Motion In a Straight Line - 11th physics

Question (3.6) A car moving along a straight highway with speed of 126 km/h is bought 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?

Solution

Given, initial velocity of car (u) = 126 km/h

⇒ Initial velocity of car (u) = 35 m/s

Final velocity of car (v) = 0 (since car is brought to a stop)

Distance (s) covered by car before stop = 200m

Thus, time (t) = ?

Calculation of Retardation (negative acceleration)

We know that, v² = u² + 2as

⇒ 0² = (35)² + 2 a × 200

⇒ 0 = 1225 + 400 a

⇒ 400 a = 0 – 1225

⇒ a = – 3.06 ms^–2

Calculation of time (t)

We know that, v = u + at

⇒ 0 = 35 + (–3.06 t)

⇒ 0 = 35 – 3.06t

⇒ 3.06 t = 35

⇒ t = 11.44 s

Thus, car will stop after a time of 11.44 second Answer

Question (3.7) Two trains A and B of length 400m 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 50s, the guard of B just brushes past the driver of A, what was the original distance between them?

Solution

For train A

Given, initial speed (u) = 72 km/h

⇒ u = 20 m/s

Time (t) = 50s

Acceleration (a) = 0 m/s/s

[Since train is moving with uniform speed]

Thus, distance covered (s) = ?

We know that, s = ut + 1/2 at²

⇒ s = 20 × 50 + 1/2 × 0 × 50

⇒ s = 1000 km

For train B

Initial velocity (u) = 72 km/h = 20m/s

Time (t) = 50s

Acceleration (a) = 1m/s²

Thus, distance (s) = ?

We know that, s = ut + 1/2 at²

⇒ s = (20 × 50) + (1/2 × 1 × (50)²)

⇒ s = 1000 + (1/2 × 2500)

⇒ s = 1000 + 1250

⇒ s = 2250 m

Now train A covers 1000 m and train B covers 2250 m when the guard of B just brushes past the driver of train A.

Thus, initial distance between engine of train A and guard of train B = Distance covered by train B – Distance covered by train A

= 2250m – 1000m

= 1250 m

And initial distance between guard of train A and engine of train B = Distance covered by train B – Distance covered by train A – sum of length of both the train

= 2250m – 1000m – 800m

= 450m

Thus, initial distance between engine of train A and guard of train B = 1250m. And initial distance between guard of train A and engine of train B = 450 m Answer

Alternate Method

Given, Length of each of the train = 400m

Initial velocity of each of the train = 72 km/h = 20m/s

Since both train is running with the same uniform speed, thus, relative velocity of train B with respect to train A (v_BA) = 0

Now, Acceleration of train B (a) = 1ms^–2

And time (t) = 50s

Thus, initial distance between two trains = ?

Let the initial distance between two trains = s

Thus, total distance covered when guard of train B just brushes past the driver of A = initial distance between two trains + sum of length of both of the trains

= s + 800m

Now, we know that, s = u_BA t + 1/2 at²

⇒ s + 800 = 0 × 50 + (1/2 × 1 × (50)²)

⇒ s + 800 = 1/2 × 2500

⇒ s + 800 = 1250

⇒ s = 1250 – 800

⇒ s = 450 m

Thus, initial distance between guard of train A and engine of train B = 450m

And initial distance between driver of train A and guard of train B = Initial distance between guard of train A and engine of train B + sum of length of both of the trains

= 450m + 800m

= 1250m

Thus, initial distance between engine of train A and guard of train B = 1250m. And initial distance between guard of train A and engine of train B = 450 m Answer

Question (3.8) On a two land road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h 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?

Solution

Given, Initial velocity of car A (u_A) = 36 km/h

⇒ Initial velocity of car A (u_A) = 10 m/s

Initial velocity of car B (u_B) = 54 km/h

⇒ Initial velocity of car B (u_B) = 15 m/s

And, ⇒ Initial velocity of car C (u_C) = 54km/h = 15 m/s

Assuming car B is coming in the same direction as car A

The relative velocity of car B with respect to car A (u_BA) = Velocity of car B – Velocity of car A

= 15 m/s – 10 m/s

⇒ u_BA = 5 m/s

And assuming car C coming in the opposite direction of car A

The relative velocity of car C with respect to car A (u_CA) = Velocity of car C + Velocity of car A

= 15 m/s + 10 m/s

⇒ u_CA = 25 m/s

As given in the question, the distance of car B and C both from car A = 1 km = 1000 m

Thus, time taken to cover the distance of 1000 m by car C, i.e to approach the car A = Distance/relative velocity (u_CA)

Now, avoid the accident, car B must have to cover the distance of 1000 m in a maximum time of 40s.

Thus, acceleration of car B = ?

