Measurement of Time Using Periodic Motion

Measurement of Time Using Periodic Motion

Any object that is moving is said to be in motion. In order to measure time, we need a motion that repeats itself in equal intervals. Such a motion is called periodic motion.

Some examples of periodic motion are the rotation of the Earth about its axis, the revolution of the Earth around the sun, the revolution of the moon around the Earth, the to-and-fro movement of a spring, and the oscillation of a pendulum. These types of periodic events are used to make clocks and calendars. Some of the first accurate clocks were based on the periodic movement of a pendulum.

Simple Pendulum

A small mass that is suspendedfrom a fixed point and allowed to swing freely under the influence of gravity is called a pendulum.

An ideal, simple pendulum consists of a small mass (like a stone or a metal ball) called a bob suspended by a string. When the bob moves from one position and returns to the same position, it is said to complete one oscillation. The time taken to complete one oscillation is called the time period of the pendulum. For example, if the bob starts from point A, goes to points B and C, and returns to point A, it completes one oscillation.

Measurement of Time Using Periodic Motion 1
A Simple Pendulum

By studying the oscillations of a simple pendulum, the observations can be summarized as follows:

  • The time taken by a pendulum to complete one oscillation (i.e., its time period) does not depend on the extent to which the bob of the pendulum is displaced.
  • The time period does not depend on the mass of the bob used (within reasonable limits).
  • The time period depends on the length of the string or wire used; greater the length of the string, greater is the time period (assuming that the string itself has negligible mass).

This means that if the length of the pendulum is fixed, its time period is constant. This was a wonderful discovery. Scientists began to build new clocks based on this property of the pendulum.
How fast or slow an object travels depend on the time it takes to travel a certain distance, it is closely related to the measurement of time. We will now discuss slow and fast motion, and how we can measure it.

Activity

Aim: To show the constancy of the time period of a pendulum.
Materials needed: A string, a small stone, a doorknob/handle, and a stopwatch.
Method:

  1. Take a string of length about 30 cm. Tie the stone to one end. Tie the other end of the string to a doorknob/handle.
  2. Displace the stone slightly and let it oscillate. Make sure it oscillates freely and does not rub against any object or the door.
  3. Use the stopwatch and note the time taken by the stone to complete ten oscillations. [Take any one point, the extreme position in the oscillation or the mean (middle) position of the stone and count the number of times the stone crosses this point, in the same direction.
  4. Divide the time taken for ten oscillations by 10. This gives the time period of the pendulum.]
  5. Repeat the experiment many times without changing the arrangement. Compare the value of the time periods obtained.

Observation: The time period is the same in each case.
Conclusion: This shows the constancy of the time period of the pendulum.

How to Find Acceleration Using a Velocity Time Graph

What Is Acceleration

 

  • Rate of change of velocity is called acceleration. It is a vector quantity
    \(i.e.\text{    }a=\frac{v-u}{t}\)
    where u is the initial velocity of the object, v is its final velocity and t is the time taken.
  • Unit of acceleration = m/s2 or ms-2
  • If the velocity of a body decreases, then it will experience a negative acceleration which is called deceleration or retardation.
  • Figure shows a car moving along a straight line. The speedometer of the car shows that it is moving with increasing velocity. The car is accelerating.
    Analysing Linear Motion 3
  • We say that an object is undergoing a deceleration or retardation when it is slowing down. The rate of change of velocity of the object then has a negative value. Figure shows a car decelerating. The speedometer of the car shows that it is moving with decreasing velocity.

Analysing Linear Motion 4

Types of acceleration

Uniform acceleration: If a body travels in a straight line and its velocity increases by equal amounts in equal intervals of time then it is said to be in state of uniform acceleration.
e.g. motion of a freely falling body.

Non uniform acceleration: A body has a non-uniform acceleration if its velocity increases by unequal amounts in equal intervals of time.

Instantaneous acceleration: The acceleration of a body at any instant is called its instantaneous acceleration.

Acceleration is determined by the slope of time-velocity graph.
\(\tan \theta =\frac{dv}{dt}\)

  1. If the time velocity graph is a straight line, acceleration remains constant.
  2. If the slope of the straight line is positive, positive acceleration occurs.
  3. If the slope of the straight line is negative, negative acceleration or retardation occurs.

Analysing MotionAnalysing Motion 1

  1. Figure shows a set-up of apparatus to analyse motion in the laboratory.
  2. (a) A ticker timer is an apparatus that gives a permanent record of motion for further analysis. When connected to an alternating current (a.c.) power supply (usually 12 V), it vibrates at a frequency of 50 Hz.
    (b) The ticker timer makes a series of dots at a rate of 50 dots per second on a piece of ticker tape as it is pulled through the timer by a trolley. Therefore, the time interval of a dot and the next dot which is also known as one tick is 1/50 or 0.02 s.
    (c) The distance between two dots is equal to the distance travelled by the trolley during the time interval between the dots.
    (d) The ticker tape can be analysed to determine the time, displacement, average velocity, acceleration and type of motion of an object.
  3. The ticker tape can be cut into strips of equal time (equal number of ticks) and pasted together to form a chart for analysing the motion of a trolley.
  4. Figure shows three charts formed from strips of ticker tape, each consisting of ten ticks.Analysing Motion 2
  5. For motion with uniform acceleration or deceleration, its value can be determined by analysing the chart. Figure shows a chart formed from strips of ticker tape with ten ticks each.
    Analysing Motion 3
    Analysing Motion 4
    Analysing Motion 5

Acceleration Using A Velocity Time Graph Example Problems With Solutions

Example 1. A van accelerates uniformly from a velocity of 10 m s-1 to 20 m s-1 in 2.5 s. What is the acceleration of the van?
Solution: Initial velocity, u = 10 ms-1
Final velocity, v = 20 ms-1
Time taken, t = 2.5 s
Analysing Linear Motion 7

Example 2. A car travelling at 24 m s-1 slowed down when the traffic light turned red. After undergoing uniform deceleration for 4 s, it stopped in front of the traffic light. Calculate the acceleration of the car.
Solution:
 Initial velocity, u = 24ms-1
Final velocity, v = 0 ms-1
Time taken, t = 4 s
Analysing Linear Motion 8

Example 3. Time-velocity graph of a body is shown in the figure. Find its acceleration in m/s2.
Solution:    As it is clear from the figure,
At t = 0 s, v = 20 m/s
At t = 4 s, v = 80 m/s
Acceleration-Example-Problem
\(therefore \text{Acceleration,}a=\frac{\text{Change}\,\text{in}\,\text{velocity}}{\text{Timeint}\,\text{erval}} \)
\( =\frac{\Delta v}{\Delta t}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}} \)
\( =\frac{(80-20)\,}{(4-0)}=15\text{ m/}{{\text{s}}^{\text{2}}} \)

Example 4. Time-velocity graph of a particle is shown in figure. Find its instantaneous acceleration at following intervals
Acceleration-Example-Problem-1
(i) at t = 3s
(ii) at t = 6s
(iii) at t = 9s
Solution:    (i) Instantaneous acceleration at t = 3s, is given by
a = slope of line AB = zero
(ii) Instantaneous acceleration at t = 6 s, is given by a = slope of line BC
\( =\frac{CM}{BM}=\frac{100-60}{8-4}=\text{ }10\text{ m/}{{\text{s}}^{\text{2}}} \)
(iii) Instantaneous acceleration at t = 9 s, is given by a = slope of line CD
\( =\frac{0-100}{10-8}=-50\text{ m/}{{\text{s}}^{\text{2}}} \)

Example 5. Starting from rest, Deepak paddles his bicycle to attain a velocity of 6 m/s in 30 seconds then he applies brakes so that the velocity of the bicycle comes down to 4 m/s in the next 5 seconds. Calculate the acceleration of the bicycle in both the cases.
Solution:    (i) Initial velocity, u = 0, final velocity,
v = 6 m/s, time, t = 30 s
Using the equation v = u + at, we have
\( a=\frac{v-u}{t} \)
substituting the given values of u, v and t in the above equation, we get
\( a=\frac{6-0}{30}=0.2\text{ m/}{{\text{s}}^{\text{2}}}\text{; }\!\!~\!\!\text{ } \)
which is positive acceleration.
(ii) Initial velocity, u = 6 m/s, final velocity,
v = 4 m/s, time, t = 5 s, then
\( a=\frac{v-u}{t}=\frac{4-6}{5}=-0.4\text{ m/}{{\text{s}}^{\text{2}}}\text{; }\!\!~\!\!\text{ } \)
which is retardation.
Note: The acceleration of the case (i) is positive and is negative in the case (ii).

