Q.A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground, (Note: more than one of the given options may be correct.)

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.To a person going towards east in a car with a velocity of 25 kmph, a train appears to move towards north with a velocity of $25\sqrt{3}$ kmph. The actual velocity of the train will be (A) 25 kmph (B) 50 kmph (C) 5 kmph (D) 53 kmph

   ›Reveal solutionSolution

   Adding the car's eastward velocity to the train's northward relative velocity (as perpendicular vector components) gives the train's actual speed of 50 kmph.

   Concept and Intuition

   Relative velocity of the train with respect to the car, $\vec v_{TC}=\vec v_T-\vec v_C$, is given as pointing due north. To recover the train's actual (ground-frame) velocity $\vec v_T$, we simply add back the car's velocity: $\vec v_T=\vec v_{TC}+\vec v_C$. Since these two components are perpendicular (east vs north), the resultant magnitude follows from Pythagoras.

   Step-by-Step Solution

   1. Take east as the $x$-axis, north as the $y$-axis.
   2. Car's velocity: $\vec v_C=(25,0)$ kmph (east).
   3. Train's velocity relative to car: $\vec v_{TC}=(0,25\sqrt3)$ kmph (north), as observed by the person in the car.
   4. Actual train velocity: $\vec v_T=\vec v_{TC}+\vec v_C=(25,\,25\sqrt3)$ kmph. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A swimmer can swim at $5\ ms^{-1}$ in still water. River flows at $3\ ms^{-1}$. To cross the river in shortest time, the angle with respect to perpendicular to the flow is (A) $0^\circ$ (B) $\sin^{-1}(3/5)$ (C) $\tan^{-1}(3/5)$ (D) $90^\circ$

   ›Reveal solutionSolution

   Classic river-crossing problem: minimum-TIME crossing (as opposed to minimum-drift crossing) requires heading straight across; answer is $0^\circ$.

   Concept and Intuition

   The time to cross the river depends ONLY on the component of the swimmer's velocity that is perpendicular to the banks (i.e., directly across the river), since that's the component that closes the river's width. The river current is parallel to the banks and only shifts the swimmer downstream — it never helps or hurts the crossing time itself.

   Step-by-Step Solution

   1. Let the river width be $d$ and the swimmer's speed relative to water be $v = 5\ ms^{-1}$, at angle $\theta$ from the perpendicular to the flow.
   2. The perpendicular (cross-river) component of the swimmer's velocity is $v\cos\theta$.
   3. Time to cross: $t = \dfrac{d}{v\cos\theta}$.
   4. This is minimized when $\cos\theta$ is maximum, i.e., $\cos\theta = 1 \Rightarrow \theta = 0^\circ$. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A projectile is projected from a moving truck. Its range depends on (A) Truck velocity only (B) Projectile velocity with respect to truck only (C) Projectile velocity with respect to ground (D) Projectile mass

   ›Reveal solutionSolution

   This tests the frame-dependence of projectile range: range is always calculated using the velocity relative to the ground (the frame in which the range/landing point is measured). The answer is (C).

   Concept and Intuition

   Projectile motion equations (such as $R=\dfrac{u^2\sin2\theta}{g}$) always use the launch velocity as measured in the frame where the trajectory and landing point are being described — normally the ground frame. If the projectile is launched from a moving platform (like a truck), its velocity relative to the ground is the vector sum of the truck's velocity and the projectile's velocity relative to the truck. It is this ground-frame velocity that determines how far it travels, relative to the ground, before landing.

   Step-by-Step Solution

   1. Range is a distance measured on the ground, so it must be computed using motion as observed from the ground frame.
   2. If the truck moves with velocity $\vec{v}_{truck}$ and the projectile is launched with velocity $\vec{v}_{rel}$ relative to the truck, then by Galilean velocity addition the projectile's velocity relative to ground is $\vec{v}_{ground}=\vec{v}_{truck}+\vec{v}_{rel}$.
   3. Using $\vec{v}_{ground}$'s horizontal and vertical components in the standard projectile equations gives the correct range as seen from the ground. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Rain is falling vertically with a speed of $2\ ms^{-1}$. A boy from rest starts moving with a constant acceleration of $2\ ms^{-2}$ along a straight road holding umbrella. For the rain to be always parallel to the axis of umbrella, the rate at which the angle of axis of umbrella with the vertical should be changed with time '$t$' is (A) $\dfrac{1+t^2}{t^3}$ (B) $\dfrac{1}{1+t^2}$ (C) $\dfrac{t^2}{1+t^4}$ (D) $\dfrac{t}{1+t^3}$

   ›Reveal solutionSolution

   The umbrella must always point along the rain's velocity relative to the boy; as the boy speeds up this relative-velocity direction tilts, and $\theta=\tan^{-1}t$ gives $d\theta/dt=1/(1+t^2)$.

