Introduction

Motion is a fundamental concept in the universe, describing the change in position of an object over time.

To describe motion, we use concepts like velocity and acceleration.

This chapter focuses on motion in a straight line, also known as rectilinear motion.

For situations involving uniform acceleration along a straight line, simple kinematic equations can be used to relate different aspects of the motion.

The concept of relative velocity is introduced to understand motion as observed from different moving frames of reference.

In discussing motion, objects are often treated as point objects or point masses. This is a valid approximation when the size of the object is much smaller than the distance it travels or the relevant distances involved.

This chapter deals with Kinematics – the description of motion – without investigating the causes of motion (which is the domain of Dynamics, discussed later).

Position, Path Length And Displacement

To precisely describe the position of an object, a reference point (origin) and a set of axes are needed. This forms a coordinate system.

Adding a clock to the coordinate system creates a frame of reference, which is necessary to specify the position of an object at a particular time.

An object is in motion if one or more of its coordinates change with time relative to a chosen frame of reference. Otherwise, it is considered at rest relative to that frame.

For motion along a straight line (one-dimensional motion), a single axis (e.g., X-axis) is sufficient. Positions are measured along this axis from an origin (O), with a convention for positive and negative directions (e.g., right of O is positive, left is negative).

Path Length

Path length is the total distance covered by an object along its actual path of motion during a given time interval.

It is a scalar quantity, meaning it only has magnitude and no direction.

Path length is always non-negative.

Example: If a car moves from point O to P, the path length is the distance OP. If it moves from O to P and then back to Q, the path length is OP + PQ.

Displacement

Displacement is defined as the change in position of an object.

If an object's position changes from $x_1$ at time $t_1$ to $x_2$ at time $t_2$, the displacement ($\Delta x$) in the time interval $\Delta t = t_2 - t_1$ is:

Displacement is a vector quantity, possessing both magnitude and direction. In one-dimensional motion, the direction can be indicated by a positive (+) or negative (-) sign.

• If $x_2 > x_1$, $\Delta x$ is positive (displacement is in the positive direction).
• If $x_2 < x_1$, $\Delta x$ is negative (displacement is in the negative direction).

Relationship between Path Length and Displacement:

• The magnitude of displacement ($|\Delta x|$) may or may not be equal to the path length.
• If the object moves along a straight line in the same direction throughout the time interval, the magnitude of displacement is equal to the path length.
• If the object changes direction during the motion, the path length is greater than the magnitude of displacement.
• The displacement can be zero if the object returns to its starting position, even though the path length covered is non-zero. (e.g., Round trip journey).

Motion can be visually represented and analyzed using graphs:

• Position-time (x-t) graph: Plots the position of an object as a function of time.
• For a stationary object, the x-t graph is a horizontal line parallel to the time axis.
• For an object in uniform motion along a straight line (equal distances in equal time intervals), the x-t graph is a straight line with a constant slope.
• For non-uniform motion, the x-t graph is a curve.

Average Velocity And Average Speed

These quantities describe how fast an object is moving over a given time interval.

Average Velocity

Average velocity ($\bar{v}$) is defined as the displacement ($\Delta x$) divided by the time interval ($\Delta t$) during which the displacement occurs.

• It is a vector quantity (sign indicates direction in 1D).
• The SI unit of velocity is metres per second (m/s or m s$^{-1}$).
• Geometrically, on a position-time (x-t) graph, the average velocity over a time interval is the slope of the straight line (secant) connecting the points corresponding to the initial and final positions at the given times.
• Average velocity can be positive, negative, or zero, depending on the sign of the displacement. It is zero if the total displacement is zero.

Average Speed

Average speed is defined as the total path length travelled divided by the total time interval taken for the motion.

• It is a scalar quantity.
• It has the same SI unit (m/s) as velocity.
• Average speed is always non-negative (greater than or equal to zero).

Relationship between Average Speed and Average Velocity:

• Average speed is generally greater than or equal to the magnitude of the average velocity over the same time interval.
• Equality holds only when the motion is along a straight line in the same direction, because in this case, the path length equals the magnitude of the displacement.

Answer:

Refer to the position diagram (Fig 3.1 from text): O is at x=0 m, P is at x=+360 m, Q is at x=+240 m.