We know that, s = ut + 1/2 at²

⇒ 1000 = 200 + 800 a

⇒ 800 a = 1000 – 200

⇒ 800 a = 800

⇒ a = 800/800

⇒ a = 1m/s²

Thus, car B has to acquire minimum acceleration of 1 m/s² to avoid accident Answer

Question (3.9) Two town 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 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?

Solution

Given, The speed of man cycling (v_m)= 20 km/h

Time interval of bus goes in the direction of cyclist = 18 m = 18/60 h

And, time interval of bus goes in the opposite direction of cyclist = 6 m = 6/60h = 1/10 h

Thus, speed of bus and time period (T) of bus = ?

Let the speed of bus running between town A and B = v_B

And time period of plying of bus = T

Let the distance covered by bus = D

In the case of bus goes in the direction of cyclist

The relative speed of bus with respect to the cyclist (v_BC) = Speed of bus – speed of cyclist

[Since bus is going in the same direction as cyclist]

⇒ v_BC = (v_B – 20) km/h

We know that Distance = Speed × Time

Thus, distance covered by bus,

D = (v_B – 20) km/h × 18/60 h - - - - - (i)

And as given in the question, one bus goes at a duration of every T minutes = T/60 h

Thus, distance covered by bus,

D = v_B × T/60 - - - - - (ii)

Now, from equation (i) and (ii), we get

v_B × T/60 = (v_B – 20) × 18/60 - - - - - (iii)

In the case when bus goes in the opposite direction of cyclist

The relative speed of bus with respect to cyclist (v_BC)

= speed of bus + speed of cyclist

= v_B + 20 km/h

Thus, the distance covered by bus,

D = (v_B + 20) × 1/10 km - - - - - (iv)

Now, from equation (ii) and (iv) we get

v_B × T/60 = (v_B + 20) × 1/10 - - - - (v)

Now, from equation (iii) and (v), we get

(v_B + 20) × 1/10 = (v_B – 20) × 18/60

⇒ (v_B + 20) × 1 = (v_B – 20) × 18/6

⇒ v_B + 20 = (v_B – 20) × 3

⇒ v_B + 20 = 3v_B – 60

⇒ 20 + 60 = 3v_B – v_B

⇒ 80 = 2v_B

⇒ v_B = 80/2 = 40 km/h

Thus, speed of bus (v_B) = 40 km/h

Now, After substituting the value of (v_B) in equation (v), we get

After cross multiplication, we get

4T = 36

⇒ T = 36/4 = 9 minute

Thus, time period T of bus service = 9 minute and speed of bus = 40 km/h Answer

Question (3.10) A player throws a ball upwards with an initial speed of 29.4 m/s.

(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 dos the ball rise and after how long does the ball return to the player's hands? (Take a = 9.8 m/s^{2 and neglect air resistance)}

Solution

(a) When a ball is thrown in upward direction, the direction of acceleration of ball is vertically downward.

Explanation

When a ball or anything is thrown in upward direction, its speed decreased continuously because of gravitational force acting upon it.

And we know that rate of change in velocity is called acceleration.

This means the direction of acceleration of a ball thrown in upward direction is in vertically downward direction.

(b) The velocity of the ball becomes zero at the highest point and acceleration of the ball is in vertically down ward direction and is equal to 9.8 m/s²

Explanation

When a ball or anything is thrown in upward direction, its velocity gradually decreases and finally it becomes zero at which the ball reaches the highest point. And the acceleration of the object thrown in vertically upward direction is 9.8 m/s² due to gravitational pull of earth.

(c) The position of ball (x) in both upward and downward direction is greater than zero, thus sign of position is positive in both the direction.

The velocity of the ball during vertically upward direction is less than zero, and thus velocity of the ball during upward direction is negative.

The velocity of the ball during vertically downward direction is greater than zero, and thus velocity of the ball during downward direction is positive.

The acceleration of the ball in both upward and downward direction is greater than zero, this means sign of acceleration during both upward and downward motion is positive.

(d) Given, initial velocity of the ball (u) = –29.4 m/s [Because ball is moving in upward direction so velocity has negative sign]

And final velocity (v) = 0 [Because ball stops after reaching the highest point. ]

Acceleration (a)= g = 9.8 m/s² [Acceleration is due to gravity of earth]

Thus, maximum height (h) = s = ?

And time (t) = ?

Calculation of height (h=s)

We know that, v² = u² + 2ah

⇒ 0 = (–29.4)² + 2 × 9.8 h

⇒ 0 = 864.36 + 19.6 h

⇒ 19.6 h = –864.36

⇒ h = –44.1 m

Calculation of time (t)

We know that, v = u + at

⇒ 0 = –29.4 + 9.8t

⇒ 9.8t = 29.4

⇒ t = 3 second

Since, ball takes 3 second to reach at the highest point, thus same time will be taken to reach the ground, i.e. in the hand of player.

Thus, total time = 3s + 3s = 6second

Thus, maximum height of ball = 44.1 m and total time of ball to reach the ground = 6 second Answer