   Concept and Intuition

   Rain "appears" to come from a different direction to a moving observer — this is relative velocity. The umbrella must be tilted so its axis is parallel to the rain's velocity as seen by the boy (not as seen from the ground), so we need $\vec v_{rain,ground}-\vec v_{boy,ground}$.

   Step-by-Step Solution

   1. Rain velocity (ground frame): vertically down, magnitude 2 m/s, i.e. $\vec v_r=(0,-2)$.
   2. Boy starts from rest with constant acceleration $2\,ms^{-2}$ along the road (x-direction): $\vec v_{boy}=(2t,0)$.
   3. Velocity of rain relative to boy: $\vec v_{rel}=\vec v_r-\vec v_{boy}=(0-2t,\,-2-0)=(-2t,-2)$.
   4. The umbrella axis must be along $\vec v_{rel}$. The angle $\theta$ this makes with the vertical satisfies $\tan\theta=\dfrac{|\text{horizontal component}|}{|\text{vertical component}|}=\dfrac{2t}{2}=t$.
   5. So $\theta=\tan^{-1}(t)$. …

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5. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A motor boat covers a given distance in 6 hours moving down stream in a river. It covers the same distance in 10 hours moving upstream. The time it takes to cover the same distance in still water is (A) 6.5 hours (B) 8 hours (C) 9 hours (D) 7.5 hours

   ›Reveal solutionSolution

   Setting up downstream/upstream distance equations and solving for boat speed

   in terms of water speed gives a still-water time of 7.5 hours. Answer: (D).

   Concept and Intuition

   For boats in a river: downstream speed $=b+w$, upstream speed $=b-w$, where

   $b$ is the boat's speed in still water and $w$ is the river's current speed.

   The distance $d$ travelled is the same in all three scenarios (downstream,

   upstream, still water) since the boat covers the "same distance" each time —

   only the effective speed differs.

   Step-by-Step Solution

   1. Downstream: $d = (b+w)\times 6$.
   2. Upstream: $d = (b-w)\times 10$.
   3. Equate: $6(b+w) = 10(b-w) \Rightarrow 6b + 6w = 10b - 10w \Rightarrow 16w = 4b \Rightarrow b = 4w$. …

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6. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A police van is moving on a highway with a speed of 36 kmph and a thief's car is speeding away in the same direction with 162 kmph. The police fired a bullet on thief's car. If the muzzle speed of bullet is 150 $\mathrm{ms^{-1}}$, then the speed with which bullet hits thief's car is (A) 145 $\mathrm{ms^{-1}}$ (B) 130 $\mathrm{ms^{-1}}$ (C) 115 $\mathrm{ms^{-1}}$ (D) 105 $\mathrm{ms^{-1}}$

   ›Reveal solutionSolution

   Add the van's velocity to the muzzle (relative) velocity to get the bullet's ground velocity, then subtract the thief's car's velocity to get the bullet's velocity relative to the target. Answer: 115 m/s.

   Concept and Intuition

   This is a straightforward relative-velocity addition problem. The "muzzle speed" of a bullet is always specified relative to the gun (and hence the vehicle carrying it), not relative to the ground. To find how fast the bullet actually strikes a moving target, we must work entirely in one frame (the ground frame), converting every given velocity into that frame first.

   Step-by-Step Solution

   1. Convert all speeds to m/s: police van $= 36 \times \frac{5}{18} = 10\ \mathrm{ms^{-1}}$; thief's car $=162\times\frac{5}{18}=45\ \mathrm{ms^{-1}}$.
   2. The bullet's muzzle speed (150 m/s) is relative to the van. Since the van moves in the same direction the bullet is fired, the bullet's speed relative to the ground is: $v_{bullet,ground} = 150 + 10 = 160\ \mathrm{ms^{-1}}$ …

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7. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A man standing on a road has to hold his umbrella at $30°$ with the vertical to keep the rain away. He throws the umbrella and starts running at 10 kmph. He finds that rain drops are hitting his head vertically. What is the speed of rain with respect to ground in kmph? (A) 10 (B) 20 (C) $\dfrac{20}{\sqrt{3}}$ (D) $\dfrac{10}{\sqrt{3}}$

   ›Reveal solutionSolution

   Using relative velocity of rain with respect to the man in the two situations (standing, then running), the true (ground-frame) speed of the rain works out to 20 kmph.

