MOTION- NCERT SOLUTIONS

Here, Side of the given square field = 10m

so, perimeter of a square = 4*side = 10 m * 4 = 40 m

Farmer takes 40 s to move along the boundary.

Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 seconds

since in 40 s farmer moves 40 m

Therefore, in 1s the distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter

= 140 m / 40 m = 3.5 round

Thus, after 3.5 round farmer will at point C of the field.

Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.

(a) It cannot be zero.

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

First statement is completely false because displacement can be zero in case of a circular path. Second statement is also false because displacement is less than or equal to the distance travelled by the object in case of motion in opposite direction.

Speed | Velocity |

Speed is the distance travelled by an object in a given interval of time. | Velocity is the displacement shown by an object in a given interval of time. |

Speed = Distance / Time | Velocity = Displacement / Time |

Speed is scalar quantity and has only magnitude. | Velocity is vector quantity and has both magnitude and direction. |

Speed= 3 × 108 m s^{-1}

Time= 5 min = 5 *60 = 300 seconds.

Distance= Speed * Time

Distance= 3 * 10^{8} m s^{-1} * 300 secs. = 9 x 10^{10}m

(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal time intervals.

(ii) A body is said to be in non-uniform acceleration if the rate of change of its velocity is not constant, that is differs in different time intervals.

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

Acceleration, a = 0.1 m/s^{2}

Time taken, t = 2 minutes = 120 s

(a) v= u + at

v= 0 + 0×1 × 120

v= 12 ms^{-1}

(b) According to the third equation of motion, v^{2} - u^{2}= 2as

s is the distance covered by the bus

(12)^{2} - (0)^{2}= 2(0.1) s

s = 720 m

Speed acquired finally by the bus is 12 m/s.

Distance travelled by the bus is 720 m.

Final speed of the train, v = 0 (finally the train comes to rest and its velocity becomes 0)

Acceleration = - 0.5 m s^{-2}

According to third equation of motion:

v^{2}= u^{2} + 2 as

(0)^{2}= (25)^{2} + 2 ( - 0.5) s

Where, s is the distance covered by the train

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

Acceleration, a= 2 cm s^{-2}

Time, t= 3 s

It is known that final velocity, v= u + at = 0 + 2*3 cms^{-1}

Therefore, the velocity of train after 3 seconds is 6 cms^{-1}

Acceleration, a= 4 m s^{-2}

Time, t= 10 s

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

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^{-2}

we know that 2 as= v^{2} - u^{2}

Diameter of circular track (D) = 200 m

Radius of circular track (r) = 200 / 2=100 m

Time taken by the athlete for one round (t) = 40 s

Distance covered by athlete in one round (s) = 2π r

= 2 * ( 22 / 7 ) * 100

Speed of the athlete (v) = Distance / Time

= (2 x 2200) / (7 x 40)

= 4400 / 7 × 40

So, **Distance covered in 140 s = Speed (s) × Time(t)**

= 4400 / (7 x 40) x (2 x 60 + 20)

= 4400 / ( 7 x 40) x 140

= 4400 x 140 /7 x 40

= 2200 m

Number of rounds in 40 s =1 round

Number of rounds in 140 s =140/40

=3 ½

After taking start from position X, the athlete will be at Y after 3 ½ rounds as shown in figure

Therefore, **Displacement of the athlete **with respect to initial position at x= xy

= Diameter of circular track

= 200 m

Total time taken = 2 x 60 + 30 s

=150 s

Therefore, Average Speed from AB = Total Distance / Total Time

=300 / 150 m s^{-1}

=2 m s^{-1}

Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s^{-1}

=2 m s^{-1}

Total Distance covered from AC = AB + BC

= 300 + 200 m

Total time taken from A to C = Time taken for AB + Time taken for BC

= (2 x 60+30) + 60 s

= 210 s

Therefore, Average Speed from AC = Total Distance /Total Time

= 400 /210 m s^{-1}

= 1.904 m s^{-1}

Displacement (S) from A to C = AB - BC

= 300-100 m

= 200 m

Time (t) taken for displacement from AC = 210 s

Therefore, Velocity from AC = Displacement (s) / Time(t)

= 200 / 210 m s^{-1}

= 0.952 m s^{-1}

Assume time taken by Abdul to cover this distance = t_{1}

Distance Abdul covers while driving from School to Home = S

Assume time taken by Abdul to cover this distance = t_{2}

Average speed from home to school v1av = 20 km h_{-1}

Average speed from school to home v2av = 30 km h_{-1}

Also Time taken from Home to School t1 =S / v_{1av}

Similarly, Time taken from School to Home t2 =S/v_{2av}

Total distance from home to school and backward = 2 S

Total time taken from home to school and backward (T) = S/20+ S/30

Therefore, Average speed (Vav) for covering total distance (2S) = Total Dostance/Total Time

= 2S / (S/20 +S/30)

= 2S / [(30S+20S)/600]

= 1200S / 50S

= 24 kmh_{-1}

Acceleration of motorboat, a = 3.0 m s^{-2}

Time under consideration, t = 8.0 s

Distance, s = ut + (1/2)at^{2}

Therefore, distance travel by motorboat = 0 x 8 + (1/2)3.0 x 8 2

= (1/2) x 3 x 8 x 8 m

= 96 m

Distance Travelled by first car before coming to rest =Area of Δ OPR

= (1/2) x OR x OP

= (1/2) x 5 s x 52 kmh_{-1}

= (1/2) x 5 x (52 x 1000) / 3600) m

= (1/2) x 5x (130 / 9) m

= 325 / 9 m

= 36.11 m

Distance Travelled by second car before coming to rest =Area of Δ OSQ

= (1/2) x OQ x OS

= (1/2) x 10 s x 3 kmh_{-1}

= (1/2) x 10 x (3 x 1000) / 3600) m

= (1/2) x 10 x (5/6) m

= 5 x (5/6) m

= 25/6 m

= 4.16 m

(a) Object B

(b) No

(c) 5.714 km

(d) 5.143 km

Therefore, Speed = slope of the graph

as slope of object B is greater than objects A and C, it is travelling the fastest.

(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.

On the distance axis:

7 small boxes = 4 km

Therefore,1 small box = 4 / 7 Km

Initially, object C is 4 blocks away from the origin.

Therefore, Initial distance of object C from origin = 16 / 7 Km

Distance of object C from origin when B passes A = 8 km

Assume, the final velocity with which the ball will strike the ground be ‘v’and time it takes to strike the ground be ‘t&Rsquo;

Initial Velocity of ball, u =0

Distance or height of fall, s =20 m

Downward acceleration, a =10 m s^{-2}

As we know, 2as =v^{2}-u^{2}

v^{2} = 2as+ u^{2}

= 2 x 10 x 20 + 0

= 400

∴ Final velocity of ball, v = 20 ms^{-1}

t = (v-u)/a

∴Time taken by the ball to strike = (20-0)/10

= 20/10

= 2 seconds

(a) Find out how far the car travels 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)

The shaded area which is equal to 1 / 2 x 4 x 6 = 12 m represents the distance travelled by the car in the first 4 s.

(b)

The part of the graph in red color between time 6 s to 10 s represents uniform motion of the car.

(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.

(a) Possible

when a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.

(b) Possible

When a car is moving in a circular track, its acceleration is perpendicular to its direction.

Radius of circular orbit, r= 42250 km

Time taken to revolve around the earth, t= 24 h

Speed of a circular moving object, v= (2π r)/t

=[2× (22/7)×42250 × 1000] / (24 × 60 × 60)

=(2×22×42250×1000) / (7 ×24 × 60 × 60) m s^{-1}

=3073.74 m s^{-1}