Q.The variation of a quantity $A$ with a quantity $B$ describes the motion of a particle along a straight line. When $A$ (vertical axis) is plotted against $B$ (horizontal axis) the graph is a straight line of positive slope that does not pass through the origin — it meets the vertical ($A$) axis at a positive intercept and then rises linearly. Choose the correct statement(s). (Note: more than one option may be correct.)

The motion of a body thrown vertically upwards is a classic example of uniformly accelerated motion under gravity. The key idea here is that for any height less than the maximum height, the body will be at that height at two distinct times: once while it is moving upwards, and again while it is falling back down. We need to use the equations of motion to find these two times and then calculate their ratio.

1. Relate maximum height to initial velocity:

   Let the body be thrown vertically upwards from the ground with an initial velocity $u$. The acceleration due to gravity is $g$, acting downwards. We take the upward direction as positive, so the acceleration is $a = -g$.

   At the maximum height $H$, the final velocity $v$ of the body is $0$.

   Using the kinematic equation $v^2 = u^2 + 2as$:

   $0^2 = u^2 + 2(-g)H$

   $0 = u^2 - 2gH$

   This gives us a crucial relationship between the initial velocity and the maximum height:

   $u^2 = 2gH$

   So, $u = \sqrt{2gH}$.

2. Set up the equation for height at time $t$:

   The height $h$ of the body at any time $t$ is given by the kinematic equation $s = ut + \frac{1}{2}at^2$.

   Substituting $s=h$ and $a=-g$:

   $h = ut - \frac{1}{2}gt^2$

3. Solve for time at height $H/2$:

   We are interested in the times when the body is at a height of $\dfrac{H}{2}$. Substitute $h = \dfrac{H}{2}$ into the equation from Step 2:

   $\dfrac{H}{2} = ut - \frac{1}{2}gt^2$

   Rearrange this into a standard quadratic equation in $t$:

   $\frac{1}{2}gt^2 - ut + \frac{H}{2} = 0$

   Multiply by $2$ to simplify:

   $gt^2 - 2ut + H = 0$

   Now, substitute the expression for $u$ from Step 1 ($u = \sqrt{2gH}$) into this quadratic equation:

   $gt^2 - 2(\sqrt{2gH})t + H = 0$

   This is a quadratic equation of the form $At^2 + Bt + C = 0$, where $A=g$, $B=-2\sqrt{2gH}$, and $C=H$.

   The solutions for $t$ are given by the quadratic formula $t = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:

   $t = \dfrac{-(-2\sqrt{2gH}) \pm \sqrt{(-2\sqrt{2gH})^2 - 4(g)(H)}}{2g}$

   $t = \dfrac{2\sqrt{2gH} \pm \sqrt{4(2gH) - 4gH}}{2g}$

   $t = \dfrac{2\sqrt{2gH} \pm \sqrt{8gH - 4gH}}{2g}$ …

Distance in the $n$th second. For uniform acceleration,

Use the given data.

Subtracting: $2a=12\Rightarrow a=6\ \text{m s}^{-2}$, and $u=25-3.5(6)=4\ \text{m s}^{-1}$.

"Next two seconds" = the 7th and 8th seconds. …

The key insight: power is the rate of doing work. If the engine delivers constant power, the product of force and velocity stays fixed. Since force causes acceleration, this links velocity and time in a way that is not linear — and that determines how distance grows.

1. Set up the constant power condition. Power $P$ is work per time: $P = F v$, where $F$ is the net force along the motion and $v$ is the instantaneous speed. Constant power means

2. Relate force to acceleration. By Newton’s second law, $F = m a = m \frac{dv}{dt}$. Substitute into the power equation:

3. Separate variables and integrate. Treat $v$ and $t$:

Integrate from $t=0$ (where $v=0$, assuming start from rest) to time $t$:

So

Velocity grows as the square root of time.

4. Find distance from velocity. Distance $s$ is the integral of velocity: …

Concept and intuition

When water drains from a small hole at the bottom of a cylindrical vessel, the speed at which it leaves is given by Torricelli’s law: $v = \sqrt{2gh}$, where $h$ is the height of the water above the hole. This means the flow rate (volume per unit time) is not constant — it depends on the instantaneous height. As water drains, $h$ decreases, so the speed drops, and the same volume of water takes longer to exit. Since the vessel is cylindrical, the cross-sectional area is constant, so the height is directly proportional to the volume remaining. Therefore, draining the first 5 litres (when the water is highest) is fastest, the next 5 litres slower, and the last 5 litres slowest.

Step-by-step reasoning

1. Set up the relation between volume and height

   Let the cross-sectional area of the cylinder be $A$. The volume of water $V$ is related to height $h$ by $V = A h$. Initially, the total volume is 15 litres, so the initial height $h_0 = 15/A$. After draining 5 litres, height becomes $h_1 = 10/A$; after another 5 litres, $h_2 = 5/A$; and after the last 5 litres, $h_3 = 0$.

2. Apply Torricelli’s law for the efflux speed

   The speed of water exiting the hole of area $a$ is $v = \sqrt{2gh}$. The volume flow rate is $\frac{dV}{dt} = -a v = -a \sqrt{2gh}$. The negative sign indicates volume decreasing.

3. Express the time to drain a given volume

   Since $V = A h$, we have $dV = A \, dh$. Substituting into the flow equation:

Rearranging:

Integrate from initial height $h_i$ to final height $h_f$:

The left side gives $2(\sqrt{h_f} - \sqrt{h_i})$. So the time to drain from height $h_i$ to $h_f$ is:

4. Compute the times for each 5-litre segment Let $k = \frac{2A}{a \sqrt{2g}}$ (a positive constant).
   • For the first 5 litres: from $h_0 = 15/A$ to $h_1 = 10/A$ $$$$

When objects fall freely under gravity, they all experience the same acceleration, $g$, directed downwards. This is a crucial insight: because their accelerations are identical, their relative acceleration is zero. If the relative acceleration between two objects is zero, it means their relative velocity remains constant. This simplifies the problem significantly, as we only need to find the relative velocity at a specific point in time (when the second ball starts moving) and then use it to calculate the time taken for the distance to reach $10$ m.

Let's use the standard approximation $g = 10 \text{ m/s}^2$ for the acceleration due to gravity.

1. Define the coordinate system and initial conditions:

   Let the starting height be the origin, and let the downward direction be positive. Both balls are dropped from rest, meaning their initial velocities are $0$.

   • Ball 1: Dropped at $t=0$.
   • Ball 2: Dropped at $t=1$ s.

2. Analyze the motion of the first ball up to $t=1$ s:

   At $t=1$ s, the first ball has been falling for $1$ second.

   • Its displacement $s_1(1)$ can be found using the equation of motion $s = ut + \frac{1}{2}at^2$:

*   Its velocity $v_1(1)$ can be found using $v = u + at$:

Substituting $g = 10 \text{ m/s}^2$:
*   $s_1(1) = \frac{1}{2}(10) = 5 \text{ m}$
*   $v_1(1) = 10 \text{ m/s}$

3. Determine the initial relative conditions at $t=1$ s:

At $t=1$ s, the second ball is just being dropped from rest.

* Its displacement $s_2(1) = 0$.

* Its velocity $v_2(1) = 0$.

At this moment ($t=1$ s), the distance between the two balls is the displacement of the first ball, as the second ball hasn't moved yet:

The relative velocity of the first ball with respect to the second ball at $t=1$ s is:

4. Use constant relative velocity to find the time: Since both balls are falling under the same acceleration $g$, their relative acceleration is $a_{rel} = g - g = 0$. This means their relative velocity remains constant at $v_{rel} = 10 \text{ m/s}$ for all times $t \ge 1$ s. …