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Chapter 3 Motion In A Straight Line
Introduction
Motion is a fundamental characteristic of the universe. From the microscopic flow of blood in our veins to the macroscopic movement of galaxies, everything is in a state of motion. Motion is defined as the change in the position of an object with time. The branch of physics that studies motion is called Mechanics, which is further divided into Kinematics and Dynamics.
In this chapter, we will focus on Kinematics, which is the study of motion without considering the forces that cause it. We will learn how to describe motion using concepts like velocity and acceleration. Our study will be confined to motion along a straight line, also known as rectilinear motion.
The Concept of a Point Object
In many situations, the size of an object is negligible compared to the distance it travels. In such cases, we can approximate the object as a point object or a particle. This simplification allows us to describe its motion without worrying about the complexities of its size, shape, or rotation.
For example, when studying the revolution of the Earth around the Sun, the Earth's diameter is insignificant compared to the radius of its orbit, so it can be treated as a point object.
Position, Path Length and Displacement
Frame of Reference
To describe the position of an object, we need a frame of reference. A frame of reference consists of:
- A reference point or origin (O).
- A set of coordinate axes (e.g., three mutually perpendicular axes X, Y, and Z).
- A clock to measure time.
An object is said to be in motion if one or more of its coordinates change with time. If its coordinates do not change with time, it is said to be at rest with respect to that frame of reference.
For rectilinear motion, we can use a single axis, say the X-axis, which coincides with the object's path. Positions to the right of the origin are taken as positive, and positions to the left are taken as negative.
Path Length
Path length is the total length of the path traversed by an object during its motion. It is the actual distance covered by the object.
- It is a scalar quantity, meaning it has only magnitude and no direction.
- It can never be negative or zero (if the object has moved).
Example: If a car moves from origin O to point P (+360 m) and then back to point Q (+240 m), the path length is:
$ \text{Path Length} = OP + PQ = 360 \text{ m} + (360 - 240) \text{ m} = 360 \text{ m} + 120 \text{ m} = 480 \text{ m} $
Displacement
Displacement is defined as the change in the position of an object. It is the shortest distance between the initial and final positions.
- It is a vector quantity, having both magnitude and direction. In one-dimensional motion, direction can be indicated by a positive (+) or negative (-) sign.
- Formula: If $x_1$ and $x_2$ are the initial and final positions at times $t_1$ and $t_2$ respectively, the displacement $\Delta x$ is:
$ \Delta x = x_2 - x_1 $
Example: Using the same car motion from O to P and then to Q:
The displacement is calculated from the initial position (O, where $x_1=0$) to the final position (Q, where $x_2 = +240$ m).
$ \text{Displacement} = \Delta x = x_2 - x_1 = (+240 \text{ m}) - (0 \text{ m}) = +240 \text{ m} $
Here, the magnitude of displacement (240 m) is not equal to the path length (480 m). The magnitude of displacement is either less than or equal to the path length.
Position-Time Graph (x-t Graph)
A graph plotted with time on the x-axis and position on the y-axis is called a position-time graph. It is a powerful tool for analyzing motion.
- Object at Rest: The position-time graph is a straight line parallel to the time axis.
- Object in Uniform Motion: The position-time graph is a straight line inclined to the time axis.
Average Velocity and Average Speed
Average Velocity
Average velocity is defined as the ratio of the displacement to the time interval in which the displacement occurs. It tells us how fast, on average, an object's position has changed and in what direction.
- It is a vector quantity.
- Formula: $ \bar{v} = \frac{\text{Displacement}}{\text{Time Interval}} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t} $
- The SI unit is m/s or m s$^{-1}$.
- It can be positive, negative, or zero.
- Graphically, the average velocity is the slope of the secant line connecting the initial and final points on a position-time graph.
Average Speed
Average speed is defined as the ratio of the total path length travelled to the total time interval.
- It is a scalar quantity.
- Formula: $ \text{Average Speed} = \frac{\text{Total Path Length}}{\text{Total Time Interval}} $
- It is always positive for a moving object.
The magnitude of average velocity is generally less than the average speed. It is equal to the average speed only when the object moves along a straight line in the same direction.
Example 1. A car is moving along a straight line. It moves from O to P in 18 s and returns from P to Q in 6.0 s. The position of P is +360 m and Q is +240 m from O. What are the average velocity and average speed of the car in going (a) from O to P? and (b) from O to P and back to Q?
Answer:
(a) Motion from O to P:
Displacement $\Delta x = 360 \text{ m} - 0 \text{ m} = +360 \text{ m}$
Path Length = $360 \text{ m}$
Time Interval $\Delta t = 18 \text{ s}$
Average Velocity $\bar{v} = \frac{\Delta x}{\Delta t} = \frac{+360 \text{ m}}{18 \text{ s}} = +20 \text{ m s}^{-1}$
Average Speed = $\frac{\text{Path Length}}{\Delta t} = \frac{360 \text{ m}}{18 \text{ s}} = 20 \text{ m s}^{-1}$
In this case, magnitude of average velocity equals average speed.
(b) Motion from O to P and back to Q:
Final Position is Q, so displacement $\Delta x = 240 \text{ m} - 0 \text{ m} = +240 \text{ m}$
Path Length = OP + PQ = $360 \text{ m} + 120 \text{ m} = 480 \text{ m}$
Total Time Interval $\Delta t = 18 \text{ s} + 6.0 \text{ s} = 24 \text{ s}$
Average Velocity $\bar{v} = \frac{\Delta x}{\Delta t} = \frac{+240 \text{ m}}{24 \text{ s}} = +10 \text{ m s}^{-1}$
Average Speed = $\frac{\text{Path Length}}{\Delta t} = \frac{480 \text{ m}}{24 \text{ s}} = 20 \text{ m s}^{-1}$
Here, the average speed is not equal to the magnitude of the average velocity.
Instantaneous Velocity and Speed
Instantaneous Velocity
The average velocity gives an overall picture of motion over an interval, but not the velocity at a specific moment. The velocity of an object at a particular instant of time is called its instantaneous velocity (or simply, velocity).
It is defined as the limit of the average velocity as the time interval $\Delta t$ approaches zero.
- Formula (Calculus): $ v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} $
- Instantaneous velocity is the first derivative of position with respect to time.
- Graphically, it is the slope of the tangent to the position-time (x-t) graph at that particular instant.
Instantaneous Speed
Instantaneous speed, or simply speed, is the magnitude of the instantaneous velocity. For example, velocities of +24.0 m/s and -24.0 m/s both correspond to a speed of 24.0 m/s.
Example 2. The position of an object moving along the x-axis is given by $x = a + bt^2$, where $a = 8.5$ m, $b = 2.5 \text{ m s}^{-2}$, and t is measured in seconds. What is its velocity at $t = 0$ s and $t = 2.0$ s? What is the average velocity between $t = 2.0$ s and $t = 4.0$ s?
Answer:
Instantaneous Velocity:
The velocity $v$ is the derivative of position $x$ with respect to time $t$.
$ v = \frac{dx}{dt} = \frac{d}{dt}(a + bt^2) = 2bt $
Substituting the value of $b = 2.5 \text{ m s}^{-2}$, we get $v = 2(2.5)t = 5.0t \text{ m s}^{-1}$.
At $t = 0 \text{ s}$, $v = 5.0(0) = 0 \text{ m s}^{-1}$.
At $t = 2.0 \text{ s}$, $v = 5.0(2.0) = 10 \text{ m s}^{-1}$.
Average Velocity:
To find the average velocity between $t=2.0$ s and $t=4.0$ s, we first find the positions at these times.
$x(t=2.0\text{ s}) = a + b(2.0)^2 = 8.5 + 2.5(4) = 18.5 \text{ m}$.
$x(t=4.0\text{ s}) = a + b(4.0)^2 = 8.5 + 2.5(16) = 48.5 \text{ m}$.
Now, use the average velocity formula:
$\bar{v} = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{48.5 - 18.5}{2.0} \text{ m s}^{-1} = \frac{30.0}{2.0} \text{ m s}^{-1} = 15 \text{ m s}^{-1}$.
