Motion

Question: 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with example.

Answer: Yes.

Suppose an object starts from point A, goes to B and return to A.

In this case distance covered by the object is equal to 70 km +70 km = 140 km, but displacement of the object will be zero, as it return to A.

Thus, even after moving through a distance displacement can be zero.

Question: 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:

Let, ABCD is the given square field.

Given, one side of the field = 10 m

Time taken to complete one round of the filed = 40 s

Displacement after 2 minute 20 second = ?

2 m 20 s = 2 × 60 +20 = 140 s

Thus, displacement after 140 s =?

We know that perimeter of a square = 4 × side

∴ perimeter of the given field = 4 × 10 m

Or, perimeter = 40 m

Since, farmer takes 40 s to complete one round i.e. = 40 m

∴ in 1 s distance covered by farmer `=40/40 = 1\ m`

∴ in 140 s distance covered by farmer `=1 \ m xx 140 = 140\ m`

Now, number of rotation = (text{Total distance})/(text{Perimeter})`

`= (140\ m)/(40\ m) =3.5\ m`

This means in the given time of 140 s farmer reached to the point C.

Now, from the figure,

AB = 10 m, BC = 10 m

Thus, displacement AC =?

Now, since given field is square, thus here AC is hypotenuse, thus by applying Pythagoras theorem, the AC can be calculated.

We know that, hypotenuse² = Base² + height²

`:. AC^2 = AB^2+BC^2`

`=>AC^2 = (10\ m)^2+(10\ m)^2`

`=>AC^2 = 100\ m^2 + 100\ m^2`

`=>AC^2 = 200\ m^2`

`=>AC = sqrt(200\ m^2)`

`=>AC = sqrt(100\ m^2 xx2)`

`=>AC = 10sqrt2\ m`

Thus, displacement after given time = `10sqrt2\ m` Answer

Question: 3. Which of the following is true for displacement?

(a) It cannot be zero.

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

Answer: Both of the statements for displacement are not true.

Explanation: Displacement can be zero and hence its magnitude will always be equal to or less than that of the distance travelled.

Question: 4. Distinguish between speed and velocity.

Answer:

(a) Speed has only magnitude, while velocity has magnitude and direction both.

(b) The magnitude of speed is always positive while the magnitude of velocity can be positive, negative or even zero.

(c) Speed = `(text{distance})/(text{time taken})` while

Velocity = `(text{displacement})/(text{time interval})`

Question: 5. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer: When an object travels along a straight line, the magnitude of average velocity is equal to the magnitude of average speed.

Or, if the total distance covered by an object is equal to the displacement, then average velocity will be equal to the average speed.

Question: 6. What does the odometer of an automobile measure?

Answer: Odometer measures the distance travelled by an automobile.

Odometer came from Greek word 'hodos+metron' in which 'hodos' means 'path' and 'metron' means 'to measure'. Hence, the device Odometer is meant to measure the length of path that is distance.

Question: 7. What does the path of an object look like when it is in uniform motion?

Answer: Path of object looks like a straight line when object is in the uniform motion.

Question: 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m s^–1.

Answer:

Given, time taken = 5 m = 5 × 60 = 300 s

Speed of light = `3xx10^8\ m\ s^(-1)`

Thus, distance (s) =?

We know that, distance = speed × time

`=> s = 3xx10^8\ m\ s^(-1) xx 300\ s`

`=>s = 9 xx 10^10\ m`

Thus, distance of spaceship from ground station = `9 xx10^10\ m` Answer

Question: 9. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Answer: (i) When a body is moving along a straight line and velocity is changes with uniform rate in equal interval of time, the body is said in uniform acceleration.

(ii) When a body is in motion along a straight path and its velocity changes at non-uniform rate the body is said to be non-uniform acceleration.

In other words, when a body is in motion along a straight path and its velocity is increasing or decreasing with unequal amounts in an equal interval of time, the body is said to be in non-uniform acceleration.

Question: 10. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Solution:

Given, Initial velocity, `u` = 80 km/h

`= (80 xx 1000\ m)/(60xx60\ s)`

`=(80000\ m)/(3600\ s)`

`=>u=22.22\ m/s`

And given, final velocity, `u`= 60 km/h

`=(60xx1000\ m)/(60xx60\ s)`

`=(60000\ m)/(3600\ s)`

`=>v=16.66\ m//s`

And given, time = 5 s

We know that, `v = u+at`

`=>16.66\ m//s = 22.22\ m//s + a xx 5\ s`

`=>axx 5\ s = (16.66-22.22)m//s`

`=>axx5\ s = -5.56\ m//s`

`=>a = (-5.56\ m//s)/(5\ s)`

`=>a = -1.112\ m\ s^(-2)`

Thus, acceleration of the bus is equal to `-1.112\ m\ s^(-2)` Answer

Question: 11. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes. Find its acceleration.

Solution:

Given, Initial velocity, `u`=0

[Because train is starting from a railway station, i.e. starts from rest]

Final velocity, `v` = 40 km/h

`=40 xx 5/18\ m//s`

`=>v=11.11\ m//s`

And given, time, `t` = 10 minute = 10 × 60 s

`=>t = 600\ s`

Thus, Acceleration, `a` =?

Now, we know that, `v = u+at`

`=>11.11\ m//s = 0+axx600\ s`

`=>a = (11.11\ m//s)/(600\ s)`

`=>a= 0.0185\ m\ s^(-2)`

Thus, acceleration of the given train `= 0.0185\ m\ s^(-2)` Answer

Question: 12. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer: The distance-time graph for object in uniform motion is a straight line.

The distance-time graph for object in non-uniform motion is a curved line.

Question: 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer: If distance time graph of an object is a straight line and parallel to the time axis, then object will be in rest. A straight line of distance axis shows that no distance is covered with time, this means object is in rest.

Question: 14. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer: If speed time graph of an object is straight line parallel to the time axis, this means object is moving with constant speed. This means that object is in uniform motion.

Question: 15. What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer: The area occupied below the velocity time graph shows the distance travelled by the object in given time interval.

Question: 16. A bus starting from rest moves with a uniform acceleration of 0.1 m s^–2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer:

Given,

Initial velocity, `u` =0 [Since, bus starting from rest]

Acceleration, `a=0.1\ m\ s^(-2)`

Time, `t` = 2 minute = 2 × 60 s = 120 s

Thus, final velocity, i.e. speed, `v` = ?

And distance, `s` = ?

We know that, `v= u +at`

`=>v = 0 + 0.1\ m\ s^(-2) xx 120\ s`

`=>v=12\ m//s`

Again, we know that, `s=ut+1/2\ at^2`

`=>s = 0 xx t + 1/2 xx 0.1\ m\ s^(-2)xx(120\ s)^2`

`=>s = 1/2xx0.1\ m\ s^(-2)xx14400\ s^2`

`=>s = 720\ m`

Thus, speed acquired = 12 m/s and distance covered = 720 m Answer

Question: 17. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of –0.5 m s^–2. Find how far the train will go before it is brought to rest.

Solution:

Given, Initial velocity, `u` = 90 km/h

`=90 xx 5/18\ m//s`

[multiplying with `5/18` changes km/h into m/s]

`=>u=25\ m//s`

And, given acceleration, `a=-0.5\ m\ s^(-2)`

Distance =?

Here final velocity, `v`=0 [Because train comes to stop after applying brake]

We know that, `v^2 = u^2 + 2as`

`=>0^2 = (25\ m//s)^2 +2xx(-0.5\ m\ s^(-2)\ s`

`=>0 = 625\ m^2\ s^2-0.1m\ s^(-2)xxs`

`=>0.1m\ s^(-2)xxs= 625\ m^2\ s^2`

`=>s = (625\ m^2\ s^2)/(0.1\ m\ s^(-2)`

`=>s = 625\ m`

Thus, train will cover a distance of 625 m before it is brought to rest. Answer

Question: 18. A trolley, while going down an inclined plane, has an acceleration of 2 cm s^–2. What will be its velocity 3 s after the start?

Answer:

Given, Initial velocity, `u`=0

Acceleration, `a=2cm\ s^(-2)`

`=>a=2/100\ m\ s^(-2)`

` =>a=0.02\ m\ s^(-2)`

And given, time, `t=3\ s`

Thus, Final velocity, `v` = ?

We know that, `v = u+at`

`=>v = 0 + 0.02\ m\ s^(-2)xx3\ s`

`=>v = 0.06\ m//s`

`=>v=0.06xx100\ cm//s`

Thus, after given time of 3 s, the velocity of trolley becomes 0.06 m/s or 6 cm/s. Answer

Question: 19. A racing car has a uniform acceleration of 4 m s^–2. What distance will it cover in 10 s after start?

Answer:

Given, Initial velocity, `u`=0

Acceleration, `a=4\ m\ s^(-2)`

Time, `t=10\ s`

Thus, distance, `s` = ?

Here, to calculate the distance covered in the given time, first, its final velocity will be calculated.

We know, that, `v = u + at`

`=>v = 0 + 4\ m\ s^(-2) xx 10\ s`

`=>v = 40\ m//s`

Now, we know that, `s = ut + 1/2 at^2`

`=>s =0xx10\ s +1/2xx 4\ m\ s^(-2) xx (10\ s)^2`

`=>s = 2\ m\ s^(-2)xx100\ s^2`

`=>s = 200\ m`

Thus, distance covered by the car in the given time of 10 s is equal to 20 m. Answer

Question: 20. A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m s^–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Solution:

Here, given, initial velocity, `u=5\ m//s`

Final velocity, `v=0` [Since after achieving a certain height stone will come into rest before falling down]

Acceleration, `a=10\ m\ s^(-2)`

Since, stone is thrown in upward direction, thus, gravity will be applied to it, and hence acceleration will be negative,

Thus, acceleration, `a = -10\ m\ s^(-2)`

Thus, height, that is, distance, `s` = ?

And, time, `t` = ?

We know that, `v=u+at`

`=>0=5\ m//s + (-10\ m\ s^(-2))xxt`

`=>10\ m\ s^(-2)xxt=5\ m//s`

`=>t = (5\ m//s)/(10\ m\ s^(-2)`

`=>t=0.5\ s`

Now, we know that, `v^2=u^2+2as`

`=>0^2=(5\ m//s)^2+2xx(-10\ m\ s^(-2))xxs`

`=>0=25\ m^2\ s^2 -20\ m\ s^(-2)xxs`

`=>20\ m\ s^(-2)xxs=25\ m^2\ s^2`

`=>s = (25\ m^2\ s^2)/(20\ m\ s^(-2)`

`=>s = 1.25\ m`

Thus, stone will achieve a height 1.25 m and time taken to achieve this height is 0.5 s. Answer