Example 6. A trolley pulled a ticker tape through a ticker timer while moving down an inclined plane. Figure 2.10 shows the ticker tape produced.
Analysing Motion 6
Determine the average velocity of the trolley.
Solution:
Analysing Motion 7

Example 7. Figure shows ticker tapes produced from the motion of a trolley.
Analysing Motion 8
Describe the type of motion of the trolley for each ticker tape.
Solution:
(a) The distances between two neighbouring dots are the same throughout the tape. Therefore, the trolley moved with uniform velocity.
(b) The distances between two neighbouring dots are increasing. Therefore, the trolley moved with increasing velocity The trolley was accelerating.
(c) The distances between two neighbouring dots are decreasing. Therefore, the trolley moved with decreasing velocity. The trolley was decelerating.
Analysing Motion 9

Example 8. Figure shows a chart representing the movement of a trolley with uniform acceleration.
Analysing Motion 10
Determine its acceleration.
Solution:
Analysing Motion 11

Example 9. A trolley travelled down an inclined plane pulling along a ticker tape. Figure shows a chart formed by cutting and arranging the ticker tape into strips of ten ticks each.
Analysing Motion 12
Determine the acceleration of the trolley.
Solution:
Analysing Motion 13

Example 10. Figure shows a strip of ticker tape depicting the motion of a toy car with uniform acceleration.
Analysing Motion 14
Determine the acceleration of the toy car.
Solution:
Analysing Motion 15

What Is The Difference Between Speed And Velocity

Difference Between Speed And Velocity

  • The ‘distance’ travelled by a body in unit time interval is called its speed. When the position of a body changes in particular direction, then speed is denoted by ‘velocity’. i.e. the rate of change of displacement of a body is called its Velocity.
  • Speed is a scalar quantity while velocity is a vector quantity.
  • \( \text{Speed}=\frac{\text{distance}}{\text{time}} \)
  • \( \text{Velocity}=\frac{\text{displacement}}{\text{time}} \)
  • Unit: In M.K.S. system = ms-1
    In C.G.S. system = cm/s
  • If time distance graph is given then speed can be given by the slope of the line, at given time
    \( \text{V}=\frac{\text{ }\!\!\Delta\!\!\text{ s}}{\text{ }\!\!\Delta\!\!\text{ t}}=\text{Slope} \)
    velocity-time-graph
  • The area of velocity time graph gives displacement travelled.

Types of speed

(a) Average and Instantaneous speed
Average speed: It is obtained by dividing the total distance travelled by the total time interval. i.e.
\( \text{Average speed}=\frac{\text{total}\,\,\text{distance}}{\text{total}\,\,\text{time}} \)
\( \text{Average}\,\text{velocity}=\frac{\text{displacement}}{\text{total}\,\,\text{time}} \)

  • Average speed is a scalar, while average velocity is a vector.
  • For a moving body average speed can never be –ve or zero (unless t → ∞), while average velocity can be i.e
    \( {{v}_{av~}}>0\text{ while }\overset{\to }{\mathop{{{v}_{av}}}}\,>=or<\text{ }0 \)
  • In general average speed is not equal to magnitude of average velocity. However it can be so if the motion is along a straight line without change in direction
  • If a particle travels distances L1, L2, L3 at speeds v1, v2, v3 etc respectively, then
    \( {{v}_{av~}}=\frac{\Delta s}{\Delta t}=\frac{{{L}_{1}}+{{L}_{2}}+…..+{{L}_{n}}}{\frac{{{L}_{1}}}{{{v}_{1}}}+\frac{{{L}_{2}}}{{{v}_{2}}}+….+\frac{{{L}_{n}}}{{{v}_{n}}}}=\frac{\sum{Li}}{\sum{\frac{{{L}_{i}}}{{{v}_{i}}}}} \)
  • If a particle travels at speeds v1, v2 etc for intervals t1, t2 etc respectively, then
    \( {{v}_{av~}}=\frac{{{v}_{1}}{{t}_{1}}+{{v}_{2}}{{t}_{2}}+….}{{{t}_{1}}+{{t}_{2}}+….}=\frac{\sum{{{v}_{1}}{{t}_{1}}}}{\sum{{{t}_{1}}}} \)