   Concept and Intuition

   Rain has a fixed true velocity with respect to the ground, with both a vertical component (due to it falling) and a horizontal component (e.g., due to wind). A stationary observer sees the rain coming from the direction of its net (relative = actual, since observer's velocity is zero) velocity vector. A moving observer instead sees the relative velocity of rain with respect to themselves — found by vector-subtracting their own velocity from the rain's true velocity. Both scenarios in this problem give us two equations to pin down the true rain velocity.

   Step-by-Step Solution

   1. Let the rain's true velocity relative to the ground have a horizontal component $v_x$ (in the direction the man will later run against) and vertical (downward) component $v_y$.
   2. Standing still: the observer's velocity is zero, so the relative velocity of rain = true velocity of rain. He must tilt the umbrella $30^\circ$ from the vertical to block it, meaning the rain vector itself makes $30^\circ$ with vertical:

   $\tan30^\circ = \frac{v_x}{v_y} \implies v_x = v_y\tan30^\circ$

   3. Running at 10 kmph: he runs in the direction that opposes the horizontal component of the rain. The relative velocity of rain w.r.t. him now has horizontal component $v_x - u$ (where $u=10$ kmph is his speed). He observes the rain falling vertically, meaning this horizontal component is zero: …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.A person walks up an escalator at rest in a time of 60 s. When standing on the same escalator now moving with a uniform speed, he is carried up in a time of 30 s. The time taken for him to walk up the moving escalator is (A) 10 s (B) 15 s (C) 20 s (D) 25 s

   ›Reveal solutionSolution

   This tests relative-rates reasoning for combined motion (walking + escalator). The answer is (C) 20 s.

   Concept and Intuition

   When two independent motions act together in the same direction (a person walking, and an escalator moving), their rates (distance covered per unit time) simply add, because each contributes independently to covering the same total distance. This is the same logic used in "pipes filling a tank" or "two workers finishing a job" problems — express each scenario as a rate, then combine rates when both act simultaneously.

   Step-by-Step Solution

   1. Let the length of the escalator (in whatever units, say number of steps or metres) be $D$.
   2. Walking up a stationary escalator takes 60 s, so the person's walking rate relative to the escalator is $D/60$ per second.
   3. Standing still on the moving escalator takes 30 s, so the escalator's own rate is $D/30$ per second. …

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9. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.Rain is falling vertically with a speed of $36 \mathrm{~m~s^{-1}}$ and a boy is moving on a bicycle with a speed of $20 \mathrm{~m~s^{-1}}$ towards south. The angle with the vertical towards south with which the boy has to hold an umbrella is (A) $\tan^{-1}\left(\dfrac{5}{9}\right)$ (B) $\sin^{-1}\left(\dfrac{5}{9}\right)$ (C) $\cos^{-1}\left(\dfrac{5}{9}\right)$ (D) $\cot^{-1}\left(\dfrac{5}{9}\right)$

   ›Reveal solutionSolution

   Using relative velocity of rain with respect to the moving boy, the umbrella must be

   tilted toward south from the vertical by $\tan^{-1}(5/9)$, obtained from the ratio of

   the boy's speed to the rain's fall speed. Answer: (A).

   Concept and Intuition

   To find how a person should orient an umbrella while moving through vertically falling

   rain, we need the velocity of the rain relative to the person, since that is the

   direction from which the rain actually appears to approach the moving observer. This is

   computed by vector subtraction: $\vec v_{\text{rain, rel to man}} = \vec v_{\text{rain}} - \vec v_{\text{man}}$. Since the man's velocity is horizontal and the rain's velocity is

   purely vertical, these two components remain perpendicular in the relative-velocity

   vector, and simple trigonometry (via a right triangle formed by the vertical fall speed

   and horizontal component) gives the tilt angle from the vertical.

