Q.Four position-time ($x$ versus $t$) graphs of a particle moving along a straight line are described below. In only ONE of them can the average velocity over the interval $(0, T)$ be made to vanish for a suitably chosen $T$. Which one is it?
Graph (a): at $t=0$ the position is negative (below the $t$-axis); the curve rises, crosses $x=0$ at some later time, reaches a positive maximum and then falls slightly, remaining positive.
Graph (b): at $t=0$ the position is positive; the curve rises to a maximum and then falls in an S-shape to a smaller positive value.
Graph (c): at $t=0$ the position is a large positive value; the curve decreases steadily (concave up) and levels off toward zero, staying positive.
Graph (d): the curve starts at the origin ($x=0$ at $t=0$), rises quickly and then flattens (concave down), approaching a constant positive value.

The problem is about relative motion in two dimensions. When you are inside a moving train and look at a motorcyclist outside, what you see is not the motorcyclist’s actual velocity, but the velocity of the motorcyclist relative to you — that is, the vector difference between the motorcyclist’s velocity and the train’s velocity.

The motorcyclist goes north at $10\ \mathrm{m/s}$. The train goes north-west at $10\sqrt{2}\ \mathrm{m/s}$. North-west means exactly halfway between north and west — a direction that makes a $45^\circ$ angle with both north and west. So the train’s velocity has equal components northward and westward.

We want the direction of the motorcyclist as seen from the train. That is the direction of the relative velocity vector $\vec{v}_{MT} = \vec{v}_M - \vec{v}_T$.

1. Set up coordinates. Let north be the $+y$ direction and east be the $+x$ direction. Then:

   • Motorcyclist’s velocity: $\vec{v}_M = 10\ \hat{j}\ \mathrm{m/s}$.
   • Train’s velocity: north-west means $45^\circ$ west of north. So its components are:
     • Northward: $10\sqrt{2} \cos 45^\circ = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10\ \mathrm{m/s}$.
     • Westward: $10\sqrt{2} \sin 45^\circ = 10\ \mathrm{m/s}$.
     • Since west is the negative $x$-direction, $\vec{v}_T = -10\ \hat{i} + 10\ \hat{j}\ \mathrm{m/s}$.

2. Find the relative velocity. The velocity of the motorcyclist as seen by the passenger in the train is:

That is, $\vec{v}_{MT} = 10\ \mathrm{m/s}$ due east.

3. Interpret the result. The relative velocity points purely east. So from the train, the motorcyclist appears to be moving directly eastward — not north, not west, not south.

The core idea here is relative motion. When an observer is moving, the rate at which they encounter objects that are also moving changes. This is because the relative speed between the observer and the objects determines how quickly the distance between them closes or opens.

Imagine a series of buses, equally spaced, moving along a road. If you stand still, you see a bus every $T$ minutes. This means the distance between any two consecutive buses is $v_b T$, where $v_b$ is the speed of the bus. Now, if you start moving, this observed time interval will change.

• If you move in the same direction as the buses, and slower than them, the buses will still overtake you, but they will appear to pass less frequently than if you were standing still. This is because the relative speed between you and the bus is reduced.
• If you move in the opposite direction to the buses, they will appear to pass more frequently. This is because the relative speed between you and the bus is increased.

We will use this principle to set up two equations based on the man's observations, and then solve for $T$.

1. Define variables and the constant spacing:

   Let $v_b$ be the speed of a bus and $v_m$ be the speed of the man.

   Buses leave every $T$ minutes. This means that at any given instant, the distance between two consecutive buses moving in the same direction is constant. Let's call this distance $L$.

   In time $T$, a bus travels a distance $v_b T$. So, the distance between two consecutive buses is $L = v_b T$. This distance $L$ is the key to relating the observed time intervals to the actual time interval $T$.

2. Consider buses moving in the same direction as the man (X to Y):

   The man is moving from X to Y. Buses also leave from X to Y.

   The man observes a bus passing him every $t_1$ minutes. For a bus to "go past him in the direction of his motion", the bus must be moving faster than the man. So, $v_b > v_m$.

   The relative speed of a bus with respect to the man is $v_{rel,1} = v_b - v_m$.

   In the time $t_1$, the man observes one bus pass, and then the next bus (which was $L$ distance behind the first one) also passes him. This means that in time $t_1$, the relative distance covered by the bus with respect to the man is $L$.

   Therefore, we can write:

   $L = v_{rel,1} \times t_1$

   Substituting $L = v_b T$ and $v_{rel,1} = v_b - v_m$:

3. Consider buses moving in the opposite direction to the man (Y to X): The man is moving from X to Y. Buses also leave from Y to X, meaning they are moving opposite to the man's direction. The man observes a bus passing him every $t_2$ minutes. The relative speed of a bus with respect to the man is $v_{rel,2} = v_b + v_m$. Similar to the previous case, in time $t_2$, the relative distance covered by the bus with respect to the man is $L$. Therefore, we can write: $L = v_{rel,2} \times t_2$ Substituting $L = v_b T$ and $v_{rel,2} = v_b + v_m$:

4. Solve the system of equations for $T$:

   We have two equations:

   1. $v_b T = v_b t_1 - v_m t_1$
   2. $v_b T = v_b t_2 + v_m t_2$

   Rearrange Equation 1 to express $v_m t_1$:

   $v_m t_1 = v_b t_1 - v_b T$

   $v_m t_1 = v_b (t_1 - T) \quad \text{(Equation 3)}$

   Rearrange Equation 2 to express $v_m t_2$:

   $v_m t_2 = v_b T - v_b t_2$

   $v_m t_2 = v_b (T - t_2) \quad \text{(Equation 4)}$

   Now, divide Equation 3 by Equation 4 to eliminate $v_m$ and $v_b$:

Cross-multiply:
$t_1 (T - t_2) = t_2 (t_1 - T)$
$t_1 T - t_1 t_2 = t_1 t_2 - t_2 T$
Group terms with $T$:
$t_1 T + t_2 T = t_1 t_2 + t_1 t_2$
$T (t_1 + t_2) = 2 t_1 t_2$
Finally, solve for $T$:

The displacement–time relation $x = t^2 + 7$ encodes all the kinematics. Velocity comes from differentiating once, acceleration from differentiating twice. Once we have acceleration, Newton's second law gives the force, and the work–energy theorem connects force and displacement to the work done. The key is to extract each piece systematically.

Finding displacement when velocity is 2 m/s

1. Differentiate to find velocity.

2. Solve for the time when $v = 2$ m/s.

3. Substitute $t = 1$ s into the displacement equation.

Finding work done during the first 10 s

The cleanest route is the work–energy theorem: work done equals the change in kinetic energy.

4. Find the velocity at $t = 0$ s and $t = 10$ s.

   At $t = 0$: $v_0 = 2(0) = 0$ m/s

   At $t = 10$: $v_{10} = 2(10) = 20$ m/s

5. Compute the change in kinetic energy.

6. Apply the work–energy theorem.

The key idea here is that average velocity is not the same as instantaneous velocity. It’s simply the straight-line rate of change of position over a finite interval:

We don’t need calculus for this — just plug the given times into the position function and compute.

1. Find the position at $t = 2.00$ s

2. Find the position at $t = 4.00$ s

3. Compute the displacement

4. Compute the time interval

5. Average velocity

The key idea is that the car leaves the cliff horizontally, so its initial vertical speed is zero. Gravity only affects the vertical motion, while horizontal speed stays constant (no air resistance). At the peak Z, the car has fallen a certain vertical distance, gaining vertical speed. The total speed is the vector sum of horizontal and vertical components. Instead of using time, we can use energy conservation: the loss in gravitational potential energy equals the gain in kinetic energy. That gives the final speed directly.

1. Identify the given data and the goal.

   Initial horizontal speed: $v_x = 10\ \mathrm{m/s}$ (constant throughout).

   Height fallen from cliff to peak Z: from the figure (not shown here, but typical in such problems) it is $h = 40\ \mathrm{m}$.

   We need the speed $v$ at Z.

2. Apply conservation of mechanical energy.

   Initial energy (at cliff X):

(taking the cliff as height $h$ above Z).

Final energy (at Z, height = 0):

Since no air resistance, $E_i = E_f$:

3. Cancel mass and solve for $v$.

Multiply by 2:

Substitute $v_x = 10\ \mathrm{m/s}$, $g = 10\ \mathrm{m/s^2}$, $h = 40\ \mathrm{m}$:

So $v = \sqrt{900} = 30\ \mathrm{m/s}$.

4. Interpret the result. The speed at Z is purely the magnitude of the velocity vector. The horizontal part is still $10\ \mathrm{m/s}$, and the vertical part is $v_y = \sqrt{2gh} = \sqrt{800} \approx 28.28\ \mathrm{m/s}$. Their vector sum gives $30\ \mathrm{m/s}$, consistent with energy conservation.

Set up the two velocity vectors (origin, $x$ horizontal, $y$ vertical)

• Body 1, straight up: $\vec v_1 = (0,\ 10)~\text{m/s}$.
• Body 2, at $60^\circ$ to the vertical: vertical component $10\cos 60^\circ = 5$, horizontal component $10\sin 60^\circ = 5\sqrt3$, so $\vec v_2 = (5\sqrt3,\ 5)~\text{m/s}$.

Use relative motion

Both bodies have identical acceleration $\vec a = (0,-g)$, so the relative acceleration is zero. The relative velocity is constant:

Separation after $t=2~\text{s}$

Since relative acceleration is zero, the distance between them grows linearly: