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Motion in a Straight Line

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5 months ago

Motion in a Straight Line Class 11th

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.

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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

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Motion in a Straight Line Class 11th

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

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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

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Motion in a Straight Line Class 11th

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

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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

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Motion in a Straight Line Class 11th

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?

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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

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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

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Motion in a Straight Line Class 11th

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?

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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

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Motion in a Straight Line Class 11th

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?

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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

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5 months ago

Motion in a Straight Line Class 11th

3.6 A car moving along a straight highway with speed of 126 km h^–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

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Vishal Baghel

Contributor-Level 10

3.6 The speed of the car = 126 kmph = (126 * 1000/3600) m/sec = 35 m/s

From the relation V² = U² + 2fs, where Final velocity V = 0, U = 35 m/s, s = 200m, we get retardation, f = -U²/2s= 3.06 m²/s

From the relation V = U – ft, we get t = U/f = 35/3.06 = 11.42 s.

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5 months ago

Motion in a Straight Line Class 11th

3.5 A jet airplane travelling at the speed of 500 km h^–1 ejects its products of combustion at the speed of 1500 km h^–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

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Vishal Baghel

Contributor-Level 10

3.5 The speed of the jet plane, Vj = 500 kmph. If the speed of the product of combustion is Vpc, then V pc = Vpc-Vj = -1500 kmph (relative to the jet plane)

Speed of the combustion product relative to the observer on the ground,

Vpc = -1500 + Vj = -1500+500 = -1000 kmph

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New answer posted

5 months ago

Motion in a Straight Line Class 11th

3.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1m long and requires 1 s. Plot the xt graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

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Vishal Baghel

Contributor-Level 10

3.4

The time taken for each step is 1s, he covers 5 m in 5 secs and goes backward by 3 m in 3 secs.

So in 8 s he covers 2 m. To cover 13m distance he needs to complete 8 m (13-5). To cover 8m, he will require 8 * 4 = 32 s. To fall in a pit, he needs to move another 5m in 5s.

So total time taken = 32 +5 = 37 s

Let's have a look at how we came to this answer.

We have to calculate how long it will take for the drunkard to fall into a pit located 13 metres from his starting point.

In each cycle of movement, the drunkard takes 5 steps forward and 3 steps backward. We calculate the net displacement per cycle.

Time for one cycle = 5 seco

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5 months ago

Motion in a Straight Line Class 11th

3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h^–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^–1. Choose suitable scales and plot the xt graph of her motion.

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Vishal Baghel

Contributor-Level 10

3.3 Office distance = 2.5 km

Walking speed = 5 kph

Time taken to reach office = 5/2.5 = 0.5 h

OA is the path to reach office

AB is the duration of office stay

Auto speed = 25 kph

Time taken by auto = 2.5/25 h = 6 minutes

BC is the path for returning

On the position-time graph, this is a straight line sloping downwards. It starts at the point representing 5:00 PM and 2.5 km, and ends at 5:06 PM at 0 km. It tells us that the woman has arrived home.

The slope of this line represents instantaneous velocity. Because the return trip is by the auto at a much higher speed (25 km/h), the slope of line BC is significantly steeper than the line

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