Q.Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of $v$ and $a$ in the three intervals. What are the accelerations at the points A, B, C and D?

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The driver of a bus moving with a velocity of 72 kmph observes a boy walking across the road at a distance of 50 m in front of the bus and decelerates the bus at $5\,\mathrm{ms}^{-2}$ by applying brakes and is just able to avoid an accident. The reaction time of the driver is (A) $4\,\mathrm{s}$ (B) $3.5\,\mathrm{s}$ (C) $0.5\,\mathrm{s}$ (D) $4.5\,\mathrm{s}$

   ›Reveal solutionSolution

   The driver’s reaction time is the delay before braking begins; during that time the bus travels at constant speed, and the remaining distance must be exactly enough for the deceleration to stop the bus. Solving gives a reaction time of 0.5 s, so option (C) is correct.

   ---

   The key idea is that the driver does not brake instantly — there is a reaction time during which the bus continues at its initial speed. Only after that delay does the bus decelerate. The total distance from the driver’s first sight of the boy to the bus’s final stop is 50 m. Part of that distance is covered at constant speed during the reaction time, and the rest is covered while braking to a halt. We need to find the reaction time such that the bus just stops before reaching the boy.

   ---

   1. Convert the initial speed to m/s The bus moves at $72\ \text{km/h}$.

   $72\ \frac{\text{km}}{\text{h}} = 72 \times \frac{1000}{3600} = 20\ \text{m/s}.$

   So $u = 20\ \text{m/s}$.

   2. Let the reaction time be $t_r$ seconds During this time, the bus travels at constant speed $u$ and covers a distance

   $d_{\text{reaction}} = u \cdot t_r = 20\,t_r\ \text{m}.$

   3. After the reaction time, the driver applies brakes The bus decelerates at $a = -5\ \text{m/s}^2$ (negative because it’s slowing down). The initial speed for the braking phase is still $u = 20\ \text{m/s}$, and the final speed is $0$. Using the equation of motion $v^2 = u^2 + 2 a s$ for the braking distance $s$:

   $0 = (20)^2 + 2(-5)\,s \quad\Rightarrow\quad 0 = 400 - 10s \quad\Rightarrow\quad s = 40\ \text{m}.$

   So the bus needs 40 m to stop once brakes are applied.

   4. The total distance from the driver’s first sight to the boy is 50 m …

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2. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.For a particle moving along a straight line path, the displacements in third and fifth seconds of its motion are 10 m and 18 m respectively. The speed of the particle at time $t = 4\,\mathrm{s}$ is (A) $32\,\mathrm{ms}^{-1}$ (B) $8\,\mathrm{ms}^{-1}$ (C) $12\,\mathrm{ms}^{-1}$ (D) $16\,\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The problem involves constant acceleration motion; using the formula for displacement in the $n$th second, we find acceleration $a = 4\,\mathrm{m/s^2}$ and initial velocity $u = 0$, so speed at $t=4\,\mathrm{s}$ is $16\,\mathrm{m/s}$ — option (D).

   We are told the particle moves along a straight line, and the displacements in the third and fifth seconds are given. This is a classic constant-acceleration kinematics problem. The key is to use the formula for displacement during a specific second, which directly relates the given data to the initial velocity and acceleration.

   Why this approach works:

   For uniformly accelerated motion, the displacement in the $n$th second is $s_n = u + \frac{a}{2}(2n-1)$. This formula comes from subtracting the distance traveled in $(n-1)$ seconds from that in $n$ seconds. It gives us two equations in $u$ and $a$, which we can solve.

   Let’s work through it step by step.

   1. Write the formula for displacement in the $n$th second. For constant acceleration $a$ and initial velocity $u$, the displacement in the $n$th second is:

   $s_n = u + \frac{a}{2}(2n - 1)$

   This is derived from $s_n = \left[ u n + \frac{1}{2} a n^2 \right] - \left[ u (n-1) + \frac{1}{2} a (n-1)^2 \right]$.