Instantaneous speed: The speed of a body at a particular instant of time is called its instantaneous speed.
\( =\underset{\Delta t\to 0}{\mathop{\lim }}\,\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt} \)

(b) Uniform and Non uniform speed
Uniform speed: If an object covers equal distance in equal interval of time, then time speed graph of an object is a straight line parallel to time axis then body is moving with a uniform speed.
Non-uniform speed: If the speed of a body is changing with respect to time it is moving with a non-uniform speed.

Speed And Velocity Example Problems With Solutions

Example 1. The distance between two points A and B is 100 m. A person moves from A to B with a speed of 20 m/s and from B to A with a speed of 25 m/s. Calculate average speed and average velocity.
Solution:    (i) Distance from A to B = 100 m
Distance from B to A = 100 m
Thus, total distance = 200 m
Time taken to move from A to B, is given by
\( {{t}_{1}}=\frac{\text{distance}}{\text{velocity}}=\frac{100}{20}=5\text{ seconds} \)
Time taken from B to A, is given by
\( {{t}_{2}}=\frac{\text{distance}}{\text{velocity}}=\frac{100}{25}=4\text{ seconds} \)
Total time taken = t1 + t2 = 5 + 4 = 9 sec.
∴ Average speed of the person
\( =\frac{\text{Total}\,\text{dis}\,\text{tan}\,\text{cecovered}}{\text{Total}\,\text{time}\,\text{taken}}=\frac{200}{9}=22.2\text{ m/s} \)
(ii) Since person comes back to initial position A, displacement will be zero, resulting zero average velocity.

Example 2. A car moves with a speed of 40 km/hr for first hour, then with a speed of 60 km/hr for next \(1\frac{1}{2}\) half hour and finally with a speed of 30 km/hr for next hours. Calculate the average speed of the car.
Solution:    Distance travelled in first hour, is given by
s1 = speed × time = 40 km/hr × 1 hr = 40 km
Distance travelled in next half an hour, is given by
s2 = speed × time = 60 km/hr × \(\frac { 1 }{ 2 }\) hr = 30 km
Distance travelled in last \(1\frac{1}{2}\) hours, is given by
s3 = speed × time = 30 km/hr × \(\frac { 3 }{ 2 }\) hr = 45 km
Thus, total distance travelled = s1 + s2 + s3
= 40 + 30 + 45 = 115 km
Total time taken = 1 + \(\frac { 1 }{ 2 }\) + \(1\frac{1}{2}\) = 3 hours
Average speed = \(\frac { Total distance covered }{ Total time taken } \) = \(\frac { 115km }{ 3hrs }\)
= 38.33 km/hr

Example 3. Figure shows time distance graph of an object. Calculate the following :
(i) Which part of the graph shows that the body is at rest ?
(ii) Average speed in first 10 s.
(iii) Speeds in different parts of motion.
Speed-And-Velocity-Example
Solution:    (i) The part BC shows that the body is at rest.
(ii) In first 10 seconds, distance travelled = 100m
\( \text{Average speed}=\frac{\text{total}\,\,\text{distance}}{\text{total}\,\,\text{time}} \)
\( =\frac{100}{10}=10\text{ m/s} \)
(iii) Speed of the object in part AB is given by slope = 100/6 = \(\frac { 50 }{ 3 }\) m/s
Speed of object in part BC = 0 m/s
Speed of the object in part CD
\( =\frac{100-40}{12-10}=\frac{60}{2}=30~\text{m/s} \)
Speed of object in part DE
\( =\frac{40-0}{14-12}=\frac{40}{2}=30~\text{m/s} \)

Example 4. Time-velocity graph of a particle is shown in Figure. Calculate the distance travelled in first seconds.
Speed-And-Velocity-Example-1
Solution:    Distance travelled in first 8s is given by area OABCG
= area of rectangle OAMG + area of triangle BMC
= 8 × 60 + \(\frac { 1 }{ 2 }\) × 4 × 40
= 480 + 80 = 560 m.

Example 5. A cow walked along a curved path from P to Q, which is 70 m away from P. Q lies to the south-west of P. The distance travelled by the cow is 240 m and the time taken is 160 s.
Analysing Linear Motion 2
Calculate the
(a) average speed,
(b) average velocity,
of the cow moving from P to Q.
Solution:
Total distance travelled = 240 m
Displacement = 70 m
Time taken = 160 s
Analysing Linear Motion 5
Analysing Linear Motion 6

What Is The Difference Between Distance And Displacement

Difference Between Distance and Displacement

What Is Distance And Displacement In Physics

 

  • The path length between the initial and final positions of the particle gives the distance covered by the particle.
    Distance
  • Distance is defined as the total length of the path travelled from one location to another.
  • The distance travelled by the car is given by the following,
    Distance travelled = (200 + 500 + 600 + 300) m = 1600 m
  • The minimum distance between the initial and final positions of a body during that time interval is called displacement.
    displacement
  • Displacement by an object is the distance travelled by the object in a specific direction.
  • It is the distance travelled by the object from its initial position to its final position in a straight line.
  • For the car travelling from Delima to Mengkibol, the displacement is determined by measuring the length of the straight line connecting Delima to Mengkibol in the direction of Mengkibol from Delima. Hence from Figure, its displacement is given by the following,
    Magnitude of the displacement
    = √ (800² + 800²) = 1131 m
    Its direction is South-West of Delima (or South 45° West of Delima)
  • Distance and displacement both are measured in meter in m.k.s. system.

Difference between distance and displacement

  • Distance travelled is a scalar quantity while displacement is a vector quantity.
  • When a body continuously moves in the same straight line and in the same direction then displacement will be equal to the distance travelled. But if the body changes its direction while moving, then the displacement is smaller than the distance travelled.
    Displacement ≤ Distance
  • Displacement in any interval of time may be zero, positive or negative where as distance cannot be negative.

Distance And Displacement Example Problems With Solutions

Example 1:    A person travels a distance of 5 m towards east, then 4 m towards north and then 2 m towards west.
(i) Calculate the total distance travelled.
(ii) Calculate the resultant displacement.
Solution:    (i) Total distance travelled by the person
= 5 m + 4 m + 2 m = 11 m
distance-and-displacement-example-1
(ii) To calculate the resultant displacement, we choose a convenient scale, where 1 cm represents 1 m. We draw a 5 cm long line AB towards east and then 4 cm long line BC towards north. Finally, a 2 cm long line CD towards west. The resultant displacement is calculated by joining the initial position A to the final position D. We measure AB = 5 cm.
Since 1 cm = 1 m
∴ 5 cm = 5 m
Hence, the displacement of the person
= 5m towards AD.

Example 2:    A body is moving in a straight line. Its distances from origin are shown with time in Fig. A, B, C, D and E represent different parts of its motion. Find the following :
(i) Displacement of the body in first 2 seconds.
(ii) Total distance travelled in 7 seconds.
(iii) Displacement in 7 seconds
distance-and-displacement-example-2
Solution:    (i) Displacement of the body in first 2s = 40m
(ii) From t = 0 to t = 7 s, the body has moved a distance of 80 m from origin and it has again come back to origin.
Therefore, the total distance covered = 80 × 2 = 160 m
(iii) Since the body has come back to its initial position, the displacement is zero.

