Motion In a Straight Line - 11th physics

Motion in A Straight Line eleventh physics NCERT Question (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

Solution

(a) True

Explanation When anything is thrown upward direction, on reaching at maximum height its speed becomes zero but acceleration due to gravity is continuously acting upon it.

Thus, a particle in one-dimensional motion with zero speed at an instant may have non-zero acceleration at that instant.

(b) False

Explanation We know that velocity is the magnitude of speed with direction. Thus it is not possible that a particle in motion in one dimension can have non-zero velocity with zero speed.

(c) True

Explanation Acceleration is the rate of change in velocity with time. Thus when a particle in motion with constant speed have zero acceleration.

(d) False

Explanation It is not necessary that with positive acceleration speed will increase. Because when an object is thrown in upward direction, the acceleration of the object is positive but velocity is decreasing upto reaching at the maximum height.

Motion in A Straight Line eleventh physics NCERT Question (3.12) A ball is dropped from a height of 90m 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.

Solution

Given, height (h) = 90 m

Time (t) = 0 to 12 second

Initial velocity (u) = 0 (Since ball is dropped from a height, thus initial velocity will be equal to zero)

Acceleration (a) = g = 9.8 m/s² (Since ball is dropped from a height, thus acceleration will be because of gravity of earth)

Thus, final velocity (v) and time (t) = ?

We know that, s = ut + 1/2 gt²

⇒ 90 m = 0 × t + (1/2 × 9.8 t²)

⇒ 90 = 1/2 × 9.8 t²

⇒ 90 = 4.9 t²

⇒ t² = 90/4.9

⇒ t² = 900/49

⇒ t = 30/7 = 4.29 second

Now, we know that, v = u + at

⇒ v = 0 + 9.8 × 4.29

⇒ v = 42.042 m/s

Thus, velocity of ball while falling on the ground = 42.04 m/s

Rebound Velocity

Since, as per question, ball loses its 1/10^th speed at each collision with ground

Thus, velocity of ball after 1^st collision (v_1c)

= velocity of ball – velocity of ball × 1/10

= 42.042 – 42.042 × 1/10

= 42.042 – 4.2042

= 37.838 m/s ≈ 37.84 m/s

⇒ v_1c = 37.84 m/s

Calculation after 1^st collision

Initial velocity (u) = 37.84 m/s

Final velocity (v) = 0

Acceleration (g) = –9.8m/s²

Thus, height (h) = ? and time (t) =?

We know that, v = u + at

⇒ 0 = 37.84 + (–9.8t)

⇒ 0 = 37.84 – 9.8t

⇒ 9.8t = 37.84

⇒ t = 3.86 second

Thus, time taken by ball to reach the maximum height = 3.86 second

Consequently ball will take same time to reach on the ground.

Thus, time taken by ball to reach on the ground for 2^nd collision = 3.86 second

Thus, total time taken upto second collision = time taken in 1^st collision + time taken to reach at maximum height after 1^st collision + time taken to reach the ground for 2^nd collision

= 4.29 s + 3.86 s + 3.86 s

Thus, total time taken upto reaching the ground after 1^st collision = 12.01 s

Now, since ball loses it's 1/10^th speed after each collision

Thus, velocity after 2^nd collision = velocity after 1^st collision – velocity after 1^st collision × 1/10

= 37.84 – 37.84 ×1/10

= 37.84 – 3.784

= 34.056 m/s ≈34.06 m/s

Thus, velocity after second collision = 34.06 m/s

Now, we have

Speed of the ball before hitting the ground = 42.04 m/s

And time taken to hit the ground = 4.29 s

Speed of ball after first collision with ground = 37.84 m/s

Time taken to reach the ground for second collision = 8.15 s

Speed of ball after second collision of ground = 34.05 m/s

Thus, speed-time graph for its motion between t = 0 to 12s is given below

Motion in A Straight Line eleventh physics NCERT Question (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 time 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 only].

Solution

(a) Magnitude of displacement is equal to the change in position of object in the given time interval.

While Magnitude of distance is equal to the total path length.

(b) Magnitude of average velocity is the displacement of the object in unit time. While; magnitude of average speed is the total distance covered by a moving object in unit time.

That is Magnitude of average velocity

And, Magnitude of average speed

Example:

Let a cyclist moves from A to B and return to C in time t

Thus, here path length = AB + BC

And displacement = AC

Thus, Magnitude of Average speed

And, Magnitude of Average velocity

The magnitude of displacement is less than the distance in this case.

That is AC < AB + BC

But if the object is moving in the one direction only along a straight line, then magnitude of distance is equal to the magnitude of displacement.

In the given example, let a cyclist moves from A to B

Then, magnitude of distance, i.e. path length = Magnitude of displacement.

That is, AB = AB

Motion in A Straight Line eleventh physics NCERT Question (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. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. 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!]

Solution

Given, Distance of home to market = 2.5 km

Speed of man while going to the market = 5 km/h

And speed of man while returning to home from market = 7.5 km/h

Now, we know that, Time = Distance/speed

Thus, Time taken while going to the market (t_m) = 2.5 km/5kmh^–1

= 1/2 hour = 30 minute

And time taken while returning from the market to home (t_h)

= 2.5 km/7.5 kmh^–1

= 1/3 h = 20 minute.

Thus, total time in to and fro walking = 30 + 20 = 50 minute

Now,

(a) Magnitude of average velocity

Here as per question, man instantly returns back to home because of closing of market

Thus, displacement = 0

We know that Average velocity (v_a) = Displacement/interval of time

Thus, v_a = 0/50 m = 0

Thus, average velocity = 0

(b) Average speed

(i) Average speed between 0 to 30 min

= 1/2 hour = 0.5 h

Distance travelled between interval of time of 0 to 30 minute = 2.5 km

Thus, Average speed = Total path length / Interval of time

= 2.5 km/0.5 h = 5 km/h

Thus, Average speed between 0 to 30 min = 5 km/h Answer

(ii) Average speed between intervals of time of 0 to 50 min

= 50/60 hour = 5/6 hour

Since time taken to go to market and return to home = 50 min = 5/6 hour

And one side distance = 2.5 km

Thus, both side distance = 2.5 km + 2.5 km = 5 km

Thus, here we have Total path length = 5 km

And, Interval of Time = 5/6 hour

Thus, Average speed = Total path length/Interval of Time

= 5 km/5/6 hour

= 6 km/h

Thus, Average speed in the time interval of 0 to 50 min = 6 km/h Answer

(iii) Average speed between time intervals of 0 to 40 min

= 40/60 h = 2/3 hour

As per above calculation, time taken to reach the market = 30 minute

And time taken to return home from market = 20 min

And distance of home to market = 2.5 km

Now, since In 20 minute distance covered = 2.5 km

Thus, in 10 minute distance covered = 2.5/2 = 1.25 km

Thus, total distance covered between interval 0 to 50 minute

= distance covered in first 30 minute + distance covered in next 10 minute

= 2.5 km + 1.25 km

= 3.75 km

Thus, here we have, interval of time = 2/3 hour

And total path length = 3.75 km

Thus, Average speed = Total path length/Interval of time

= 3.75 km/2/3 h

= 5.625 km/h

Thus, average speed between interval of time of 0 to 40 min = 5.625 km/h Answer

Motion in A Straight Line eleventh physics NCERT Question (3.15) In Exercis 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?

The Velocity of a moving object at particular instants of time interval is called the Instantaneous Velocity.

That is the velocity (v) at an instant (t) is termed as Instantaneous Velocity.

The velocity at an instant is defined as the limit of the average velocity as the time interval Δt becomes infinitesimally small.

In other words, instantaneous velocity,

Since, in the case of instantaneous velocity the change in time is so small that the magnitude of displacement is equal to the distance covered by moving object in that instant of time.

Thus, magnitude of speed is always equal to the magnitude of instantaneous velocity.