   2. Apply to the third second ($n=3$). Given $s_3 = 10\,\mathrm{m}$:

   $10 = u + \frac{a}{2}(2 \cdot 3 - 1) = u + \frac{a}{2}(5)$

   So:

   $10 = u + \frac{5a}{2} \quad \text{(Equation 1)}$

   3. Apply to the fifth second ($n=5$). Given $s_5 = 18\,\mathrm{m}$:

   $18 = u + \frac{a}{2}(2 \cdot 5 - 1) = u + \frac{a}{2}(9)$

   So:

   $18 = u + \frac{9a}{2} \quad \text{(Equation 2)}$

   4. Solve the system of equations. Subtract Equation 1 from Equation 2:

   $(18 - 10) = \left(u + \frac{9a}{2}\right) - \left(u + \frac{5a}{2}\right)$

   $8 = \frac{4a}{2} = 2a$

   Hence:

   $a = 4\,\mathrm{m/s^2}$

   5. Find the initial velocity $u$. …

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3. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.For a particle moving along a straight line path, the displacements in third and fifth seconds of its motion are 10 m and 18 m respectively. The speed of the particle at time $t = 4$ s is (A) $32~\mathrm{ms}^{-1}$ (B) $12~\mathrm{ms}^{-1}$ (C) $16~\mathrm{ms}^{-1}$ (D) $8~\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   Using $s_{n}=u+\tfrac{a}{2}(2n-1)$: the $3^{\text{rd}}$ and $5^{\text{th}}$-second displacements give $a=4\ \mathrm{ms^{-2}},\ u=0$, so $v(4)=u+at=16\ \mathrm{ms^{-1}}$.

   The displacement in the $n^{\text{th}}$ second of uniformly accelerated motion is

   $s_{n}=u+\frac{a}{2}(2n-1).$

   Third second ($n=3$): $\;s_{3}=u+\dfrac{a}{2}(5)=u+2.5a=10.$

   Fifth second ($n=5$): $\;s_{5}=u+\dfrac{a}{2}(9)=u+4.5a=18.$

   Subtracting the two equations: …

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4. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.When a bullet is fired with a velocity of $150\,\mathrm{ms}^{-1}$ at a target of thickness $50\,\mathrm{cm}$, it emerges with a velocity of $100\,\mathrm{ms}^{-1}$. If another bullet of same mass is fired with same velocity at a second target of thickness $80\,\mathrm{cm}$, then the velocity with which the bullet emerges from the second target is (Retarding forces are equal in both the cases) (A) $60\,\mathrm{ms}^{-1}$ (B) $75\,\mathrm{ms}^{-1}$ (C) $50\,\mathrm{ms}^{-1}$ (D) $40\,\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The retarding force does constant work per unit thickness, so the loss in kinetic energy is proportional to the thickness. Using the work–energy theorem for both targets gives the emerging speed as $50\,\mathrm{m/s}$.

   The key idea is that the retarding force is the same in both cases, and it acts over a known distance (the thickness of the target). The work done by this force equals the loss in kinetic energy of the bullet. Since the force is constant, the work done is simply force times distance, so the kinetic energy lost is directly proportional to the thickness of the target.

   We don’t need to know the mass or the force explicitly — we can work with the change in $v^2$, which is proportional to the distance travelled under a constant retarding force.

   1. First target — thickness $d_1 = 50\ \text{cm} = 0.5\ \text{m}$. Initial speed $u = 150\ \text{m/s}$, final speed $v_1 = 100\ \text{m/s}$. Loss in kinetic energy:

   $\frac{1}{2}m(150^2 - 100^2) = \frac{1}{2}m(22500 - 10000) = \frac{1}{2}m(12500)$

   This loss equals the work done by the retarding force $F$ over distance $d_1$:

   $F d_1 = \frac{1}{2}m(12500)$

   2. Second target — thickness $d_2 = 80\ \text{cm} = 0.8\ \text{m}$. Same initial speed $u = 150\ \text{m/s}$, let the emerging speed be $v_2$. Loss in kinetic energy:

   $\frac{1}{2}m(150^2 - v_2^2) = \frac{1}{2}m(22500 - v_2^2)$

   This equals the work done over distance $d_2$:

   $F d_2 = \frac{1}{2}m(22500 - v_2^2)$

   3. Divide the two equations to eliminate $F$ and $m$: $$$$

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5. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.A person walks in such a way that he covers equal distance in each step. The person takes 2 steps forward towards east, then takes a right turn and walks 4 steps towards south, then takes a right turn and walks 6 steps towards west and then takes a right turn and walks further. The direction of his final position after a total of 20 steps walk with respect to his initial position is (A) North-West (B) $60^\circ$ West of South (C) $60^\circ$ South of West (D) South-East

   ›Reveal solutionSolution

   Track the four legs on a grid; the walker ends at $(-4,+4)$ — equal parts west and north — i.e. North-West.

   Set up coordinates. Take east as $+x$, north as $+y$; each step is one unit. He starts facing east, and each "right turn" rotates his heading east→south→west→north.

   Steps used so far: $2+4+6=12$, so the fourth leg is $20-12=8$ steps, heading north.

   Leg-by-leg positions:

   • Start: $(0,0)$
   • East 2 steps: $(2,0)$
   • South 4 steps: $(2,-4)$
   • West 6 steps: $(-4,-4)$
   • North 8 steps: $(-4,\,+4)$ …

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6. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.A body starts from the rest and acquires a velocity of 10 m/s in 2s. What is the acceleration of the body and the distance travelled (A) $5\ \mathrm{m/s^2}$ and $10\ \mathrm{m}$ (B) $5\ \mathrm{m/s^2}$ and $5\ \mathrm{m}$ (C) $5\ \mathrm{m/s^2}$ and $6\ \mathrm{m}$ (D) $6\ \mathrm{m/s^2}$ and $5\ \mathrm{m}$

   ›Reveal solutionSolution

   Using the equations of motion for constant acceleration from rest, the acceleration is $5\ \mathrm{m/s^2}$ and the distance travelled is $10\ \mathrm{m}$. The correct option is (A).

   The problem gives a body starting from rest (initial velocity $u = 0$), reaching a velocity $v = 10\ \mathrm{m/s}$ in time $t = 2\ \mathrm{s}$. We need the acceleration and the distance travelled. Since the body starts from rest and we assume constant acceleration (the most natural reading for such a problem), we can use the standard kinematic equations.

   Why this works:

   When acceleration is constant, velocity changes uniformly with time, and distance is the area under the velocity–time graph (or the average velocity times time). Starting from rest simplifies the formulas because $u = 0$.

   1. Find acceleration The definition of acceleration (for constant acceleration) is

   $a = \frac{v - u}{t}$

   Here $u = 0$, $v = 10\ \mathrm{m/s}$, $t = 2\ \mathrm{s}$:

   $a = \frac{10 - 0}{2} = 5\ \mathrm{m/s^2}$

   2. Find distance travelled Using the equation $s = ut + \frac{1}{2} a t^2$ with $u = 0$:

   $s = 0 \cdot 2 + \frac{1}{2} \cdot 5 \cdot (2)^2 = \frac{1}{2} \cdot 5 \cdot 4 = 10\ \mathrm{m}$

   Alternatively, average velocity is $\frac{u+v}{2} = \frac{0+10}{2} = 5\ \mathrm{m/s}$, and distance = average velocity × time = $5 \times 2 = 10\ \mathrm{m}$. …

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7. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
   Q.The speed distance graph is shown below. At what instant of time (in sec) the speed becomes 4 m/s? [FIGURE] (A) $t = \ln(2)$ (B) $t = \ln(4)$ (C) $t = \ln(8)$ (D) $t = \ln(6)$

   ›Reveal solutionSolution

   The speed–distance graph is a straight line of slope 1, so $\frac{dv}{dx} = 1$. Using the chain rule $\frac{dv}{dt} = \frac{dv}{dx}\cdot v = v$ and integrating gives $v = 4$ m/s at $t = \ln 2$, option (A).