What Are The Equations Of Motion

Equations Of Motion

Motion under uniform acceleration

1st Equation of motion
For an object moving with uniform velocity, v, its displacement, s after time, t is given by:
s = v × t
Consider a body having initial velocity ‘u’. Suppose it is subjected to a uniform acceleration ‘a’ so that after time ‘t’ its final velocity becomes ‘v’. Now we know,
\( \text{ }\!\!~\!\!\text{ Acceleration}=\frac{\text{change}\,\,\text{in}\,\text{velocity}}{\text{Time}} \)
\( a=\frac{v-u}{t} \)
or     v = u + at          …..(i)

2nd Equation of motion
Suppose a body has an initial velocity ‘u’ and uniform acceleration ‘a’ for time ‘t’ so that its final velocity becomes ‘v’. The distance travelled by moving body in time ‘t’ is ‘s’ then the average velocity = (v + u)/2.
Distance travelled = Average velocity × time
\( \text{S}=\left( \frac{u+v}{2} \right)t\text{ } \)
\( \Rightarrow \text{S}=\left( \frac{u+u+at}{2} \right)t\text{ }\left( as\text{ }v=u+at \right)~ \)
\( \Rightarrow \text{S}=\left( \frac{2u+at}{2} \right)t \)
\( \Rightarrow \text{S}=\frac{2ut+a{{t}^{2}}}{2} \)
\( \text{S}=ut+\frac{1}{2}a{{t}^{2}}\text{ }……\text{ (ii)} \)

3rd Equation of motion
Distance travelled = Average velocity x time
\( S=\left( \frac{u+v}{2} \right)t\text{ }………\text{ (iii)} \)
\( \text{from equation }\left( \text{i} \right)\text{ }t=\frac{v-u}{a} \)
Substituting the value of t in equation (iii),
\( \text{we get }S=\left( \frac{v-u}{a} \right)\left( \frac{v+u}{2} \right) \)
\( S=\left( \frac{{{v}^{2}}-{{u}^{2}}}{2a} \right) \)
⇒ 2as = v2 – u2 or
v2 = u2 + 2as            ….(iv)

The equations of motion under gravity can be obtained by replacing acceleration by acceleration due to gravity (g) and can be written as follows

  • When the body is coming towards the centre of earth
    (a) v = u + gt                  (b) h = ut + \(\frac { 1 }{ 2 }\) gt2               (c) v2 = u2 + 2gh
  • When a body is thrown upwards with some initial velocity, then a retardation produced due to attraction of the earth. In equations of motion, a is replaced by (–g) and thus equations become.
    (a) v = u – gt                 (b) h = ut – \(\frac { 1 }{ 2 }\) gt2                (c) v2 = u2 – 2gh

Body Falling Freely Under Gravity

Assuming u = 0 for a freely falling body
Body-Falling-Freely-Under-Gravity

Body is projected vertically up:
Taking initial position as origin and direction of motion (i.e. vertically up) as positive.
(a) At the highest point v = 0
(b) a = – g
Body-Falling-Freely-Under-Gravity-1
It is clear that in case of motion under gravity
(a) Time taken to go up is equal to the time taken to fall down through the same distance.
(b) The speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.
(c) The body returns to the starting point with the same speed with which it was thrown.

Equations Of Motion Example Problems With Solutions

Example 1. A car is travelling with a uniform velocity of 80 km h-1 northward from Johor Bahru.
What is its displacement after 15 minutes?
Solution:
What Are The Equations Of Motion 1

Example 2. A train travelling in a straight line at 30 ms-1accelerates uniformly to 54 ms-1 in 3.0 seconds. Calculate the distance travelled by the train during that time.
Solution:
What Are The Equations Of Motion 2

Example 3. A school bus accelerates with an acceleration of 4.0 m s-2 after picking up some students at a bus stop.
What Are The Equations Of Motion 3
Calculate the
(a) velocity of the bus after 5 s.
(b) distance travelled by the bus after 5 s.
Solution:
Initial velocity, u = 0 ms-1
Acceleration, a = 4.0 ms-2
Time, t = 5 s
What Are The Equations Of Motion 4

Example 4. A long jumper was running at a velocity of 5 m s-1 towards the long jump pit. He needed to achieve a velocity of 10 m s-1 after covering a distance of 4.5 m before lifting himself off the ground from the jumping board.
What Are The Equations Of Motion 5
(a) Calculate the required acceleration for him to do so.
(b) Calculate the time taken for him to cover the horizontal distance of 4.5 m.
Solution:
What Are The Equations Of Motion 6
Example 5. 10 A body starts moving with an initial velocity 50 m/s and acceleration 20 m/s2. How much distance it will cover in 4s ? Also, calculate its average speed during this time interval.
Solution:    Given: u = 50 m/s, a = 20 m/s2,  t = 4s, s = ?
s = ut + \(\frac { 1 }{ 2 }\) at2 = 50 × 4 + \(\frac { 1 }{ 2 }\) × 20 × (4)2
= 200 + 160 = 360 m
Average speed during this interval,
\(\overline{V}=\frac{\text{distance travelled}}{\text{time interval}}=\frac{360}{4}=90\text{ m/s}\)

Example 6. A body is moving with a speed of 20 m/s. When certain force is applied, an acceleration of 4 m/s2 is produced. After how much time its velocity will be 80 m/s ?
Solution:    Given: u = 20 m/s, a = 4 m/s2, v = 80 m/s, t = ?
Using equation, v = u + at, we get
80 = 20 + 4 × t
or     4t = 80 – 20 = 60
or      t = 15 s
Therefore, after 15 seconds, the velocity of the body will be 80 m/s.

Example 7. A body starts from rest and moves with a constant acceleration. It travels a distance s1 in first 10 s, and a distance s2 in next 10 s. Find the relation between sand s1.
Solution:    Given : u = 0, t1 = 10 s
∴ Distance travelled in first 10 seconds, is given by
s1 = ut + \(\frac { 1 }{ 2 }\) at2 = 0 + \(\frac { 1 }{ 2 }\) × a × (10)2
s1 = 50a              …(1)
To calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s.
s = ut + \(\frac { 1 }{ 2 }\) at2 = 0 + \(\frac { 1 }{ 2 }\) × a × (20)2
s = 200a           …(2)
∴ Distance travelled in 10th second interval,
s2 = s – s1 = 200a – 50a          …(3)
or    s2 = 150a
\(\text{Now,    }\frac{{{s}_{2}}}{{{s}_{1}}}=\frac{150a}{50a}=\frac{3}{1}\)
or      s2 = 3s1

Example 8. A train is moving with a velocity 400 m/s. With the application of brakes a retardation of 10 m/s2 is produced. Calculate the following
(i) After how much time it will stop ?
(ii) How much distance will it travel before it stops ?
Solution:    (i) Given: u = 400 m/s, a = –10 m/s2, v = 0, t = ?
Using equation, v = u + at, we get
0 = 400 + (–10) × t
or t = 40 s
(ii) For calculating the distance travelled, we use equation,
v2 = u2 + 2as, we get
(0)2 = (400)2 + 2 × (–10) × s
or      20s = 400 × 400
or       s = 8000 m = 8 km

Example 9. A body is thrown vertically upwards with an initial velocity of 19.6 m/s. If g = –9.8 m/s2. Calculate the following
(i) The maximum height attained by the body.
(ii) After how much time will it come back to the ground ?
Solution:    (i) Given: u = 19.6 m/s, g = –9.8 m/s2, v = 0, h = ?
Using equation v2 = u2 + 2gh, we get
(0)2 = (19.6)2 + 2(–9.8) × h
\(h=\frac{19.6\times 19.6}{2\times 9.8}=19.6\text{ m}\)
(ii) Time taken to reach the maximum height can be calculated by the equation,
v = u + gt
0 = 19.6 + (–9.8) × t
t = 2s
In the same time, it will come back to its original position.
∴ Total time = 2 × 2 = 4s

Example 10. From the top of a tower of height 490 m, a shell is fired horizontally with a velocity 100 m/s. At what distance from the bottom of the tower, the shell will hit the ground ?
Solution:    We know that the horizontal motion and the vertical motion are independent of each other. Now for vertical motion, we have u = 0,
h = 490 m, g = 9.8 m/s2, t = ?
Using equation, h = ut + \(\frac { 1 }{ 2 }\) gt2, we get
490 = 0 + \(\frac { 1 }{ 2 }\) × 9.8 × t2
or     t2 = \(\frac { 490 }{ 4.9 }\) = 100
or     t = 10 s
∴ It takes 10 seconds to reach the ground.
Now, horizontal distance = horizontal velocity × time
= 100 m/s × 10 s = 1000 m
∴ The shell will strike the ground at a distance of 100 m from the bottom of the tower.