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(a) Motion from O to P:

• Initial position $x_O = 0$ m at $t_O = 0$ s.
• Final position $x_P = +360$ m at $t_P = 18$ s.
• Displacement $\Delta x = x_P - x_O = +360 \text{ m} - 0 \text{ m} = +360$ m.
• Time interval $\Delta t = t_P - t_O = 18 \text{ s} - 0 \text{ s} = 18$ s.
• Path length = Distance OP = 360 m.

Average Velocity $ \bar{v} = \frac{\Delta x}{\Delta t} = \frac{+360 \text{ m}}{18 \text{ s}} = +20 \text{ m/s}$.

Average Speed $= \frac{\text{Path length}}{\Delta t} = \frac{360 \text{ m}}{18 \text{ s}} = 20 \text{ m/s}$.

In this case (motion in one direction), average speed equals the magnitude of average velocity.

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(b) Motion from O to P and back to Q:

• Initial position $x_O = 0$ m at $t_O = 0$ s.
• Intermediate position $x_P = +360$ m at $t_P = 18$ s.
• Final position $x_Q = +240$ m at $t_Q = 18 \text{ s} + 6.0 \text{ s} = 24$ s.
• Total time interval $\Delta t = t_Q - t_O = 24 \text{ s} - 0 \text{ s} = 24$ s.
• Displacement $\Delta x = x_Q - x_O = +240 \text{ m} - 0 \text{ m} = +240$ m.
• Path length = Distance OP + Distance PQ. Distance OP = 360 m. Distance PQ = $|x_Q - x_P| = |+240 - (+360)| = |-120|$ m = 120 m. Total path length = 360 m + 120 m = 480 m.

Average Velocity $ \bar{v} = \frac{\Delta x}{\Delta t} = \frac{+240 \text{ m}}{24 \text{ s}} = +10 \text{ m/s}$.

Average Speed $= \frac{\text{Path length}}{\Delta t} = \frac{480 \text{ m}}{24 \text{ s}} = 20 \text{ m/s}$.

In this case (motion involves change in direction), average speed (20 m/s) is greater than the magnitude of average velocity (10 m/s).

Instantaneous Velocity And Speed

Instantaneous Velocity

The average velocity over a time interval only provides an overall rate of displacement. To know how fast and in what direction an object is moving at a specific moment, we need the instantaneous velocity (v).

Instantaneous velocity is defined as the limit of the average velocity as the time interval ($\Delta t$) approaches zero:

In calculus terms, instantaneous velocity is the derivative of the position ($x$) with respect to time ($t$). It represents the rate of change of position at that specific instant.

Geometrically, on a position-time (x-t) graph, the instantaneous velocity at a specific time is the slope of the tangent line to the graph at that point in time.

The concept of the limit as $\Delta t \to 0$ means we are calculating the average velocity over increasingly smaller time intervals around the desired instant. As the interval shrinks, the average velocity approaches the instantaneous velocity.

As $\Delta t$ gets smaller, the calculated average velocity $\Delta x/\Delta t$ approaches a specific value, which is the instantaneous velocity at $t=4$ s (in this example, approximately 3.84 m/s).

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Instantaneous Speed

Instantaneous speed (or simply speed) is defined as the magnitude of the instantaneous velocity at a given instant.

Speed is a scalar quantity and is always non-negative.

At any instant, instantaneous speed is always equal to the magnitude of the instantaneous velocity.

This is because at an infinitesimally small time interval around an instant, the direction of motion does not change, so the magnitude of displacement is equal to the path length covered in that tiny interval.

The velocity-time (v-t) graph in the figure above shows how the velocity of the car from Fig 3.3 changes over time. Note the periods of increasing velocity, constant velocity, and decreasing velocity.

Acceleration

When the velocity of an object changes over time, the object is said to be accelerating.

Acceleration is defined as the rate of change of velocity with respect to time.

Galileo's studies on falling objects led to the understanding that acceleration is best described as the rate of change of velocity *with time*, not with distance.

Average Acceleration

Average acceleration ($\bar{a}$) over a time interval ($\Delta t = t_2 - t_1$) is defined as the change in velocity ($\Delta v = v_2 - v_1$) divided by the time interval:

• Here $v_1$ and $v_2$ are the instantaneous velocities at times $t_1$ and $t_2$.
• It is a vector quantity (sign indicates direction in 1D).
• The SI unit of acceleration is metres per second squared (m/s$^2$ or m s$^{-2}$).
• Geometrically, on a velocity-time (v-t) graph, the average acceleration over a time interval is the slope of the straight line (secant) connecting the points corresponding to velocities $v_1$ and $v_2$ at times $t_1$ and $t_2$.

Instantaneous Acceleration

Instantaneous acceleration (a) is the acceleration at a specific instant in time.

It is defined as the limit of the average acceleration as the time interval ($\Delta t$) approaches zero:

Instantaneous acceleration is the derivative of the velocity ($v$) with respect to time ($t$). Since velocity is the derivative of position, acceleration is the second derivative of position with respect to time ($a = \frac{d^2x}{dt^2}$).

Geometrically, on a velocity-time (v-t) graph, the instantaneous acceleration at a specific time is the slope of the tangent line to the graph at that point in time.

The acceleration-time (a-t) graph in the figure above shows the acceleration of the car over time. Note the periods of non-uniform acceleration, zero acceleration, and constant negative acceleration (deceleration).

Acceleration can be positive, negative, or zero.

• Positive acceleration means the velocity is increasing in the positive direction or decreasing in the negative direction.
• Negative acceleration (often called deceleration) means the velocity is decreasing in the positive direction or increasing in the negative direction.
• Zero acceleration means the velocity is constant (uniform motion).

The sign of acceleration depends on the chosen positive direction.

Whether an object is speeding up or slowing down depends on the signs of *both* velocity and acceleration:

• Speeding up: Velocity and acceleration have the same sign.
• Slowing down (decelerating): Velocity and acceleration have opposite signs.

On a position-time (x-t) graph:

• Positive acceleration: The curve is concave up (bends upwards).
• Negative acceleration: The curve is concave down (bends downwards).
• Zero acceleration: The graph is a straight line (uniform velocity).

On a velocity-time (v-t) graph for constant acceleration, the graph is a straight line with a constant slope equal to the acceleration. Different scenarios for constant acceleration:

An important property of the velocity-time graph is that the area under the curve between two time instants represents the displacement of the object during that time interval.

For constant velocity ($v=u$), the v-t graph is a horizontal line. The area under the curve from time 0 to T is a rectangle of height $u$ and width $T$. Area = $u \times T$. Displacement = $uT$. The units confirm this: (m/s) $\times$ (s) = m, which are the units of displacement.

For motion with varying velocity, integral calculus is needed to find the area under the curve, but the concept holds true.

In realistic motion, velocity and acceleration are continuous functions of time, meaning they do not change values abruptly. Graphs of velocity and acceleration would be smooth curves or lines, without sharp kinks.

Kinematic Equations For Uniformly Accelerated Motion

For motion along a straight line with constant acceleration (a), we can use a set of simple equations to relate initial velocity ($v_0$), final velocity ($v$), time interval ($t$), and displacement ($x$ or $\Delta x$). Assuming the motion starts at $t=0$ with initial velocity $v_0$ and initial position $x_0=0$, these equations are:

1. Velocity-time relation: Derived from the definition of constant acceleration ($\bar{a} = a = (v-v_0)/t$). $$ v = v_0 + at $$
2. Position-time relation: Derived from the area under the v-t graph (a trapezoid for constant acceleration, or the sum of a rectangle and a triangle). $$ x = v_0 t + \frac{1}{2}at^2 $$ This equation can also be understood by noting that for constant acceleration, the average velocity is $\bar{v} = (v_0 + v)/2$. Since $x = \bar{v}t$, substituting $v = v_0 + at$ gives $x = \frac{v_0 + (v_0+at)}{2}t = \frac{2v_0+at}{2}t = v_0t + \frac{1}{2}at^2$.
3. Velocity-displacement relation: Derived by eliminating time ($t$) from the first two equations. From $v = v_0 + at$, $t = (v-v_0)/a$. Substituting this into the second equation: $$ x = v_0 \left(\frac{v-v_0}{a}\right) + \frac{1}{2} a \left(\frac{v-v_0}{a}\right)^2 $$ $$ x = \frac{v_0 v - v_0^2}{a} + \frac{1}{2} a \frac{(v-v_0)^2}{a^2} $$ $$ x = \frac{v_0 v - v_0^2}{a} + \frac{v^2 - 2v_0 v + v_0^2}{2a} $$ Multiplying by $2a$: $$ 2ax = 2v_0 v - 2v_0^2 + v^2 - 2v_0 v + v_0^2 $$ $$ 2ax = v^2 - v_0^2 $$ Rearranging gives: $$ v^2 = v_0^2 + 2ax $$

