1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Explanation:
Yes, an object can have zero displacement even though it has moved through a distance.
Displacement is the straight-line distance between the initial and final position of an object, measured along a particular direction.
So, if an object after moving comes back to its initial position then its displacement is zero.
For example, suppose an object moves from point A to point B, then back to point A along the same path. The distance covered by the object is the total length of the path it followed, but the object ended up at the same position as it started, so its displacement is zero.
2:A farmer moves along the boundary of a square field of side 10m 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?
2:A farmer moves along the boundary of a square field of side 10m 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?
Explanation:
We have given,
Length of side of the field = 10m
So, the perimeter of the square = 40 m
Time taken by the farmer to cover perimeter of 40 m = 40 s
So, the farmer cover 1 m in 1 s
Now,
Distance covered by the farmer in 2 min 20 sec-
Total time 2 min 20 sec = (2×60)+20=140s,
=1×140 = 140m
Total number of rotations taken by the farmer to cover a distance of 140 m =distance/perimeter =40=3.5
Then, the displacement from the pythagoras theorem is, s =√ (102+102)
s = 10√2
s = 14.14 m
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.
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.
Explanation:
Neither (a) nor (b) is true for displacement.
a) Displacement can be zero if the object returns to its initial position.
For example, if an object moves forward 5 meters and then moves back 5 meters to its original position, its displacement is zero.
b) Displacement is the shortest distance between the initial and final positions of an object. Therefore, the magnitude of displacement is always less than or equal to the distance travelled by the object.
4.Distinguish between speed and velocity.
4.Distinguish between speed and velocity.
Explanation:
Speed and velocity are both measures of an object's motion, but they have different meanings.
Speed refers to the distance travelled by an object per unit of time, regardless of the direction of motion. It is a scalar quantity and is measured in units of distance divided by time, such as meters per second (m/s) or kilometers per hour (km/h).
Velocity, on the other hand, refers to the rate at which an object changes its position in a particular direction. It is a vector quantity that takes into account both the magnitude (speed) and direction of the object's motion. Velocity is measured in units of distance divided by time, with a direction component, such as meters per second to the east (m/s east) or kilometers per hour to the north (km/h north).
5.Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
5.Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Explanation:
The magnitude of average velocity of an object is equal to its average speed when the object travels in a straight line and its motion is in a single direction.
In other words, if an object moves along a straight path without changing direction, then the magnitude of the average velocity and the average speed will be the same. This is because, in this case, the direction of motion is the same as the direction of displacement, so the sign of the velocity and speed will be the same, and there will be no change in direction to affect the overall magnitude of displacement of the motion.
6.What does the odometer of an automobile measure?
6.What does the odometer of an automobile measure?
Explanation:
The odometer of an automobile is a device that measures the total distance travelled by the vehicle. It does this by tracking the number of rotations of the vehicle's wheels and its perimeter then converting this information into distance traveled.
7.What does the path of an object look like when it is in uniform motion?
7.What does the path of an object look like when it is in uniform motion?
Explanation:
The path of the object in uniform motion is a straight line, as the object moves in the same direction with the same speed throughout its motion. This is true whether the object is moving horizontally, vertically, or at any angle in between, as long as its speed and direction of motion remain constant.
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.
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.
Explanation:
To calculate this we can use the formula:
Speed=Distance/Time ……………(1)
The speed of the signal given is = 3×108 m/s
And the time is = 5 mins = 5 × 60 = 300s
Putting these values in equation (1), we get,
Distance=(3×108)300=900×108=9 × 1010 m
So, the distance of the spaceship from the ground station is 9×1010 m.
9.When will you say a body is in
a) uniform acceleration?
b) non-uniform acceleration?
9.When will you say a body is in
a) uniform acceleration?
b) non-uniform acceleration?
Explanation:
a) A body is said to be in uniform acceleration if it undergoes a constant change in velocity over equal intervals of time. In other words, its acceleration remains constant over time. This means that the velocity of the object increases or decreases at a steady rate.
For example, an object falling freely under the influence of gravity experiences uniform acceleration, as the acceleration due to gravity is constant and the object's velocity increases at a steady rate.
b) A body is said to be in non-uniform acceleration if its acceleration changes over time. This means that the velocity of the object increases or decreases at a changing rate. Non-uniform acceleration can be either increasing or decreasing, and the acceleration can be either positive or negative.
For example, a car accelerating from rest experiences non-uniform acceleration, as the acceleration increases as the car's speed increases.
10.A bus decreases its speed from 80 km h^–1 to 60 km h^–1 in 5 s. Find the acceleration of the bus.
10.A bus decreases its speed from 80 km h^–1 to 60 km h^–1 in 5 s. Find the acceleration of the bus.
Explanation:
The initial speed of the bus is
u = 80 km/h = m/s
and its final speed is v = 60 km/h = m/s
The time taken for the bus to decrease its speed from 22.22 m/s to 16.67 m/s is t = 5 s.
The acceleration of the bus can be calculated using the formula:
a = (v - u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time taken.
Substituting the given values, we get:
a = (16.67 m/s - 22.22 m/s) / 5 s
a = -5.55 m/s / 5 s
a = -1.11
Therefore, the acceleration of the bus is -1.11 . The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
11.A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.
11.A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.
Explanation:
Initial speed, u = 0 km/h (since the train starts from rest)
Final speed, v = 40 km/h = 40 5/18=11.1 m/s
Time taken, t = 10 minutes = 600 seconds
We know that the acceleration of the train, a is given by:
a = (v - u) / t
Substituting the values, we get:
a = (11.1 m/s - 0 m/s) / 600 s
a = 0.0185 m/s2.
12.What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
12.What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Explanation:
The distance-time graph for uniform motion of an object is a straight line with a constant slope, indicating a constant speed.
The distance-time graph for non-uniform motion of an object is a curved line, indicating a changing speed
13.What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
13.What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Explanation:
If the distance-time graph of an object is a straight line parallel to the time axis, then the object is at rest or not moving.
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?
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?
Explanation:
If the speed-time graph of an object is a straight line parallel to the time axis, it means that the speed of the object is constant with time. This indicates that the object is moving with a uniform velocity.
15.What is the quantity which is measured by the area occupied below the velocity-time graph?
15.What is the quantity which is measured by the area occupied below the velocity-time graph?
Explanation:
The velocity-time graph is
The area below the velocity-time is represented by the rectangle OABC.
Where ,
OA or BC represents the velocity and
AB or OC represents the time
The are of the rectangle can be calculated as OA ×OC
Which is velocity ×time
And we know velocity ×time = displacement
Therefore, the area under the velocity-time graph gives the displacement.
16.A bus starting from rest moves with a uniform acceleration of 0.1 m/s2 for 2 minutes.
Find (a) the speed acquired, (b) the distance travelled.
16.A bus starting from rest moves with a uniform acceleration of 0.1 m/s2 for 2 minutes.
Find (a) the speed acquired, (b) the distance travelled.
Explanation:
Given:
Initial velocity, u = 0 m/s
Acceleration, a = 0.1 m/s2
Time, t = 2 minutes = 120 seconds
(a) To find the speed acquired by the bus, we can use the equation:
v = u + at
where v is the final velocity.
Substituting the values, we get:
v = 0 + 0.1 120
v = 12 m/s
Therefore, the speed acquired by the bus is 12 m/s.
(b) To find the distance travelled by the bus, we can use the equation:
s = ut +1/2 at2
where s is the distance travelled.
Substituting the values, we get:
s = 0 ×120 + (1/2) ×0.1 ×(120)2
s = 720 m
Therefore, the distance travelled by the bus is 720 m.
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.
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.
Explanation:
First, the initial velocity of the train is:
90 km/h 5/18 = 25 m/s
Next, we need to use the kinematic equation that relates the final velocity (when the train comes to rest) to the initial velocity, acceleration, and distance traveled:
v2-u2=2as
where v is the final velocity (0 m/s in this case), u is the initial velocity (25 m/s), a is the acceleration (-0.5 m/s2), and s is the distance traveled.
Solving for s, we get:
s = v2-u2/2as
= 0-252/2(-0.5)
= 625 m
Therefore, the train will travel 625 meters before coming to a stop when the brakes are applied with a uniform acceleration of -0.5 m/s2.
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?
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?
Explanation:
First we need to convert the acceleration from cm/s2 to m/s2:
2 cm/s2 = 0.02m/s2
We can use the kinematic equation that relates velocity, acceleration, and time:
v = u + at
where v is the final velocity, u is the initial velocity (which is zero since the trolley starts from rest), a is the acceleration, and t is the time elapsed.
Substituting the given values into the equation, we get:
v = 0 + (0.02 m/s2) 3 s
= 0.06 m/s
Therefore, the velocity of the trolley 3 seconds after the start is 0.06 m/s.
19.A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
19.A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Explanation:
We can use the kinematic equation that relates distance, acceleration, and time:
s = ut + 12at2
where s is the distance covered, u is the initial velocity (which we assume to be zero since the car starts from rest), a is the acceleration, and t is the time elapsed.