Acceleration
When the velocity of an object changes with time, the object is said to be accelerating. Acceleration is the rate of change of velocity with time.
Average Acceleration
Average acceleration is the change in velocity divided by the time interval over which the change occurs.
- Formula: $ \bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t} $
- SI unit is m/s$^2$ or m s$^{-2}$.
- It is a vector quantity.
- Graphically, it is the slope of the secant line on a velocity-time (v-t) graph.
Instantaneous Acceleration
Instantaneous acceleration is the acceleration of an object at a particular instant of time.
- Formula (Calculus): $ a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{dx}{dt}\right) = \frac{d^2x}{dt^2} $
- It is the first derivative of velocity with respect to time, or the second derivative of position with respect to time.
- Graphically, it is the slope of the tangent to the velocity-time (v-t) graph at that instant.
Acceleration can be positive, negative, or zero.
- Positive acceleration: Velocity is increasing in the positive direction.
- Negative acceleration (retardation or deceleration): Velocity is decreasing in the positive direction or increasing in the negative direction.
- Zero acceleration: Velocity is constant (uniform motion).
Area under the Velocity-Time Graph
An important feature of a v-t graph is that the area under the curve represents the displacement of the object over that time interval.
For an object moving with constant velocity $u$, the v-t graph is a horizontal line. The area under the graph from $t=0$ to $t=T$ is the area of a rectangle, which is $u \times T$. This is precisely the displacement ($x = ut$).
Kinematic Equations for Uniformly Accelerated Motion
For motion in a straight line with constant acceleration ($a$), a set of simple equations, known as the equations of motion, can be derived. These equations relate displacement ($x$), time ($t$), initial velocity ($v_0$), final velocity ($v$), and acceleration ($a$).
Derivation of the Equations
Let's assume an object starts at position $x_0=0$ at time $t=0$ with an initial velocity $v_0$. After time $t$, its final velocity is $v$ and its position is $x$.
1. First Equation of Motion: $v = v_0 + at$
By definition, constant acceleration is $a = \frac{v - v_0}{t - 0}$.
$at = v - v_0$
$v = v_0 + at$
2. Second Equation of Motion: $x = v_0t + \frac{1}{2}at^2$
The displacement $x$ is the area under the velocity-time graph. For uniform acceleration, this graph is a straight line, and the area is a trapezium.
Area = Area of rectangle OACD + Area of triangle ABC
$x = (v_0)(t) + \frac{1}{2}(v - v_0)(t)$
From the first equation, we know $(v - v_0) = at$. Substituting this:
$x = v_0t + \frac{1}{2}(at)(t)$
$x = v_0t + \frac{1}{2}at^2$
3. Third Equation of Motion: $v^2 = v_0^2 + 2ax$
We can also express the displacement $x$ as the area of the trapezium using the formula for average velocity:
$x = (\text{average velocity}) \times t = \left(\frac{v_0 + v}{2}\right)t$
From the first equation, we can write $t = \frac{v - v_0}{a}$. Substituting this for $t$:
$x = \left(\frac{v + v_0}{2}\right) \left(\frac{v - v_0}{a}\right) = \frac{v^2 - v_0^2}{2a}$
$2ax = v^2 - v_0^2$
$v^2 = v_0^2 + 2ax$
General Equations of Motion
If the initial position is $x_0$ instead of 0, the equations are modified by replacing $x$ with $(x - x_0)$:
- $v = v_0 + at$
- $x = x_0 + v_0t + \frac{1}{2}at^2$
- $v^2 = v_0^2 + 2a(x - x_0)$
Example 3. A ball is thrown vertically upwards with a velocity of 20 m s$^{-1}$ from the top of a multistorey building. The point of launch is 25.0 m from the ground. (a) How high will the ball rise? (b) How long will it be before the ball hits the ground? Take $g = 10 \text{ m s}^{-2}$.
Answer:
Let's set the origin (y=0) at the ground and take the upward direction as positive.
Given: Initial position $y_0 = +25$ m, initial velocity $v_0 = +20 \text{ m s}^{-1}$, acceleration $a = -g = -10 \text{ m s}^{-2}$.
(a) How high will the ball rise?
At the maximum height, the final velocity $v=0$. Using the third equation of motion:
$v^2 = v_0^2 + 2a(y - y_0)$
$0^2 = (20)^2 + 2(-10)(y - 25)$
$0 = 400 - 20(y - 25)$
$20(y - 25) = 400 \implies y - 25 = 20$
So, the height risen from the launch point is 20 m. The maximum height from the ground is $y = 25 + 20 = 45$ m.
(b) How long before the ball hits the ground?
When the ball hits the ground, its final position is $y = 0$. Using the second equation of motion:
$y = y_0 + v_0t + \frac{1}{2}at^2$
$0 = 25 + (20)t + \frac{1}{2}(-10)t^2$
$0 = 25 + 20t - 5t^2$
Dividing by 5, we get $t^2 - 4t - 5 = 0$.
Factoring the quadratic equation: $(t-5)(t+1) = 0$.
The solutions are $t=5$ s and $t=-1$ s. Since time cannot be negative, the answer is t = 5 s.
Example 4. Galileo’s law of odd numbers: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [1: 3: 5: 7…...]." Prove it.
Answer:
Let a body fall from rest, so $v_0 = 0$. The distance traversed in time $t$ is given by $y = \frac{1}{2}gt^2$ (taking downward as positive for simplicity).
Let's calculate the distance traversed in successive equal time intervals of $\tau$.
Distance in 1st interval (from $t=0$ to $t=\tau$):
$d_1 = y(\tau) - y(0) = \frac{1}{2}g\tau^2 - 0 = \frac{1}{2}g\tau^2$
Distance in 2nd interval (from $t=\tau$ to $t=2\tau$):
$d_2 = y(2\tau) - y(\tau) = \frac{1}{2}g(2\tau)^2 - \frac{1}{2}g\tau^2 = \frac{1}{2}g(4\tau^2 - \tau^2) = 3\left(\frac{1}{2}g\tau^2\right)$
Distance in 3rd interval (from $t=2\tau$ to $t=3\tau$):
$d_3 = y(3\tau) - y(2\tau) = \frac{1}{2}g(3\tau)^2 - \frac{1}{2}g(2\tau)^2 = \frac{1}{2}g(9\tau^2 - 4\tau^2) = 5\left(\frac{1}{2}g\tau^2\right)$
The ratio of the distances traversed in successive intervals is:
$d_1 : d_2 : d_3 : \dots = \frac{1}{2}g\tau^2 : 3\left(\frac{1}{2}g\tau^2\right) : 5\left(\frac{1}{2}g\tau^2\right) : \dots$
$d_1 : d_2 : d_3 : \dots = 1 : 3 : 5 : \dots$
This proves Galileo's law of odd numbers.
Relative Velocity
The concept of velocity is relative. The velocity of an object depends on the frame of reference from which it is observed. To understand this, we introduce the concept of relative velocity.
Consider two objects A and B moving along a straight line (x-axis) with uniform velocities $v_A$ and $v_B$ with respect to the ground.
Their positions at time $t$ are:
$x_A(t) = x_A(0) + v_At$
$x_B(t) = x_B(0) + v_Bt$
The displacement from object A to object B at time $t$ is:
$x_{BA}(t) = x_B(t) - x_A(t) = [x_B(0) - x_A(0)] + (v_B - v_A)t$
This equation shows that, as seen from object A, object B's position changes by an amount $(v_B - v_A)$ each second. Therefore, the velocity of B relative to A is:
$v_{BA} = v_B - v_A$
Similarly, the velocity of A relative to B is:
$v_{AB} = v_A - v_B$
From this, we can see that $v_{BA} = -v_{AB}$.
Special Cases
- If $v_A = v_B$: Then $v_{BA} = 0$. The two objects remain at a constant distance from each other. To an observer on A, B appears to be at rest. Their position-time graphs are parallel straight lines.
- If $v_A > v_B$ (moving in same direction): Then $v_{BA}$ is negative. Object A will eventually overtake object B.
- If $v_A$ and $v_B$ have opposite signs: The magnitude of relative velocity, $|v_B - v_A|$, will be greater than both $|v_A|$ and $|v_B|$. The objects will appear to move very fast with respect to each other.
Example 5. Two parallel rail tracks run north-south. Train A moves north with a speed of 54 km h$^{-1}$, and train B moves south with a speed of 90 km h$^{-1}$. What is the (a) velocity of B with respect to A? (b) velocity of ground with respect to B? and (c) velocity of a monkey running on the roof of train A against its motion (with a velocity of 18 km h$^{-1}$ with respect to train A) as observed by a man standing on the ground?
Answer:
Let's choose the north direction as positive.
First, convert velocities to m/s:
$v_A = +54 \text{ km/h} = +54 \times \frac{5}{18} \text{ m/s} = +15 \text{ m/s}$
$v_B = -90 \text{ km/h} = -90 \times \frac{5}{18} \text{ m/s} = -25 \text{ m/s}$
(a) Velocity of B with respect to A ($v_{BA}$):
$v_{BA} = v_B - v_A = (-25) - (15) = -40 \text{ m/s}$.
Train B appears to A to move with a speed of 40 m/s towards the south.
(b) Velocity of ground with respect to B ($v_{GB}$):
Velocity of ground, $v_G = 0$.
$v_{GB} = v_G - v_B = 0 - (-25) = +25 \text{ m/s}$.
The ground appears to move with a speed of 25 m/s towards the north for an observer on train B.
(c) Velocity of the monkey with respect to ground ($v_M$):
Let the velocity of the monkey with respect to train A be $v_{MA}$. Since it's against the motion of A, it is in the south direction.
$v_{MA} = -18 \text{ km/h} = -18 \times \frac{5}{18} \text{ m/s} = -5 \text{ m/s}$.
We know that $v_{MA} = v_M - v_A$.
So, $v_M = v_{MA} + v_A = (-5) + (15) = +10 \text{ m/s}$.
The man on the ground observes the monkey moving with a velocity of 10 m/s towards the north.
Elements of Calculus
Calculus is a powerful branch of mathematics that deals with continuous change. It has two major branches: Differential Calculus and Integral Calculus. These tools are essential for defining and working with physical quantities like instantaneous velocity and acceleration.
Differential Calculus
Differential calculus is concerned with the rate at which quantities change. The core concept is the derivative.
If a quantity $y$ is a function of another quantity $x$, written as $y = f(x)$, the derivative of $y$ with respect to $x$ (denoted as $dy/dx$) represents the instantaneous rate of change of $y$ with respect to $x$.
Graphical Interpretation of the Derivative
Graphically, the derivative of a function at a point is the slope of the tangent line to the function's graph at that point. It is found by taking the limit of the slope of a secant line between two points as the distance between the points approaches zero.
The derivative is formally defined as the limit:
$ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} $
Application in Physics
In kinematics, derivatives are used to define instantaneous velocity and acceleration:
- Instantaneous Velocity ($v$): It is the time derivative of position ($x$).
- Instantaneous Acceleration ($a$): It is the time derivative of velocity ($v$), or the second time derivative of position ($x$).
$ v = \frac{dx}{dt} $
$ a = \frac{dv}{dt} = \frac{d^2x}{dt^2} $
Common Derivatives
Below are some fundamental rules and derivatives of common functions. Here, $u$ and $v$ are functions of $x$, and $a$ and $n$ are constants.
Rules of Differentiation:
- Constant Rule: $\frac{d}{dx}(a) = 0$
- Product Rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$
- Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
| Function $f(x)$ | $x^n$ | $\sin(x)$ | $\cos(x)$ | $\tan(x)$ | $e^x$ | $\ln(x)$ |
| Derivative $\frac{df}{dx}$ | $nx^{n-1}$ | $\cos(x)$ | $-\sin(x)$ | $\sec^2(x)$ | $e^x$ | $\frac{1}{x}$ |
Integral Calculus
Integral calculus is, in essence, the reverse process of differentiation. It is primarily used to find the area under a curve. This is useful for calculating quantities that are the result of accumulation, such as displacement from velocity or work done by a variable force.
