The document NCERT Solutions - Motion, Class 9, Science | EduRev Notes is a part of the UPSC Course NCERT Textbooks (Class 6 to Class 12).

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**Q.1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.****Ans. **Yes, an object can have zero displacements even if it has moved through a distance.**Example:** If an athlete runs around a circular path of radius 'r' and comes back to the initial point, then distance covered = 2πr, displacement = zero.**Q.2. A farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?****Ans. **

Distance covered by farmer in 40 s = 40 m, i.e. perimeter of the square field. If he completes one round in 40 s, then in 2 min 20s, i.e. 140s, he completes 140/40 = 3.5 rounds. Thus, if he starts from point B (see figure), and reaches point D.

Displacement, BD = length of diagonal

Diagonal of square = side √2 = 10√2

= 10 x 1.414

= 14.14 m**Q.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.****Ans.** **(a)** False; because the displacement is zero if an object returns to its initial position after completing the trip.**(b)** False; the displacement always has equal or less magnitude than the distance travelled.

**Page No. 102**

**Q.1. Distinguish between speed and velocity.****Ans. ****Q.2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?****Ans. **When the object travels along a straight-line path, then the distance travelled = displacement hence, average speed = average velocity.**Q.3.** **What does the odometer of an automobile measure? ****Ans.** The odometer of an automobile measures the **distance covered** by it.**Q.4.** **What does the path of an object look like when it is in uniform motion?****Ans. **The path of the object will be a **straight line **at the instant of measurement when it is in uniform motion.**Q.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} ms^{-}^{1}.**

Time in which signal reaches ground = 5 min = 5 × 60 = 300 s

Distance of spaceship from the ground level = speed × time = 3 × 10

**Q.1. When will you say a body is in****(i) Uniform acceleration.****(ii) Non-uniform acceleration?****Ans. ****(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 body is said to be in uniform acceleration. - The motion of a
**freely falling body**is an example of uniformly accelerated motion.

**(ii)** If an object travels in a straight line and its velocity changes by unequal amounts in equal intervals of time, then the body is said to be in non-uniform acceleration. **Example: **If a car is travelling along a straight road and passes through a crowd, suffers an unequal change in velocity, in equal intervals of time, so the car is moving with non-uniform acceleration.

**Q.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.**

**Ans. **

(We know, 1km = 1000m ; 1hr = 3600s)

Given, Initial speed of the bus,u = 80 kmh^{-1}

ms^{-1}

The final speed of the bus,v = 60 kmh^{-1}

= ms^{-1}

Time is taken, t= 5s

We know, v = u+at

a = -1.11 ms^{-2}

The negative sign implies retardation.

Therefore, the acceleration of the bus is **-1.11 ms**^{-2}

Or, the retardation(de-acceleration) of the bus is 1.11 ms^{-2}

**Q.3. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.****Ans.** Initial velocity, u = 0 km h^{-1}, Final velocity,

v = 40 kmh^{-1}

Time, t = 10 min = 1/6 h

Acceleration,

**Q.1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?****Ans.** When the motion is uniform, the distance-time graph is a straight line with a slope.

When the motion is non-uniform, the distance-time graph is not a straight line. It can be any curve.

**Q.2. What can you say about the motion of an object whose distance time-graph is a straight line parallel to the time axis?****Ans.** If the distance-time graph is a straight line parallel to the time axis, the body is at rest.

**Q.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?****Ans.** If the speed-time graph is a straight line parallel to the time axis, the object is moving uniformly.**Q.4.** **What is the quantity which is measured by the area occupied below the velocity-time graph? ****Ans.** The area under velocity-time graph = displacement of the body.

**Q.1. A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find****(a) the speed acquired.****(b) the distance travelled.****Ans. **Initial velocity, u = 0 ms^{-1}, Final velocity, v = ?, Acceleration, a = 0.1 ms^{-2}

Time, t = 2 min = 120 s**(a)** Speed, v = u + at = 0 + 0.1 x 120

= 12 ms^{-1}**(b)** Distance, s

= 720 m**Q.2.** **A train is travelling at a speed of 90 km/h ^{-1}. Brakes are applied so as to produce a uniform acceleration of 0.5 ms^{-}^{2}. Find how far the train will go before it is brought to rest.**

Final speed of the train, v = 0 (finally the train comes to rest)

Acceleration = - 0.5 m s

According to third equation of motion:

v

(0)

Where, s is the distance covered by the train

The train will cover a distance of 625 m before it comes to rest.

Acceleration, a= 2 cm s

Time, t= 3 s

We know that final velocity, v= u + at = 0 + 2 x 3 cms

Therefore, The velocity of train after 3 seconds = 6 cms

Acceleration, a= 4 m s

Time, t= 10 s

We know Distance, s= ut + (1/2)at

Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 × 102

= 0 + (1/2) × 4 × 10 × 10 m

= (1/2) × 400 m

= 200 m

Downward of negative Acceleration, a= 10 m s

We know that 2as= v

⇒ 0 = (5)^{2 }+ 2 x(-10) x s

⇒ 0 = 25 - 20s

⇒ s = 25/20 m

⇒ s = 1.25

Height attained by stone , s = 1.25 m

We know that , v = u + at

⇒ 0 = 5 + (–10) × t

⇒ 0 = 5 − 10t

⇒ t = 5/10 s

⇒ t = 0.5 s

Thus, stone will attain a height of 1.25 m and time taken to attain the height is

0.5 s.

**Q.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?****Ans. **Here, the diameter of circular track = 200 m.

Radius of circular track, r = 100 m.

Let the athlete start moving from A, which is treated as a reference point.

The distance covered by the athletes in 40 s

= circumference of the circle

= 2πr = 2π x 100 m = 200π m

The distance traversed in one second = 200π/40 = 5π

Distance covered by athletes in 2 minutes and 20 seconds, i.e., 140 s

Number of complete rounds done by the athlete

Therefore, the final position of athletes at the end of 2 minutes and 20 seconds or just after three and a half rounds is B.

The displacement at the end of 2 minutes and 20 seconds.

= AB; the shortest distance between initial and final position.

= diameter of track = 200 m**Q.2.** **Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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? ****Ans. **Let Joseph jogs from A to B and back to C as shown.**(a)** From A to B,

distance covered = displacement = 300 m

Time taken = 2 minutes 30 s = 150 s

∴ Average speed = average velocity**(b)** From A to C,

distance covered = 300 m + 100 m

= 400 m

Total time = 150 s + 60 s = 210 s

Average speed = 400m/210s = 1.9 ms^{-1}

However,

displacement = 300 m -100 m = 200 m

Total time = 210 s

Average velocity = 200m/210s = 0.952 ms^{-1}**Q.3. Abdul while driving to school computes the average speed for his trip to be 20 kmh ^{-}^{1}. On his return trip along the same route, there is less traffic and the average speed is 40 kmh^{-}^{1}. What is the average speed for Abdul's trip? **

Time taken by Abdul from home to school,

Time taken by Abdul from school to home,

Total time, t = t

Total distance, s = x + x = 2x km

Average speed =

a = 3 ms

t = 8 s

Using, s = ut + 1/2 at

s = 0 x 8 + 1/2 x 3 x 8

**Ans.**

Speed of car A = 52 km/h

Speed of car B = 3 km/h

= 3 ∗ 5/18 = 0.833 m/s

Distance covered by car 1 = area of ΔABC

Distance covered by car 2 = area of ΔBDE

= 4.166m

Thus, car 1 travelled farther after the brakes were applied.**Q.6. Figure below 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?**

**Ans.** **(a)** The slope of graph of object B is maximum. Thus object B is travelling the fastest.**(b)** No, all three are never at the same point on the road as the graphs are not intersecting at a single point.**(c)** When B crosses A, i.e. at point D, then the corresponding position of C, i.e. at point E will be 7 km from the start.**(d)** B crosses C at point E Corresponding position on graph is 4.5 km from the start.**Q.7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms ^{-}^{2}, with what velocity will it strike the ground? After what time will it strike the ground?**

Height, h = 20 m, Acceleration, a = 10 ms

Final velocity, v = ?, Time, t = ?

v

Here, s = h

∴ v

or Final velocity, v = 20 ms

v = u + at

20 = 0 + 10t

∴ t = 20/10

or time, t = 2 s

**Ans.** **(a)**

In velocity-time graph,

Distance = Area of v-t graph

Here we approximate the area using area of triangle,

Distance travelled = Area of triangle AOB

The shaded area, which is equal to 1/2 × 4 × 6 = 12 m represents the distance travelled by the car in the first 4 s.**(b)**

The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.**Q.9. State which of the following situations are possible and given an example for each of these.****(a) An object with a constant acceleration but with zero velocity.****(b) An object moving in a certain direction with acceleration in the perpendicular direction.****(b) an object moving with an acceleration but with uniform speed.****Ans.** **(a)** When an object is thrown upwards, it comes to a momentary rest at the highest point. Thus velocity is zero, but acceleration due to gravitational pull of the earth still acts upon it.**(b)** In uniform circular motion, the speed remains constant, but there is varying velocity as it changes it’s direction, so there always be acceleration which is given by centripetal force.**(c)** When an object is thrown in the forward direction, then during its motion in horizontal direction, the acceleration due to gravity of earth acts in vertically downward direction.**Q.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.****Ans.** Satellite completes one round in 24 hours

t = 24 h = 24 x 60 min. = 24 x 3600 s

Angle traversed by the satellite in completing one round = 2π

Linear speed of satellite =

(∵ π = 3.14)

= 3.1 km/s

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