maximum height:

   $H = \frac{u^2 \sin^2\theta}{2g}.$

   For P: $\sin 30^\circ = \frac{1}{2}$, so

   $H_P = \frac{u_P^2 \cdot \left(\frac{1}{2}\right)^2}{2g} = \frac{u_P^2}{8g}.$

   For Q: $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so

   $$$$

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.A solid sphere rolls down without slipping from the top of an inclined plane of height 28 m and angle of inclination $30^\circ$. The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity $= 10\, \mathrm{ms}^{-2}$) (A) $20\, \mathrm{ms}^{-1}$ (B) $28\, \mathrm{ms}^{-1}$ (C) $10\, \mathrm{ms}^{-1}$ (D) $14\, \mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   For a rolling rigid body, gravitational potential energy converts into both translational and rotational kinetic energy. Using conservation of energy and the rolling condition $v = \omega R$, the final speed is $v = \sqrt{\frac{10}{7} g h}$. With $g = 10\, \mathrm{m/s^2}$ and $h = 28\, \mathrm{m}$, we get $v = 20\, \mathrm{m/s}$. The correct option is (A).

   ---

   Concept and intuition

   When a sphere rolls without slipping, its motion is a combination of translation of the center of mass and rotation about the center. Friction does no net work because the point of contact is instantaneously at rest — so mechanical energy is conserved. The key is that the kinetic energy has two parts: $\frac{1}{2} m v^2$ for translation and $\frac{1}{2} I \omega^2$ for rotation. For a solid sphere, the moment of inertia is $I = \frac{2}{5} m R^2$. The no-slip condition links $\omega$ and $v$ by $\omega = v / R$. This lets us write the total kinetic energy as a single term proportional to $v^2$, and then equate it to the loss in gravitational potential energy $mgh$.

   ---

   Step-by-step solution

   1. Set up conservation of energy The sphere starts from rest at height $h = 28\, \mathrm{m}$. At the bottom, all potential energy has become kinetic energy (translational + rotational).

   $mgh = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$

   2. Insert the moment of inertia and rolling condition For a solid sphere, $I = \frac{2}{5} m R^2$. Rolling without slipping gives $\omega = v / R$. Substitute:

   $mgh = \frac{1}{2} m v^2 + \frac{1}{2} \left( \frac{2}{5} m R^2 \right) \left( \frac{v}{R} \right)^2$

   3. Simplify the rotational term

   $\frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2} = \frac{1}{5} m v^2$

   So the energy equation becomes:

   $mgh = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2$

   4. Solve for $v$ Cancel $m$ (mass doesn’t matter) and rearrange: $$$$

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9. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Pebbles are dropped freely for every half second into a river from a bridge above it. When the first pebble is about to strike the surface of the water, the fifth pebble is dropped. Then the height from the surface of the water from which the pebbles are dropped is (Acceleration due to gravity $= 10\, \mathrm{ms}^{-2}$) (A) 20 m (B) 31.25 m (C) 125 m (D) 80 m

   ›Reveal solutionSolution

   The key is that the first pebble has been falling for 2 seconds when the fifth is dropped, so the height is the distance fallen in that time: $h = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 10 \cdot 4 = 20$ m. The correct option is (A).

   The problem describes pebbles dropped at regular intervals of half a second. The crucial insight is that the first pebble has been falling for a certain total time by the moment the fifth pebble is released. That total time determines the height of the bridge above the water.

   Why this approach works:

   If pebbles are dropped every 0.5 s, then when the fifth pebble is just released, the first pebble has already been falling for the time interval between the first and fifth drops. That interval is four half-second gaps, so $4 \times 0.5 = 2$ seconds. The distance fallen under constant gravity from rest is given by $s = \frac{1}{2} g t^2$. So the height is simply the distance the first pebble has fallen in those 2 seconds.

   Step-by-step reasoning:

   1. Determine the time gap between the first and fifth drops.

      The drops occur every 0.5 seconds. From the first drop to the fifth drop, there are $5 - 1 = 4$ intervals.

      Total time elapsed = $4 \times 0.5 = 2$ seconds.

   2. Interpret the condition.

      The problem says: "When the first pebble is about to strike the surface of the water, the fifth pebble is dropped." This means that at the exact moment the first pebble hits the water, the fifth pebble is released. Therefore, the first pebble has been falling for exactly 2 seconds.

   3. Apply the free-fall distance formula.

      For an object dropped from rest, the distance fallen in time $t$ is:

      $h = \frac{1}{2} g t^2$ …

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10. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.A cannon fires two similar shells one after the other each with a velocity of $100~\mathrm{ms}^{-1}$ at angles $60^\circ$ and $30^\circ$ respectively with the horizontal such that they both hit the target at the same time. The time interval between the firing of the two shells is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $6.57~\mathrm{s}$ (B) $7.32~\mathrm{s}$ (C) $8.28~\mathrm{s}$ (D) $3.14~\mathrm{s}$

    ›Reveal solutionSolution

    Time interval $=T_{60^\circ}-T_{30^\circ}=17.32-10=7.32$ s.

    Concept: Projectiles launched at complementary angles (30° and 60°) with the same speed have equal horizontal range, so both can hit the same target. Their times of flight differ, with $T=\dfrac{2u\sin\theta}{g}$.

    Calculation ($u=100$ m/s, $g=10$ m/s$^2$):

    • $T_{60^\circ}=\dfrac{2\times100\times\sin60^\circ}{10}=20\times0.866=17.32$ s
    • $T_{30^\circ}=\dfrac{2\times100\times\sin30^\circ}{10}=20\times0.5=10$ s …

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The ratio of the displacements of a freely falling body during first, second and third seconds of its motion is (A) $1:1:1$ (B) $1:3:5$ (C) $1:2:3$ (D) $1:4:9$

    ›Reveal solutionSolution

    For a freely falling body starting from rest, the distances covered in successive seconds follow the odd-number ratio $1:3:5$, so the correct option is (B).

    The key concept here is uniformly accelerated motion under gravity. When an object falls freely from rest, its velocity increases linearly with time because the acceleration $g$ is constant. The distance traveled in any given second is not the same as the total distance fallen; it’s the difference between the total distances at the end of that second and the end of the previous second. That difference turns out to follow a simple pattern: the odd numbers.

    Let’s work it out step by step.

    1. Recall the formula for distance under constant acceleration from rest. For a body starting from rest ($u = 0$) with constant acceleration $a = g$, the distance fallen in time $t$ is

    $s(t) = \frac{1}{2} g t^2.$

    2. Find the distance covered during the first second. This is simply the total distance fallen from $t=0$ to $t=1$:

    $s_1 = s(1) - s(0) = \frac{1}{2} g (1)^2 - 0 = \frac{1}{2} g.$

    3. Find the distance covered during the second second. This is the distance from $t=1$ to $t=2$:

    $s_2 = s(2) - s(1) = \frac{1}{2} g (2)^2 - \frac{1}{2} g (1)^2 = \frac{1}{2} g (4 - 1) = \frac{3}{2} g.$

    4. Find the distance covered during the third second. This is the distance from $t=2$ to $t=3$:

    $s_3 = s(3) - s(2) = \frac{1}{2} g (3)^2 - \frac{1}{2} g (2)^2 = \frac{1}{2} g (9 - 4) = \frac{5}{2} g.$

    5. Form the ratio. …

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12. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
    Q.The relationship between the final velocity $v$ and the distance $x$ travelled by a bus moving with uniform acceleration is $v = \sqrt{256 - 10x}$. The acceleration of the bus is (All quantities are given in SI units) (A) $-5\ \mathrm{m/s^2}$ (B) $-10\ \mathrm{m/s^2}$ (C) $-20\ \mathrm{m/s^2}$ (D) $-15\ \mathrm{m/s^2}$

    ›Reveal solutionSolution

    The given relation $v = \sqrt{256 - 10x}$ is a velocity–distance equation. By comparing it with the standard kinematic formula $v^2 = u^2 + 2ax$, we find the acceleration is $-5\ \mathrm{m/s^2}$, so the correct option is (A).

    We are told the bus moves with uniform acceleration, so the standard kinematic equations apply. The relation given is $v = \sqrt{256 - 10x}$, where $v$ is final velocity and $x$ is distance travelled. Squaring both sides gives $v^2 = 256 - 10x$. This looks very much like the familiar equation $v^2 = u^2 + 2ax$, where $u$ is initial velocity and $a$ is constant acceleration. The trick is to match the two forms term by term.

    1. Square the given relation

    $v = \sqrt{256 - 10x} \quad\Rightarrow\quad v^2 = 256 - 10x.$

    2. Recall the standard kinematic equation for motion under uniform acceleration:

    $v^2 = u^2 + 2ax.$

    Here $u$ is the initial velocity (when $x = 0$), and $a$ is the constant acceleration.

    3. Compare the two expressions
       • In $v^2 = 256 - 10x$, the constant term is $256$. In the standard form, the constant term is $u^2$. So we identify:

    $u^2 = 256 \quad\Rightarrow\quad u = 16\ \mathrm{m/s} \ (\text{since velocity is positive}).$

    • The coefficient of $x$ in the given equation is $-10$. In the standard form, the coefficient of $x$ is $2a$. Therefore: $2a = -10 \quad\Rightarrow\quad a = -5\ \mathrm{m/s^2}.$ …

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