Substituting the given values into the equation, we get:
s = 0+ (1/2)4m/s2(10s)2
= 200 m
Therefore, the racing car will cover a distance of 200 meters in 10 seconds after the start, assuming a uniform acceleration of 4 m/s2.
20.A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. 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?
20.A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. 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?
Explanation:
First find the time taken by the stone to reach its maximum height:
v = u + at
where v is the final velocity (which is zero at the maximum height), u is the initial velocity (5 m/s in the upward direction), a is the acceleration (-10 m/s2 in the downward direction), and t is the time taken.
Substituting the given values into the equation, we get:
0 = 5 m/s + (-10 m/s2) ×t
t = 0.5 s (positive value as time can never be negative)
Therefore, the time taken by the stone to reach its maximum height is 0.5 seconds.
Now, let's find the maximum height reached by the stone:
s = ut + 1/2at2
where s is the displacement (height), u is the initial velocity (5 m/s in the upward direction), a is the acceleration (-10 m/s2 in the downward direction), and t is the time taken (0.5 s).
Substituting the given values into the equation, we get:
s = 5 m/s ×0.5 s + 1/2 ×(-10 m/s2)× (0.5 s)2
= 1.25 m
Therefore, the height attained by the stone is 1.25 meters, and it takes 0.5 seconds to reach there, assuming an acceleration of 10 m/s2 in the downward direction.
21.An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
21.An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Explanation:
The distance covered by the athlete after 2 minutes 20 seconds is equal to the distance covered in one round multiplied by the number of rounds completed in 2 minutes 20 seconds:
2 minutes 20 seconds = 140 s
Number of rounds completed = (140 s) / (40 seconds per round)=3.5 rounds
Distance covered = 3.5 rounds circumference of the circular track
= 3.5 (π diameter)
= 3.5 (22/7× 200 m)
= 2200 m
As the number of rounds completed by the athlete in 140 s is 3.5, the athlete is going to end up at the side opposite (point B) from the initial position (point A).
Therefore, the displacement is 200 m.
22.Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute.
What are Joseph’s average speeds and velocities in jogging
(a) from A to B and (b) from A to C?
22.Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute.
What are Joseph’s average speeds and velocities in jogging
(a) from A to B and (b) from A to C?
Explanation:
(a) To find the average speed and velocity of Joseph from A to B, we can use the formula:
Average speed = Total distance traveled / Total time taken
Average velocity = Total Displacement / Total time taken
A to B is a distance of 300 meters.
The time taken by Joseph to jog from A to B is 2 minutes 30 seconds = 150 seconds.
The average speed of Joseph from A to B is:
Average speed = 300 m / 150 s
= 2 m/s
The average velocity of Joseph from A to B is:
Average velocity = 300 m / 150 s
=2 m/s
(a) The average speed and velocity of Joseph from A to C is given as:
Distance A to C = (A to B) + (B to C) = 300 + 100 = 400 m
Displacement A to C = (A to B) - (B to C) = 200 m
Time taken to jog from A to C = 150 + 60 = 210 s
The average speed of Joseph from A to C is:
Average speed = 400 m / 210 s
= 1.9 m/s
The average velocity of Joseph from A to C is:
Average velocity = 200 m / 210 s
= 0.95 m/s
23.Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
23.Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
Explanation:
We can start by using the formula for average speed, which is:
average speed = total distance / total time
Let's assume that the distance from Abdul's home to school is "d" km. Then the total distance for the whole trip is 2d km (since he traveled the same distance twice). Let's also assume that the time it took Abdul to travel to school at 20 km/h is "t1" hours, and the time it took him to return at 30 km/h is "t2" hours.
We know that:
distance = speed time
So, we can rearrange the formulas for distance to get:
time = distance / speed
Using this formula, we can express the time it took Abdul to travel to school and back as:
t1 = d / 20
t2 = d / 30
The total time for the whole trip is the sum of these two times:
total time = t1 + t2
Substituting the expressions we just found for t1 and t2, we get:
total time = (d / 20) + (d / 30)
To find the average speed for the whole trip, we need to divide the total distance (2d km) by the total time.
So, we get:
average speed = total distance / total time
average speed = 2d / [(d/20) + (d/30)]
Simplifying this expression, we get:
average speed = 2d / [(3d + 2d) / 60]
average speed = 2d / (5d / 60)
average speed = 24 km/h
Therefore, the average speed for Abdul's trip is 24 km/h.
24.A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s^2 for 8.0 s. How far does the boat travel during this time?
24.A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s^2 for 8.0 s. How far does the boat travel during this time?
Explanation:
We can use the kinematic equation to calculate the distance traveled by the boat during this time. The equation we need is:
s = ut + 1/2at2
where s is the distance traveled, u is the initial velocity (which is 0 in this case), a is the acceleration, t is the time
Putting in the given values, we get:
s = 0 + 1/2 (3 m/s2) (8 s)2
s = 96 meters
Therefore, the boat travels a distance of 96 meters during the 8.0 seconds of acceleration.
25.A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
25.A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Explanation:
The total distance can be calculated by calculating the area under each graph
Distance travelled by first car:
Area of triangle AOB
= 1/2×OB×OA
OB = 5 s and OA = 52 km/h = 14.4 m/s
= 1/2×5×14.4
= 36 m
Distance travelled by second car:
Area of triangle COD
= 1/2×OD×OC
OD = 10 s and OC = 3 km/h = 0.83 m/s
= 1/2×10×0.83
= 4.15 m
Therefore, the first car (traveling at 52 km/h) traveled farther than the second car (traveling at 3 km/h) after the brakes were applied.
26. shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
26. shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Explanation:
(a) B is travelling at the fastest speed because it has the greatest slope among the three lines.
(b) The three objects never meet at the same point on the road because the three lines do not intersect at a single point.
(c) One graph unit is equal to 4/7 km because there are 7 unit areas of the graph between 0 and 4 on the Y axis. The initial distance of object C from the origin is 16/7 km because its initial point is 4 graph units away from the origin.
When B passes A, the distance between the origin and C is 8km.
Therefore, the total distance travelled by C in this time is
8 - (16/7) = 5.71 km
(d) The distance covered by object B at the point where it passes C is 9 graph units.
Thus, the total distance travelled by B when it crosses C is
9 (4/7) = 5.14 km
27.A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
27.A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
Explanation:
Let's denote the initial velocity of the ball as u, the final velocity as v, the acceleration as a, the distance traveled as s, and the time taken as t.
From the problem statement, we know that u = 0 (the ball is dropped from rest), a = 10 m/s2 , and s = 20 m (the height from which the ball is dropped).
Using the equation:
s = ut + 12at2
We can solve for t:
20 m = 0t + 1/2 (10 m/s2) t2
Simplifying and solving for t, we get:
t =√4 = 2 seconds
So it takes 2 seconds for the ball to reach the ground.
Now, using the equation:
v = u + at
We can solve for v:
v = 0 + (10 m/s2)(2 s)
v = 20 m/s
Therefore, the ball will strike the ground with a velocity of 20 m/s.
28.The speed-time graph for a car is shown in Fig. 8.12
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
28.The speed-time graph for a car is shown in Fig. 8.12
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Explanation:
a)
The car's distance over a period of 4 seconds is represented by the shaded area, which can be determined using the formula:
distance = (1/2) ×time ×speed
Plugging in the values for time (4 seconds) and speed (6 m/s), we get:
distance = (1/2) ×4×6 = 12 meters
Hence, we can conclude that the car covers a total distance of 12 meters during the first four seconds of travel.
(b) Since the speed of the car remains constant from the points (x=6) to (x=10), it can be inferred that the car is moving in uniform motion during this time period.
29.State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.
29.State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.
Explanation:
(a) It is possible for an object to have constant acceleration but zero velocity. This occurs when an object is thrown vertically upwards and reaches its highest point, at which its velocity is zero while its acceleration remains constant (due to the force of gravity pulling it back down).
(b) It is possible for an object to move with an acceleration but with uniform speed.
This occurs when an object moves in a circular path with uniform speed, it experiences acceleration. This is because the direction of the object's velocity is constantly changing due to its circular motion, resulting in a continuous acceleration towards the center of the circle (known as centripetal acceleration).
(c) It is possible for an object to move in a certain direction with an acceleration in the perpendicular direction. This occurs when an object is moving in a curved path, and its acceleration is directed towards the center of the curve (centripetal acceleration). For example, a car moving around a circular track experiences an acceleration towards the center of the track, even though it is moving in a forward direction along the track.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
We can use the formula for the speed of an object in circular motion to calculate the speed of the artificial satellite:
v = (2 ×π× r) / T
where v is the speed, r is the radius of the orbit, and T is the time period for one revolution.
In this case, the radius of the orbit is 42250 km, and the time period for one revolution (the satellite's orbital period) is 24 hours or 86400 seconds.
Putting these values into the formula, we get:
v = (2×π×42250) / 86400
v = 3.074 km/s
Therefore, the speed of the artificial satellite is approximately 3.074 km/s.
Also Read: Motion Class 9 Extra Questions and Answers
