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Total distance traveled = actual path covered = OC CO = 25 25 = 50m

Total displacement = shortest distance between final position and initial position = 0m

Vector Quantities: Quantities that require both magnitudes and direction to specify them are called vector quantities or vectors. Displacement, velocity, force, momentum, weight etc. are the examples of vectors.

Given, length of each side = 10m

Distance covered in 1 lap = Perimeter of ABCD = 4 x 10 = 40m

Time taken by farmer to cover 1 lap = 40s

Speed of farmer = Distance ÷ Time Taken for one lap = 40/40s = 1m/s

Distance covered by farmer in 2min 20 secs = Speed x Time = 1 x 140s = 140m

Number of laps covered = 140 ÷ 40 = 3.5 laps.

⇒ After 140s, the farmer will be at position C (i.e. 3 and ½ laps).

Displacement = AC = (AB2 BC2)½ (Pythagoras theorem)= (100 100)½=10√2 = 10 x 1.414 = 14.14m .... (answer)

(a) It cannot be zero.

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

(a) False. Displacement can be zero. (See Q1).

(b) False. Displacement is less than or equal to the distance travelled by the object.

(a) False. Displacement can be zero. (See Q1).

(b) False. Displacement is less than or equal to the distance travelled by the object.

(a) A = 4m, B = 6m

(b) A = −4m, B = −2m

(c) A = −4m, B = −8m

(d) A = −4m, B = 2m

*Speed = distance ÷ time ; Velocity = displacement ÷ time

*Speed is scalar quantity i.e. it has magnitude only. VelocityI is vector quantity i.e. has both magnitude and

direction.

Let us assume time taken = T secs.

Average velocity (v) = Total Displacement ÷ Time Taken = 0 ÷ T = 0 m/sec

While solving numericals, focus on such words. These two words help in understand we need to compute displacement not distance. If average speed is being asked, we need to compute distance traveled.

Speed of signal (v) = 3 × 108 m s–1

Distance of spaceship = v x t = 3 × 108 m s–1 x 300 = 9 x 1010 m= 9 x 107 km

Case 1: Arithmetic mean i.e. avg speed = (20 30)/2 = 25 km/hr

Case 2: Harmonic mean i.e. avg speed = 2/(1/20 1/30) = 24 km/hr

Alternately,

Total Time taken = 2t seconds

Total Distance covered = 3t 5t = 8t m

Avg. Speed = Total Distance / Total Time = 8t/2t = 4m/s ... (answer)

(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant in equal intervals of time.

Initial speed of bus (u) = 80 km/h = 80 x 1000/(60 x 60) = 800/36 m/s

Final speed of bus (v) = 60 km/h = 60 x 1000/3600 = 600/36 m/s

Time taken (t) = 5s

Since v = u at

a = (v u)/

t = (800600)/(

36 x 5) = 1.11 m s–2 ...(answer)

Initial speed of train (u) = 0 km/hr

Final Speed of train (v) = 40 km/hr

Time taken = 10 min = 10/60 = 1/6 hr

Since v = u at

a = (vu)/

t = (400)/(

1/6) = 40 x 6 = 240 kmhr–2 ...(answer)

Initial speed (u) = 0 m/s

Acceleration of bus (a) = 0.1 ms–2

Time taken (t) = 2 min = 2 x 60 = 120 s

Final velocity = ?

Distance Covered (S) = ?

v = u at = 0 0.1 x 120 = 12 m/s ... (answer)

S = ut ½at2 = 0 (0.1 x 1202 )/2 = 0.1 x 14400 /2 = 720m ...(answer)

Initial Velocity of train (u) = 90 kmh–1 = 90 x 1000/ 3600 = 25 ms–1

Final velocity (v) = 0 ms–1

acceleration (a) = 0.5 ms–2

v2-u2= 2aS

⇒ S = (v2-u2) / 2a = 625/1= 625m ... (answer)

Initial velocity (u) = 0 cm s–1.

acceleration (a) = 2 cm s–2.

time taken (t) = 3 s

v = u at

⇒ v = 0 2 x 3 = 6 cm s–1... (answer)

Initial velocity of car (u) = 0 m s–1.

acceleration (a) = 4 m s–2.

time taken (t) = 10 s

Distance covered (S) = ?

S = ut ½at2= 0 4x100/2 = 400 / 2 = 200 m ... (answer)

Speed of the train (v) = 72kmph = 72 * 1000/3600 = 20 m s–1.

Length of bridge = 1/2 km = 500 m ...(I)

Time taken by train to cross bridge = 1min = 60s

Distance travelled by the train in 60s = 20 x 60 = 1200m ....(II)

Length of train = (II)

(I) = 1200-500 = 700m ...(answer)

(I) = 1200-500 = 700m ...(answer)

(b) For an object moving in a non uniform motion will cover unequal distances in equal intervals of time. It will be a curve on a distancetime graph.

Scene II is another common scenario. e.g. a free falling body, the velocity of the falling object increases every second but acceleration due to gravity is fixed.

Scene IV is possible: An object is in uniform circular motion i.e. it has fixed speed but direction of the object is changing every moment (direction of centripetal acceleration changes.)

Car A: Initial speed (uA) = 52km/h = 52 x 1000/3600 = 14.4 m/s

Final Speed (vA) = 0 m/s

Time taken (tA) = 5s

Car B: Initial speed (uB) = 3km/h = 3 x 1000/3600 = 0.83 m/s

Final Speed (vB) = 0 m/s

Time taken (tB) = 10s

Distance covered by car A = Area of △ AOB = ½ x AO x OB = ½ x 14.4 x 5 = 36m

Distance covered by car B = Area of △XOY = ½ x OX x OY = ½ x 0.83 x 10 = 4.15 m

⇒ Car A covers more distance as compared to car B.

(a) Object B is travelling the fastest. We can determine it in two ways. Slope of the distance time line represents the speed of the object. In this case, object B has the highest slope.

(b) No all objects never meet at the same point.

(c) When B and A meets at point m, the Object C is at n, which is approximately 8km on distance axis.

(d) B meets C at point k. The position on distance axis is 9 units away from origin O.

One unit on distance axis = 4/7 (7 units = 4km). ∴ Object B is = 9 x 4/7 = 5.14 km.

One unit on distance axis = 4/7 (7 units = 4km). ∴ Object B is = 9 x 4/7 = 5.14 km.

(a) The green shaded area in the graph above shows the distance travelled by the car in first 4 seconds. Taking approximation, the shaded area is a triangle. Distance covered in first 4s = Area of green shaded △ = 4 x 6 /2 = 12m.

(b) The line segment AB represents uniform motion of the car i.e. car after 6s attains

uniform motion.

We know that v = u at

and 2aS = v2-u2

⇒ 2aS = (v-u)(v u) [∵ Algebraic identity (a2-b2)= (a b)(a-b)]

⇒ 2aS = (v u)at [using v-u= at]

⇒ 2S = (v u)t or S = ½(u v)t

⇒ (u v)/2 is the average velocity.

Given, 42,260 km, t = 24 h

⇒ v = 2 ✕ 3.14 ✕ 42260 ÷ 24 = 11058.03 km/hr

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