1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

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Given, dia. of circular track = 200 m

So, circumference = \( \pi × d \ = \ 3.14 × 200 \ = \ 628 m \)

So, the athlete completes 628 m in 40 s

Then, 1 s in \( 628 \over 40 \) \( = 15.7 m \)

Now, after 2 min 20 sec that is 140 s he would have covered \( 15.7 × 140 \ = \ 2198 m \)

So, he would by 140 sec he would have covered \( 140 \over 40 \) \( =3.5 \) rounds

So, displacement will be equal to dia. of the circular track , i.e., 200 m

Hence, the net distance covered by the athlete is 2198 meters and the total displacement of the athlete is 200 m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

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(a) Given, distance covered from point A to point B = 300 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Displacement from A to B = 300 meters

We know that, Average speed = \( \frac{distance}{time taken} \)

Therefore, the average speed while travelling from A to B = \( \frac{300}{150} \) = 2 m/s

and, Average velocity = \( \frac{displacement}{time taken} \)

Average velocity while travelling from A to B = \( \frac{300}{150} \) = 2 m/s

(b) Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to C = 300m – 100m = 200 meters

Similarly, Average speed while travelling from A to C = \( \frac{400}{210} \) = 1.9 m/s

Average velocity while travelling from A to C = \( \frac{200}{210} \) = 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 40 km/h. What is the average speed for Abdul’s trip?

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Let distance travelled to reach the school = d km = distance traveled to reach home

and time taken to reach school = \( t_1 \) and time taken to reach home =\( t _2 \)

Average speed while going to school = 20 kmph

∴ \( 20 = \) \( d \over t_1 \) => \( t_1 = \) \( d \over 20 \) -(i)

Average speed while going home = 30 kmph

∴ \( 40 = \) \( d \over t_1 \) => \( t_1 = \) \( d \over 40 \) -(ii)

Now, the average speed for the entire trip = \( Total \ distance \ covered \ in \ the \ trip \ \over Total \ time \ taken \)

∴ \( v_{av. trip} = \) \( d+d \over t_1 + t_2 \) {∵ total distance = distance travelled to reach the school + distance traveled to reach home }

=> \( v_{av. trip} = \) = \( d+d \over ( \frac{d}{20} ) + ( \frac{d}{40} ) \) \( = 80 \over 3 \)

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s^{2} for 8.0 s. How far does the boat travel during this time?

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Given, initial velocity of the boat(u) = 0 m/s

Acceleration of the boat \( = 3 m/s^2 \)

By second motion equation, \( s= ut + \frac{1}{2}at^2 \)

=> \( s = \frac{1}{2}3 × 8^2 \) => \( s = 96 \ m \)

5. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

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The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB = (1/2) &$215; (OB) &$215; (OA)

But OB = 5 seconds and OA = 52 km/h = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2) &$215; (5s) &$215; (14.44m/s) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD = (1/2) &$215; (OD) &$215; (OC)

But OC = 10 seconds and OC = 3km/h = 0.83 m/s

Therefore, area of triangle COD = (1/2) &$215; (10s) × (0.83m/s) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car travelled farther post the application of brakes.

6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

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(a) since the slope of line B is the greatest, B is travelling at the fastest speed.

(b) since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.

(c) since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.

Since the initial point of object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km

When A passes B, the distance between the origin and C is 8 km

Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km

(d) the distance that object B has covered at the point where it passes C is equal to 9 graph units.

Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^{-2}, with what velocity will it strike the ground? After what time will it strike the ground?

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Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance traveled by the ball (s) = 20m

Acceleration (a) = 10 ms^{-2}

By third motion equation, \( v^2 - u^2 = 2 a s \)

Therefore,

= 2*(10)*(20) + 0

v^{2 }= 400

Therefore, v= 20m/s

The ball hits the ground with a velocity of 20 meters per second.

By first motion equation, \( v = u + at \)

Therefore, \( 20 = 0 + 10 t \)

=> t = 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

8. The speed-time graph for a car is shown is Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

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(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6^{th} to the 10^{th} second.

9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

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(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

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Given, radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2*?*42250km = 265571.42 km

Time taken for the orbit = 24 hours

Therefore, speed of the satellite = 11065.4 km.h^{-1}

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

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Yes, an object which has moved certain distance can have zero displacement as we know displacement is the shortest distance between the starting or initial point and end or final point of an object which has moved. For eg. if a person starts moving from point A and comes back to point A then his displacement is zero.

2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

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Given, a farmer covers 40 m in 40 s .

So, he will cover 1 m in \( 40 \over 40 \) \( = 1 s \)

Now, after 2 min 20 s which is equal to 140 s , he would have covered \( 140 \over 1 \) \( = 140 m \)

Case 1: When the farmer starts moving from corner of the field

Then, after 140 s he would have covered \( 140 \over 40 \) \( = 3.5 \) rounds of the field.

So, displacement would have been, \( \sqrt{10^2 + 10^2} = 14.1 \ m \) { By Pythagoras theorem}

Case 2: When the farmer starts moving from middle of any side

Then, after 140 s he will be at the middle of the opposite side of his starting side

So, his displacement will be 10 m.

Case 3: When the farmer starts moving from any random point.

Then his displacement will be in between 10 m and 14.1 m

3. Which of the following is true for displacement?

(a) It cannot be zero.

(b) Its magnitude is greater than the distance travelled by the object.

(a) It cannot be zero.

(b) Its magnitude is greater than the distance travelled by the object.

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(a) False because displacement can be zero if an object returns to its initial position after covering certain distance.

(b) False because the displacement of an object can be equal to distance travelled, but not greater as displacement is the shortest distance between initial and final position of an object, so how can it be greater then distance travelled.

1. Distinguish between speed and velocity.

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2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

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As, Average speed = \( Total distance travelled \over Total time taken \) and, <.br>

Average velocity = \( Displacement \over Total time taken \)

So, when \( total distance travelled = Displacement \) then only its magnitude of average velocity is equal to its average speed

3. What does the odometer of an automobile measure?

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The odometer measures the total distance travelled by the automobile.

4. What does the path of an object look like when it is in uniform motion?

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The path of an object look like a straight line when it is in uniform motion.

5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, \( 3 × 10^8\) m/s.

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Given, time taken by signal to reach ground = 5 min = 5×60 = 300 s

and speed of signal was \( 3 × 10^8\) m/s.

We know that, speed = \( Distance \over time \) => distance = speed × time

So, Distance \( = 3×10^8 × 300 = 9 × 10^{10} \) m

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

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(i) If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. For example the motion of a freely falling body.

(ii) an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration

2. A bus decreases its speed from 80 km h^{–1} to 60 km h^{–1} in 5 s. Find the acceleration of the bus.

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Given, initial speed of bus(u) = 80 km h^{–1} \( = 80 × \) \( 5 \over 18 \) \( = 22.22 \) m/s

final speed of bus(v) = 60 km h^{–1} \( = 60 × \) \( 5 \over 18 \) \( = 16.66 \) m/s

Time taken = 5 s

We know that, a = \( v - u \over t \) = \( 16.66 - 22.22 \over 5 \) \( = - 1.112 \ m/s^2 \)

∴ \( a = - 1.112 \ m/s^2 \)

The negative sign indicates that the bus is actually not accelerating but retarding.

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^{–1} in 10 minutes. Find its acceleration.

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Given, initial velocity of train(u) = 0 m/s { As the train starts from rest}

final velocity(v) = 40 km h^{–1} = \( 40 × \) \( 5 \over 18 \) = \( 11.11 \) m/s

Time taken(t) = 10 min \( = \ 10 × 60 \ = \ 600 \) s

So, a = \( v - u \over t \) = \( 11.11 - 0 \over 600 \) = \( 0.0185 \ m/s^2 \)

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

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The distance-time graph for an uniform motion is a straight line.

The distance-time graph for a non-uniform motion is a curve.

2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

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Since there is no change in the distance travelled by the object at any point in the X-Axis (time), therefore, the object is at rest.

3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

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Since velocity does not change with time therefore, the object is in uniform motion.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

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Let us consider an object in uniform motion, its velocity-time graph can be represented as:

We know that area of rectangle \( = l × b \)

Here, area under the velocity-time graph = velocity × time = displacement { Velocity = \( displacement \over time \) }

So, displacement is measured by the area occupied below the velocity-time graph.

1. A bus starting from rest moves with a uniform acceleration of 0.1 m s^{-2} for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

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Given, bus starts from rest, so its initial speed = 0 m/s

a = 0.1 m s^{-2}

t = 2 min = 120 s

(a) Now, a = \( v-u \over t \) => \( v = at + u \) => \( v = 0.1 × 120 + 0 \)

∴ speed acquired \( = \ 12 \ m/s \)

(b) We know that, \( v^2 - u^2 = 2as \)

So, \( 12^2 - 0^2 = 2 × 0.1 × s \) => \( s = \) \( 144 \over 0.2 \)

∴ distance travelled = 720 m

2. A train is travelling at a speed of 90 km /h. Brakes are applied so as to produce a uniform acceleration of \( –0.5 m /s ^2 \) . Find how far the train will go before it is brought to rest.

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Given, initial velocity= 90 km/h = \( 90 × \) \( 5 \over 18 \) \( = 25 \ m/s \)

final velocity = 0 m/s { as train is brought to rest}

a = \( –0.5 m /s ^2 \)

By the third motion equation,

\( v^2 - u^2 = 2as \) => \( 0^2 \ - \ 25^2 \ = \ 2 × (-0.5) s \) => \( s = 625 \ m \)

∴ The train will travel 625 meters before it is brought to rest.

3. A trolley, while going down an inclined plane, has an acceleration of \( 2 \ cm /s^2 \) . What will be its velocity 3 s after the start?

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Given, the trolley starts from rest. so its initial velocity = 0 m/s

a = \( 2 \ cm /s^2 \) = \( 0.02 \ m/s^2 \)

t= 3 s

By the first motion equation,

\( v \ = \ u \ + \ at \) => \( v = 0 + (0.02 × 3 ) \) => \( v = 0.06 \ m/s^2 \)

Therefore, the velocity of the trolley after 3 seconds will be \( 0.06 \ m/s^2 \)

4. A racing car has a uniform acceleration of \( 4 \ m/s^2 \) . What distance will it cover in 10 s after start?

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Given, the car is initially at rest ∴ initial velocity (u) = 0 m/s

a = \( 4 \ m/s^2 \)

t = 10 s

By the second motion equation,

\( s = ut + \) \( 1\over 2 \) \( at^2 \) => \( s = 200 \ m \)

Therefore, the distance covered by the car is 200 meters.

5. A stone is thrown in a vertically upward direction with a velocity of \( 5 \ m/s \) . If the acceleration of the stone during its motion is \( 10 \ m /s^2 \) in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

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Given, initial velocity = 5 m/s

Final velocity = 0 m/s {since the stone will stop at maximum height and then again starts falling down}

\( a = 10 \ m/s^2 \) in the direction opposite to the trajectory of the stone \( = -10 \ m/s^2 \)

By the third motion equation,

\( v^2 - u^2 = 2as \) => \( s = 1.25 m \)

∴ The distance travelled by the stone = 1.25 meters

Now, using the first motion equation, we get

\( v = u + at \) => \( t = 0.5 s \)

Therefore, time taken by the stone to reach a maximum height = 0.5 seconds