These three equations are:

$$ \begin{align*} v &= v_0 + at \\ x &= v_0 t + \frac{1}{2}at^2 \\ v^2 &= v_0^2 + 2ax \end{align*} $$

If the initial position at $t=0$ is $x_0$ instead of 0, the displacement is $(x - x_0)$. The equations become:

$$ \begin{align*} v &= v_0 + at \\ x - x_0 &= v_0 t + \frac{1}{2}at^2 \\ v^2 &= v_0^2 + 2a(x - x_0) \end{align*} $$

These equations relate the five quantities ($v_0, v, a, t, \Delta x$ or $x-x_0$) and are valid only for uniformly accelerated rectilinear motion. Quantities ($v_0, v, a, x, x_0$) are algebraic (can be positive or negative) and must be substituted with their proper signs according to the chosen positive direction.

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Let's set up a coordinate system. Choose the y-axis pointing vertically upwards, with the origin ($y=0$) at the ground. The point where the ball is thrown is at an initial height $y_0 = 25.0$ m.

The initial velocity of the ball is upwards, so $v_0 = +20$ m/s.

The acceleration due to gravity $g$ acts downwards. Since our positive y-direction is upwards, the acceleration $a = -g = -10$ m/s$^2$. This acceleration is constant throughout the motion (neglecting air resistance).

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(a) How high will the ball rise?

The ball rises until its instantaneous velocity becomes zero at the highest point. Let the highest point reached be at position $y$. At this point, the velocity $v = 0$ m/s.

We can use the kinematic equation relating $v, v_0, a,$ and displacement $(y-y_0)$: $v^2 = v_0^2 + 2a(y - y_0)$.

Substitute the known values:

$$ (0 \text{ m/s})^2 = (20 \text{ m/s})^2 + 2(-10 \text{ m/s}^2)(y - 25.0 \text{ m}) $$ $$ 0 = 400 \text{ m}^2/\text{s}^2 - 20 \text{ m/s}^2 (y - 25.0 \text{ m}) $$ $$ 20 (y - 25.0) = 400 $$ $$ y - 25.0 = \frac{400}{20} = 20 \text{ m} $$

The displacement from the launch point to the highest point is 20 m upwards.

The highest point reached from the ground is $y = y_0 + 20 \text{ m} = 25.0 \text{ m} + 20 \text{ m} = 45.0$ m.

The ball will rise 20 m above the point of launch, reaching a height of 45.0 m from the ground.

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(b) How long will it be before the ball hits the ground?

The ball hits the ground when its position is $y = 0$ m.

We can use the kinematic equation relating $y, y_0, v_0, a,$ and $t$: $y = y_0 + v_0 t + \frac{1}{2}at^2$.

Substitute the known values:

$$ 0 \text{ m} = 25.0 \text{ m} + (20 \text{ m/s})t + \frac{1}{2}(-10 \text{ m/s}^2)t^2 $$ $$ 0 = 25 + 20t - 5t^2 $$

Rearrange into a standard quadratic equation ($At^2 + Bt + C = 0$):

$$ 5t^2 - 20t - 25 = 0 $$

Divide by 5:

$$ t^2 - 4t - 5 = 0 $$

Factor the quadratic equation:

$$ (t - 5)(t + 1) = 0 $$

The possible solutions for $t$ are $t = 5$ s and $t = -1$ s.

Since time cannot be negative in this context (motion starts at $t=0$), the physically meaningful solution is $t = 5$ s.

The ball will hit the ground after 5 seconds.

Note: This method accounts for the entire motion (up and down) in a single step because the acceleration is constant throughout.

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Free fall describes the motion of an object when the only force acting on it is gravity. This is an idealization where air resistance is neglected.

Near the Earth's surface, the acceleration due to gravity ($g$) is approximately constant. Assuming the height of the fall is much smaller than the Earth's radius, we can consider free fall as motion with uniform acceleration.

Let's choose the y-axis to describe vertical motion. It is conventional to choose the positive y-direction upwards. Since gravity always pulls downwards, the acceleration due to gravity is in the negative y-direction.

Acceleration $a = -g$. The standard value of $g$ is approximately 9.8 m/s$^2$. So, $a = -9.8$ m/s$^2$.

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Consider an object released from rest from a height (let's set the initial position $y_0 = 0$ and initial velocity $v_0 = 0$). The kinematic equations for uniformly accelerated motion ($v = v_0 + at$, $y = y_0 + v_0 t + \frac{1}{2}at^2$, $v^2 = v_0^2 + 2a(y - y_0)$) can be applied with $v_0=0$, $y_0=0$, and $a=-g$:

1. Velocity-time: $v = 0 + (-g)t \implies \mathbf{v = -gt}$
2. Position-time: $y = 0 + (0)t + \frac{1}{2}(-g)t^2 \implies \mathbf{y = -\frac{1}{2}gt^2}$
3. Velocity-displacement: $v^2 = (0)^2 + 2(-g)(y - 0) \implies \mathbf{v^2 = -2gy}$

These equations show how the velocity and position of a freely falling object change with time, and how velocity changes with vertical position.

Since positive y is upwards, the negative sign in these equations indicates downward direction. Velocity is negative (downwards) and increasing in magnitude as time increases ($v$ becomes more negative). Position is negative (below the starting point $y=0$) and becomes more negative as time increases (object falls downwards).

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The graphs of acceleration, velocity, and position versus time for free fall (with $v_0=0, y_0=0$, and positive y upwards) are shown below:

Note that the acceleration is constant and negative (straight horizontal line on a-t graph). The velocity is linearly decreasing (becoming more negative) with time (straight line with negative slope on v-t graph). The position-time graph is a downward-opening parabola.

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The reaction time is the time it takes for your brain to process the visual signal of the ruler dropping and send the signal to your hand to catch it. During this reaction time ($t_r$), the ruler is falling freely under the influence of gravity.

Let's assume the ruler starts from rest ($v_0 = 0$) when it is dropped. The acceleration is due to gravity, acting downwards. If we take the downward direction as positive, then $a = +g = +9.8$ m/s$^2$. Let the distance the ruler falls during the reaction time be $d$.

We can use the kinematic equation relating displacement, initial velocity, acceleration, and time: $d = v_0 t_r + \frac{1}{2}at_r^2$.

Substitute $v_0=0$ and $a=g$ (taking downward as positive):

$$ d = (0)t_r + \frac{1}{2}gt_r^2 $$ $$ d = \frac{1}{2}gt_r^2 $$

Solve for the reaction time $t_r$:

$$ t_r^2 = \frac{2d}{g} $$ $$ t_r = \sqrt{\frac{2d}{g}} $$

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Given the distance fallen $d = 21.0$ cm. Convert this to meters: $d = 21.0 \times 10^{-2}$ m = 0.210 m.

Use $g = 9.8$ m/s$^2$.

$$ t_r = \sqrt{\frac{2 \times 0.210 \text{ m}}{9.8 \text{ m/s}^2}} $$ $$ t_r = \sqrt{\frac{0.420}{9.8}} \text{ s} $$ $$ t_r = \sqrt{0.042857...} \text{ s} $$ $$ t_r \approx 0.2069... \text{ s} $$

Rounding to two significant figures (based on 21.0 cm having 3 sig figs, but 9.8 having 2), or perhaps 3 based on 21.0, let's keep 3 sig figs for the result since 21.0 suggests precision to 0.1 cm.

$$ t_r \approx 0.207 \text{ s} $$

So, the estimated reaction time is about 0.207 seconds, or roughly 210 milliseconds.

Relative Velocity

The velocity of an object is always measured with respect to a specific frame of reference. Often, we are interested in the velocity of one object as observed from another moving object. This is described by the concept of relative velocity.

Consider two objects, A and B, moving along a straight line (say, the x-axis) with constant velocities $v_A$ and $v_B$, respectively, as measured with respect to a stationary frame of reference (like the ground).

If their initial positions at time $t=0$ are $x_A(0)$ and $x_B(0)$, their positions at time $t$ are:

The position of object B relative to object A is given by $x_{BA}(t) = x_B(t) - x_A(t)$.

This equation shows that the position of B relative to A changes linearly with time, with a constant rate $(v_B - v_A)$. This rate of change is the velocity of object B relative to object A.

The velocity of object B relative to object A ($v_{BA}$) is defined as:

Similarly, the velocity of object A relative to object B ($v_{AB}$) is:

From these definitions, it's clear that $v_{BA} = -v_{AB}$.

Special Cases for Relative Velocity in 1D:

Assume motion is along the x-axis. $v_A$ and $v_B$ have signs indicating their direction along the axis.

1. Objects moving in the same direction:
   • If $v_A > 0$ and $v_B > 0$.
   • If $v_A = v_B$, then $v_{BA} = v_B - v_A = 0$. The relative velocity is zero. The distance between the objects remains constant. Their position-time graphs are parallel straight lines.
   • If $v_A > v_B$ (A is faster than B), then $v_{BA} = v_B - v_A$ is negative. B appears to move backward relative to A. A will eventually overtake B. Their x-t graphs are straight lines with different slopes, with A's slope greater than B's, and they intersect when A overtakes B.
   • If $v_B > v_A$ (B is faster than A), then $v_{BA} = v_B - v_A$ is positive. B appears to move forward relative to A. B will eventually overtake A.
2. Objects moving in opposite directions:
   • If $v_A > 0$ and $v_B < 0$. The objects are approaching each other.
   • Then $v_{BA} = v_B - v_A = (\text{negative value}) - (\text{positive value})$, which results in a larger negative magnitude. The magnitude of the relative velocity is the sum of their speeds ($|v_B| + |v_A|$). They will meet at some point. Their x-t graphs are straight lines with slopes of opposite signs, which intersect at the time and position where they meet.

The concept of relative velocity is useful in analyzing situations where motion is viewed from a moving reference frame.

The relative velocity definition ($v_{BA} = v_B - v_A$) also applies to instantaneous velocities if $v_A$ and $v_B$ are the instantaneous velocities of objects A and B.

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Question 3.2. The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below ;

(a) (A/B) lives closer to the school than (B/A)

(b) (A/B) starts from the school earlier than (B/A)

(c) (A/B) walks faster than (B/A)

(d) A and B reach home at the (same/different) time

(e) (A/B) overtakes (B/A) on the road (once/twice).

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Question 3.16. Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

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Question 3.17. Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0 ? If not, suggest a suitable physical context for this graph.

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Question 3.19. Suggest a suitable physical situation for each of the following graphs (Fig 3.22):

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Question 3.20. Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.

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Question 3.21. Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.

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Question 3.22. Figure 3.25 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 ?

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Question 3.25. On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h$^{–1}$ (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h$^{–1}$. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt ?.

(b) speed of the child running opposite to the direction of motion of the belt ?

(c) time taken by the child in (a) and (b) ?

Which of the answers alter if motion is viewed by one of the parents ?

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Question 3.26. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s$^{–1}$ and 30 m s$^{–1}$. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s$^{–2}$. Give the equations for the linear and curved parts of the plot.

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Question 3.27. The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b) ?

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Question 3.28. The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 :

Which of the following formulae are correct for describing the motion of the particle over the time-interval t$_1$ to t$_2$ :

(a) $x(t_2) = x(t_1) + v(t_1)(t_2 – t_1) + (1/2) a (t_2 – t_1)^2$

(b) $v(t_2) = v(t_1) + a(t_2 – t_1)$

(c) $v_{average} = (x(t_2) – x(t_1))/(t_2 – t_1)$

(d) $a_{average} = (v(t_2) – v(t_1))/(t_2 – t_1)$

(e) $x(t_2) = x(t_1) + v_{average}(t_2 – t_1) + (1/2) a_{average} (t_2 – t_1)^2$

(f) $x(t_2) – x(t_1)$ = area under the v-t curve bounded by the t-axis and the dotted line shown.

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