   Reading the graph

   The graph plots speed $v$ against distance $x$ as a straight line: the speed starts at $v = 2$ m/s at $x = 0$ and rises with slope 1, so

   $v = 2 + x, \qquad \frac{dv}{dx} = 1.$

   The chain-rule connection

   Speed, distance and time are linked by

   $\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v,$

   since $\frac{dx}{dt} = v$.

   Step-by-step solution

   1. From the graph $\frac{dv}{dx} = 1$, so …

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8. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
   Q.The length of minute hand in a clock is 4.5 cm. If the tip of the minute hand moves from 6 AM to 6.30 AM, the average velocity of the tip is. (A) $5 \times 10^{-3}$ cm/s (B) $50 \times 10^{-3}$ cm/s (C) $0.5 \times 10^{-3}$ cm/s (D) $0.005 \times 10^{-3}$ cm/s

   ›Reveal solutionSolution

   Average velocity is displacement divided by time. The tip’s displacement is the straight‑line distance from start to end (the chord of a 180° arc), not the path length. With radius 4.5 cm and time 30 min = 1800 s, the average velocity is $\frac{2 \times 4.5}{1800} = 0.005$ cm/s = $5 \times 10^{-3}$ cm/s. The correct option is (A).

   Concept & Intuition

   Many students mistakenly compute average speed (total path length / time) instead of average velocity (displacement / time). Velocity is a vector; its magnitude depends only on the straight‑line distance between the initial and final positions, not on how far the tip actually travelled along the arc. From 6:00 to 6:30, the minute hand rotates exactly 180°, so the tip’s start and end points are opposite ends of a diameter. The displacement is therefore the diameter of the circle: $2 \times \text{radius}$.

   Step‑by‑step solution

   1. Identify the time interval From 6:00 AM to 6:30 AM is exactly 30 minutes. Convert to seconds:

   $\Delta t = 30 \times 60 = 1800 \text{ s}.$

   2. Determine the displacement At 6:00, the minute hand points straight up (12 o’clock). At 6:30, it points straight down (6 o’clock). These two positions are diametrically opposite. The displacement is the straight‑line distance between them, which is the diameter of the circle traced by the tip:

   $\text{displacement} = 2 \times \text{radius} = 2 \times 4.5 = 9 \text{ cm}.$

   3. Compute average velocity Average velocity is a vector quantity:

   $\text{average velocity} = \frac{\text{displacement}}{\text{time}}.$

   Its magnitude is:

   $$$$

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9. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
   Q.A body moves along the sides of an equilateral triangle of side 20 cm and comes back to the initial point after one round. Then the distance and displacement of the body respectively are (A) 60 cm and 20 cm (B) 60 cm and 0 cm (C) 0 cm and 60 cm (D) 60 cm and 60 cm

   ›Reveal solutionSolution

   Distance is the total path length traveled (60 cm), while displacement is the net change in position from start to finish (0 cm because the body returns to its starting point). The correct option is (B).

   The key idea here is the difference between distance (a scalar quantity that measures the total ground covered) and displacement (a vector quantity that measures the straight-line change in position from start to end). When an object returns to its starting point, the displacement is zero regardless of the path taken.

   1. Understand the motion: The body moves along the sides of an equilateral triangle with each side = 20 cm. It starts at one vertex, goes along one side, then the next, then the third, and returns exactly to the starting vertex after one full round.

   2. Calculate the distance: Distance is the sum of the lengths of all sides traveled. Since the body goes around all three sides:

   $\text{Distance} = 20\ \text{cm} + 20\ \text{cm} + 20\ \text{cm} = 60\ \text{cm}.$

   3. Calculate the displacement: Displacement depends only on the initial and final positions. The body starts at a point and ends at the same point after one round. Therefore, the net change in position is zero: $$\text{Displacement} = 0\ \text{cm}. …$$

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10. TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
    Q.The length of minute hand in a clock is 4.5 cm. If the tip of the minute hand moves from 6 AM to 6.30 AM, the average velocity of the tip is, (A) $5 \times 10^{-3}$ cm/s (B) $50 \times 10^{-3}$ cm/s (C) $0.5 \times 10^{-3}$ cm/s (D) $0.005 \times 10^{-3}$ cm/s

    ›Reveal solutionSolution

    Average velocity is displacement divided by time, not distance. The minute hand tip moves in a circular arc from 6 AM to 6:30 AM, so its displacement is the straight-line distance between the two positions — the diameter of the circle. With radius 4.5 cm, displacement = 9 cm, time = 1800 s, so average velocity = 0.005 cm/s = $5 \times 10^{-3}$ cm/s. The correct option is (A).

    ---

    Concept & Intuition

    The classic pitfall here is confusing average speed with average velocity.

    • Average speed = total path length / time. The tip traces a semicircle of radius 4.5 cm, so path length = $\pi r \approx 14.14$ cm.
    • Average velocity = displacement / time. Displacement is the straight-line vector from start to finish — here, from the 6 AM position (pointing straight down) to the 6:30 AM position (pointing straight up). That’s exactly the diameter of the clock face: $2r = 9$ cm.

    Since the question asks for average velocity, we must use displacement, not distance.

    ---

    Step-by-step solution

    1. Identify the positions

       At 6:00 AM, the minute hand points directly at the 6 (straight down). At 6:30 AM, it points directly at the 12 (straight up). These two positions are opposite ends of a vertical line through the center.

    2. Find the displacement

       Displacement is the straight-line distance between start and end points. Since they are diametrically opposite,

    $\text{displacement} = 2 \times \text{radius} = 2 \times 4.5\ \text{cm} = 9\ \text{cm}.$

    3. Find the time interval From 6:00 AM to 6:30 AM is exactly 30 minutes. Convert to seconds:

    $\Delta t = 30 \times 60 = 1800\ \text{s}.$

    4. Compute average velocity $$$$

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11. TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
    Q.A body moves along the sides of an equilateral triangle of side 20 cm and comes back to the initial point after one round. Then the distance and displacement of the body respectively are (A) 60 cm and 20 cm (B) 60 cm and 0 cm (C) 0 cm and 60 cm (D) 60 cm and 60 cm

    ›Reveal solutionSolution

    Distance is the total path length traveled (60 cm), displacement is the net change in position from start to finish (0 cm because the body returns to its starting point). The correct option is (B).

    Concept and Intuition

    The key distinction here is between distance (a scalar: how much ground was covered, regardless of direction) and displacement (a vector: the straight‑line separation between the starting point and the ending point, with direction).

    When a body moves along a closed path and returns exactly to where it began, the displacement is always zero — no matter how long or twisty the path. The distance, however, is simply the sum of the lengths of all the sides traversed.

    Step‑by‑Step Reasoning

    1. Identify the path

       The body moves along the sides of an equilateral triangle. Each side is given as $20\ \text{cm}$. It goes around once and comes back to the initial point.

    2. Compute the distance

       Distance is the total length of the path traveled. The body covers all three sides:

    $\text{Distance} = 20\ \text{cm} + 20\ \text{cm} + 20\ \text{cm} = 60\ \text{cm}.$

    3. Compute the displacement Displacement is the vector from the starting point to the ending point. Since the body returns to the exact same point after one round, the start and end coincide. Therefore the displacement vector has zero magnitude: $$$$

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12. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.Consider a series of measurements of the length of a box in an experiment. The readings are $2.4$ m, $2.5$ m, $2.6$ m, $2.8$ m, $3.0$ m. What would be the relative error? (A) $0.110$ (B) $0.089$ (C) $0.079$ (D) $0.072$

    ›Reveal solutionSolution

    Mean $=2.66$ m; mean absolute error $=0.192$ m; relative error $=\dfrac{0.192}{2.66}\approx 0.072$.

    Mean reading.

    $\bar{x}=\frac{2.4+2.5+2.6+2.8+3.0}{5}=\frac{13.3}{5}=2.66\ \text{m}.$

    Absolute deviations from the mean.

    $0.26,\ 0.16,\ 0.06,\ 0.14,\ 0.34.$

    Mean absolute error: …

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