   Step-by-Step Solution

   1. Set up components: rain velocity is vertical, magnitude 36 m/s downward, with no horizontal component. The boy's velocity is horizontal, magnitude 20 m/s, directed south.
   2. Relative velocity of rain with respect to the boy: subtract the boy's velocity vector from the rain's velocity vector. Since the boy moves south, subtracting his southward velocity is equivalent to adding an equal northward vector to the rain's vertical fall — resulting in the relative velocity having a vertical (downward) component of 36 m/s and a horizontal component of 20 m/s directed opposite to the boy's motion, i.e. pointing back toward where the boy came from (north), which means the rain appears to approach him from the south (the direction he is heading into).
   3. Because the rain appears to come from the south (the direction of his motion), the …

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10. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.A swimmer who can swim in still water at a speed of 20 kmph wants to cross a river flowing at speed of 10 kmph along shortest path, then the angle with the direction of flow in which he has to swim is (A) $120^\circ$ (B) $150^\circ$ (C) $30^\circ$ (D) $60^\circ$

    ›Reveal solutionSolution

    To cross by the shortest (straight-across) path, the swimmer must angle upstream so the current's effect is exactly cancelled; solving the vector equation gives $120^\circ$ with the flow direction.

    Concept and Intuition

    "Shortest path" across a river means the net displacement is straight across the current — the resultant velocity (swimmer's velocity relative to water + river's velocity) must have zero component along the flow. This forces the swimmer to swim at an angle upstream (angled against the current) rather than straight across, so that the current's push is exactly cancelled by the upstream component of the swimmer's own velocity.

    Step-by-Step Solution

    1. Let the flow direction (downstream) be the positive x-axis. River speed $u = 10$ kmph.
    2. Let the swimmer's speed relative to water be $v_s = 20$ kmph, directed at angle $\phi$ measured from the direction of flow.
    3. The swimmer's velocity components relative to ground are: $v_x = v_s\cos\phi + u$, $v_y = v_s\sin\phi$.
    4. For the shortest (straight-across) path, drift must be zero: $v_x = 0$.

    $v_s\cos\phi + u = 0 \implies \cos\phi = -\frac{u}{v_s} = -\frac{10}{20} = -0.5$

    5. $\phi = \cos^{-1}(-0.5) = 120^\circ$. …

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11. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.A man walks up a stationary escalator in 80 s. When this man stands on the moving escalator, he goes up in 20 s. The time taken by the man to walk up on the moving escalator in seconds is (A) 5 (B) 16 (C) 36 (D) 10

    ›Reveal solutionSolution

    Combine the man's own walking rate and the escalator's rate (as fractions of the trip per second); total time is 16 s.

    Concept and Intuition

    This is a 'rates add' problem, like two people/machines doing a job together. Treat the escalator length as one unit of 'work'. The man's walking contributes a rate, and the escalator's motion contributes another rate; when both act together the rates simply add.

    Step-by-Step Solution

    1. Man walking on a stationary escalator covers the length in 80 s → his walking rate $= \dfrac{1}{80}$ (length/s).
    2. Standing still on the moving escalator, he's carried in 20 s → escalator's rate $= \dfrac{1}{20}$ (length/s). …

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12. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.A boat man finds that he can save 8 s in crossing a river by quickest path than by the shortest path. If the velocity of the boat and the river flow are $13\ \mathrm{ms}^{-1}$ and $12\ \mathrm{ms}^{-1}$ respectively, then the width of the river is (A) $65\ \mathrm{m}$ (B) $56\ \mathrm{m}$ (C) $75\ \mathrm{m}$ (D) $57\ \mathrm{m}$

    ›Reveal solutionSolution

    Comparing quickest-crossing time ($d/13$) and shortest-path time ($d/5$) with an 8 s difference gives river width $d=65$ m.

    Concept and Intuition

    In river-crossing problems: the 'quickest path' means the boat points straight across (minimizes time, but drifts downstream), taking time $d/u$. The 'shortest path' means the boat is angled upstream so the resultant path is a straight line straight across, taking longer since the effective forward speed is reduced to $\sqrt{u^2-v^2}$.

    Step-by-Step Solution

    1. Quickest-path time: $t_1 = d/u = d/13$.
    2. Shortest-path (straight-line) time: effective speed across $=\sqrt{u^2-v^2}=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5$ m/s, so $t_2=d/5$. …

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