The area under the curve of a function $f(x)$ from $x=a$ to $x=b$ is found by summing the areas of an infinite number of infinitesimally thin rectangles. This limit of a sum is called a definite integral and is denoted by:
$ \text{Area} = \int_{a}^{b} f(x)dx $
The symbol $\int$ is an elongated 'S', signifying a sum.
The Fundamental Theorem of Calculus
This theorem establishes the inverse relationship between differentiation and integration. It states that if the derivative of a function $g(x)$ is $f(x)$, then the definite integral of $f(x)$ from $a$ to $b$ can be easily evaluated:
If $\frac{d}{dx}g(x) = f(x)$, then:
$ \int_{a}^{b} f(x)dx = [g(x)]_{a}^{b} = g(b) - g(a) $
The function $g(x)$ is called the indefinite integral of $f(x)$.
Example. Calculate the value of the definite integral $\int_{1}^{2} x^2 dx$.
Answer:
First, we need to find the indefinite integral of $x^2$. The function whose derivative is $x^2$ is $g(x) = \frac{x^3}{3}$.
Now, we apply the fundamental theorem:
$\int_{1}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_{1}^{2} = g(2) - g(1)$
$\int_{1}^{2} x^2 dx = \frac{(2)^3}{3} - \frac{(1)^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$
Common Integrals
| Function $f(x)$ | $x^n$ (for $n \neq -1$) | $x^{-1}$ or $\frac{1}{x}$ | $\sin(x)$ | $\cos(x)$ | $e^x$ |
| Indefinite Integral $\int f(x)dx$ | $\frac{x^{n+1}}{n+1} + C$ | $\ln|x| + C$ | $-\cos(x) + C$ | $\sin(x) + C$ | $e^x + C$ |
Note: The constant 'C' is the constant of integration, which is added because the derivative of any constant is zero. For definite integrals, this constant cancels out.
Exercises
Question 3.1. In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
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).
Answer:
Question 3.3. A woman starts from her home at 9.00 am, walks with a speed of 5 km h$^{–1}$ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h$^{–1}$. Choose suitable scales and plot the x-t graph of her motion.
Answer:
Question 3.4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
Question 3.5. A jet airplane travelling at the speed of 500 km h$^{–1}$ ejects its products of combustion at the speed of 1500 km h$^{–1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
Answer:
Question 3.6. A car moving along a straight highway with speed of 126 km h$^{–1}$ is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?
Answer:
Question 3.7. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h$^{–1}$ in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s$^{–2}$. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?
Answer:
Question 3.8. On a two-lane road, car A is travelling with a speed of 36 km h$^{–1}$. Two cars B and C approach car A in opposite directions with a speed of 54 km h$^{–1}$ each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?
Answer:
Question 3.9. Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h$^{–1}$ in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:
Question 3.10. A player throws a a ball upwards with an initial speed of 29.4 m s$^{–1}$.
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s$^{–2}$ and neglect air resistance).
Answer:
Question 3.11. Read each statement below carefully and state with reasons and examples, if it is true or false ;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
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Question 3.12. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
Question 3.13. Explain clearly, with examples, the distinction between :
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
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Question 3.14. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h$^{–1}$. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h$^{–1}$. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Answer:
Question 3.15. In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ?
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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.
Answer:
Question 3.18. A police van moving on a highway with a speed of 30 km h$^{–1}$ fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h$^{–1}$. If the muzzle speed of the bullet is 150 m s$^{–1}$, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car).
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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.
Answer:
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.
Answer:
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 ?
Answer:
Additional Exercises
Question 3.23. A three-wheeler starts from rest, accelerates uniformly with 1 m s$^{–2}$ on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n$^{th}$ second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ?
Answer:
Question 3.24. A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s$^{–1}$. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s$^{-1}$ and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ?
Answer:
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 ?
Answer:
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) ?
Answer:
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.
Answer: