Tags Class 11th

Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.16 Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

0 Follower 378 Views

Vishal Baghel

Contributor-Level 10

3.16

(a) Figure (a) shows two positions at the same time, which is not possible for a one-dimensional motion

(b) A particle cannot have velocity in two directions at the same time in one-dimensional motion

(c) The graph shows negative speed, which is not possible because speed is always positive

(d) Total path length decreases, which is not possible for a one – dimensional motion

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.15 In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

0 Follower 14 Views

Vishal Baghel

Contributor-Level 10

3.15 Instantaneous speed and velocity are applicable for a small interval of time because the magnitude of the displacement is effectively equal to the distance travelled by the particle.

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.14 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^–1.

What is the

(a) Magnitude of average velocity, and

(b) Average speed of the man over the interval of time

(i) 0 to 30 min,

(ii) 0 to 50 min,

(iii) 0 to 40 min?

[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

0 Follower 1k Views

Vishal Baghel

Contributor-Level 10

3.14

The distance from home to market = 2.5 km = 2500 m

The speed of the walking while going = 5 kmph = 1.388 m/s

The speed of walking back from market to home = 7.5 kmph = 2.083 m/s

(a) Magnitude of the average velocity = Displacement / time = 0, since the total displacement is zero.

(b)

i. Time taken to reach the market = distance / onward speed = 2500/1.388 = 1801 s = 30 minutes. So the average speed over 0 – 30 min is 5 kmph

ii. Time taken to reach back home = 2500 / 2.083 = 20 minutes. So the average speed = total distance covered / total time taken = (2500 +2500)/ (30 +20) m/min = 100 m/min = 6 kmph

iii. Average speed ove

...more

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.13 Explain clearly, with examples, the distinction between:

(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval

(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]

Show in both (a) and (b) that the second quantity is either greater than or equal to the first

When is the equality sign true? [For simplicity, consider one-dimensional motion

...more

0 Follower 147 Views

Vishal Baghel

Contributor-Level 10

3.13

(a) Let us take the example of football world cup 2022. During Argentina match, Di Mario (A) passed the ball to Messi (B). Messi instantly passed the ball to Di Mario. Now the magnitude of displacement of the ball is 0 as the ball returns to Di Mario at the initial position. But the total length covered by the ball is AB + BA = 2AB. Hence total length covered by the particle (the ball) is more than the magnitude.

(b) If t is the time taken to cover the entire distance, the magnitude of the average velocity of the ball over time interval t = Magnitude of displacement / t = 0 / t = 0

The average speed of the ball = Total lengt

...more

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.12 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

0 Follower 517 Views

Vishal Baghel

Contributor-Level 10

Given data:

Initial height, h = 90 m

Acceleration due to gravity, g = 9.8 ms2

Speed lost at each collision = 1/10 of its speed

time interval to be considered: 0 to 12 secs.

Step 1: Time to fall from 90 m

h = ½ gt2 ⇒ 90 = 21 * 9.8 * t2 ⇒ t2 = 9.8180 ≈ 18.37 ⇒ t ≈ 4.29 seconds

Step 2: Speed before first impact

v = gt = 9.8 * 4.29 ≈ 42 m/s

After hitting the floor with 42 m/s, ball rebound with;

vrebound = 0.9*42 = 37.8m/s

Step 3:

Max height after rebound:

Use:

t = g / v = 37.8 / 9.8 ≈ 3.86s

Again, the b

...more

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.11 Read each statement below carefully and state with reasons and examples, if it is true or false;

A particle in one-dimensional motion

(a) With zero speed at an instant may have non-zero acceleration at that instant

(b) With zero speed may have non-zero velocity

(c) With constant speed must have zero acceleration

(d) With positive value of acceleration must be speeding up

0 Follower 161 Views

Vishal Baghel

Contributor-Level 10

3.11

(a) True – zero speed may have a non zero instantaneous non-zero acceleration

(b) False – zero speed will always make zero velocity

(c) True – constant speed will have zero acceleration

(d) False – If the positive value of acceleration is in the direction of its motion then it can speed up

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.10 A player throws a ball upwards with an initial speed of 29.4 m s^–1.

(a) What is the direction of acceleration during the upward motion of the ball?

(b) What are the velocity and acceleration of the ball at the highest point of its motion?

(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion

(d) To what height does the ball rise and after how long does the ball return to the player's hands

...more

0 Follower 469 Views

Vishal Baghel

Contributor-Level 10

3.10

(a) The direction of the acceleration is downward during the upward motion of the ball because the acceleration is due to gravity. Gravity always pulls the object toward the centre of the earth.

(b) Velocity: At the peak the velocity is zero because the ball momentarily stops before falling downwards.

Acceleration: g = 9.8 ms^–2

(c) If we consider the highest point x = 0 m, t= 0 s,

During upward motion of the ball before it reaches the maximum height, x = +ve, velocity = -ve, acceleration = +ve. During downward motion, x = +ve, velocity = +ve, acceleration = +ve

(d) At the highest point, final velocity v = 0, initial vel

...more

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.9 Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^–1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

0 Follower 79 Views

Vishal Baghel

Contributor-Level 10

3.9 Cycling speed $V_{c}$ = 20 kmph

Let as assume speed of the bus is $V_{b}$

The relative velocity of the buses in the direction of motion of the cyclist = $V_{b}$ - $V_{c}$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_{b} - V_{c}$ ) * 0.3

Since the buses leave every T minutes, ( $V_{b} - V_{c}$ ) * 0.3 = $V_{b}$ * (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_{b}$ + $V_{c}$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_{b}$ + $V_{c}$ ) * 1/10

Since the buses leave every T minute, we get ( $V_{b}$ + $V_{c}$ ) * 1/10

...more

3.9 Cycling speed $V_{c}$ = 20 kmph

Let as assume speed of the bus is $V_{b}$

The relative velocity of the buses in the direction of motion of the cyclist = $V_{b}$ - $V_{c}$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_{b} - V_{c}$ ) * 0.3

Since the buses leave every T minutes, ( $V_{b} - V_{c}$ ) * 0.3 = $V_{b}$ * (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_{b}$ + $V_{c}$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_{b}$ + $V_{c}$ ) * 1/10

Since the buses leave every T minute, we get ( $V_{b}$ + $V_{c}$ ) * 1/10 = $V_{b}$ * T/60 …….(2)

From equations (1) & (2), we get

( $V_{b}$ + $V_{c}$ ) / 10 = ( $V_{b} - V_{c}$ ) * 0.3

$V_{b}$ + $V_{c}$ = 3( $V_{b}$ - 3 $V_{c}$ )

2 $V_{b}$ = 4 $V_{c}$ , $V_{b}$ = 2 $V_{c}$ So $V_{b}$ = 2 * 20 = 40 kmph

From the equation (1), we get

(40-20) * 0.3 = 40 * T/60, T = 9 min

less

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.8 On a two lane road, car A is travelling with a speed of 36 km h^–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^–1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

3.8

Speed of car A, Va = 36 kmph = 10m/s

Speed of car B, Vb = 54 kmph = 15 m/s

Speed of car C, Vc = -54 kmph = -15 m/s

Relative speed of A, w.r.t. C = Va – (-Vb) = 10 +15 = 25 m/s

Relative speed of B, w.r.t. A = Vb-Va = 15-10 = 5 m/s

Distance between AB & AC = 1 km = 1000m

Time taken by the car C to cover the distance AC, from s = ut

We get t = 1000/25 = 40s

To avoid collision with C, the minimum acceleration the vehicle B must have, from the relation, s = ut + (1/2) ft²

We get 1000 = 5 * 40 + (1/2) f * 40²f = 1 m/s²

0 0

New answer posted

8 months ago

Class 11th Motion in a Straight Line

3.7 Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s^–2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

0 Follower 342 Views

Vishal Baghel

Contributor-Level 10

3.7 The uniform speed of A and B = 72 kmph = 20 m/s

The acceleration f = 1 m/s², time t = 50 s

Distance covered by train B = ut + 0.5 * ft²= 20 * 50 + 0.5 * 1 * 50 ² = 2250 m

Distance covered by train A in the same period = ut = 50 * 20 = 1000 m

So the initial distance train A and train B was = 2250 – 1000 $- (400 + 400)$ = 450 m

1 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
684k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 11th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience