NCERT Class 9 Science Solutions: Motion
Question:
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?
Concept in a Minute:
Circular motion, distance, displacement, speed, time, unit conversion. Distance is the total path length covered. Displacement is the shortest straight-line distance from the initial to the final position. For circular motion, distance is circumference times the number of rounds. Displacement is zero if the object completes full rounds, otherwise it’s the chord length.
Explanation:
The athlete completes one round of a circular track in 40 seconds.
The diameter of the track is 200 m, so the radius is 100 m.
The circumference of the track is C = 2 * pi * r = 2 * pi * 100 m = 200 * pi m.
We need to find the distance covered and displacement at the end of 2 minutes 20 seconds.
First, convert the total time into seconds:
Total time = 2 minutes + 20 seconds = (2 * 60) seconds + 20 seconds = 120 seconds + 20 seconds = 140 seconds.
Now, calculate the number of rounds completed by the athlete:
Number of rounds = Total time / Time per round
Number of rounds = 140 s / 40 s = 14 / 4 = 7 / 2 = 3.5 rounds.
Distance covered:
The distance covered in one round is the circumference of the track.
Distance covered = Number of rounds * Circumference
Distance covered = 3.5 * (200 * pi) m
Distance covered = 700 * pi m.
Using pi ≈ 3.14, Distance covered ≈ 700 * 3.14 m ≈ 2198 m.
Displacement:
The athlete completes 3.5 rounds. This means the athlete starts at a point on the circle and ends at a point diametrically opposite to the starting point after completing 3 full rounds and then half of the fourth round.
After 3 full rounds, the displacement is zero as the athlete returns to the starting point.
After completing half of the fourth round, the athlete will be at the point diametrically opposite to the starting point.
Therefore, the displacement is the diameter of the circular track.
Displacement = Diameter = 200 m.
Question:
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?
Concept in a Minute:
The relationship between speed, time, and distance is governed by the equations of motion for uniformly accelerated or decelerated objects. For a speed-time graph, the distance traveled is represented by the area under the curve. Uniform acceleration means the speed changes at a constant rate.
Explanation:
Step 1: Convert given speeds from km/h to m/s.
Car 1: 52 km/h = 52 * (1000 m / 3600 s) = 14.44 m/s (approximately)
Car 2: 3 km/h = 3 * (1000 m / 3600 s) = 0.83 m/s (approximately)
Step 2: For each car, determine the initial speed (u), final speed (v), and time (t).
Car 1: u = 14.44 m/s, v = 0 m/s, t = 5 s
Car 2: u = 0.83 m/s, v = 0 m/s, t = 10 s
Step 3: Plot the speed-time graphs.
For both cars, the y-axis represents speed (m/s) and the x-axis represents time (s).
Car 1: The graph will be a straight line starting from (0, 14.44) and ending at (5, 0). The slope of this line represents the acceleration.
Car 2: The graph will be a straight line starting from (0, 0.83) and ending at (10, 0). The slope of this line represents the acceleration.
Step 4: Calculate the distance traveled by each car after the brakes were applied. The distance traveled is the area under the speed-time graph, which will be a triangle in both cases.
Area of a triangle = 0.5 * base * height
Distance for Car 1 = 0.5 * t1 * u1 = 0.5 * 5 s * 14.44 m/s = 36.1 m
Distance for Car 2 = 0.5 * t2 * u2 = 0.5 * 10 s * 0.83 m/s = 4.15 m
Step 5: Compare the distances traveled.
Car 1 traveled 36.1 meters.
Car 2 traveled 4.15 meters.
Therefore, Car 1 travelled farther after the brakes were applied.
Question:
A bus decreases its speed from 80 km h−1 to 60 km h−1 in 5 s. Find the acceleration of the bus.
Concept in a Minute:
The key concept is acceleration, which is the rate of change of velocity. The formula for acceleration is a = (v – u) / t, where ‘v’ is the final velocity, ‘u’ is the initial velocity, and ‘t’ is the time taken. We also need to ensure all units are consistent, so we’ll need to convert km/h to m/s.
Explanation:
The problem asks us to calculate the acceleration of a bus that changes its speed over a specific time interval. We are given the initial speed (u), the final speed (v), and the time taken (t).
Step 1: Identify the given values.
Initial speed (u) = 80 km/h
Final speed (v) = 60 km/h
Time taken (t) = 5 s
Step 2: Convert speeds from km/h to m/s.
To convert km/h to m/s, we multiply by 5/18.
u = 80 km/h * (5/18) m/s per km/h = 80 * 5 / 18 m/s
v = 60 km/h * (5/18) m/s per km/h = 60 * 5 / 18 m/s
Calculate the values:
u = 400 / 18 m/s = 200 / 9 m/s
v = 300 / 18 m/s = 50 / 3 m/s
Step 3: Use the acceleration formula.
Acceleration (a) = (v – u) / t
Substitute the values:
a = ( (50/3) m/s – (200/9) m/s ) / 5 s
Step 4: Calculate the acceleration.
First, find a common denominator for the velocities:
v = 50/3 = 150/9 m/s
a = ( (150/9) m/s – (200/9) m/s ) / 5 s
a = ( (150 – 200) / 9 ) m/s / 5 s
a = ( -50 / 9 ) m/s / 5 s
a = -50 / (9 * 5) m/s²
a = -50 / 45 m/s²
a = -10 / 9 m/s²
Step 5: Express the answer in decimal form if required.
a ≈ -1.11 m/s²
The negative sign indicates that the acceleration is deceleration, meaning the bus is slowing down.
Final Answer: The acceleration of the bus is -10/9 m/s² or approximately -1.11 m/s².
Question:
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration.
Concept in a Minute:
Uniform acceleration problems involve the relationship between initial velocity, final velocity, acceleration, and time. The relevant kinematic equation is v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. It’s crucial to ensure all units are consistent before applying the formula.
Explanation:
The train starts from a railway station, which means its initial velocity (u) is 0.
The final speed attained is 40 km h⁻¹.
The time taken to attain this speed is 10 minutes.
To find the acceleration, we need to convert all units to be consistent. Let’s convert the final speed to meters per second (m/s) and the time to seconds (s).
Final speed (v) = 40 km h⁻¹
To convert km h⁻¹ to m s⁻¹, we multiply by (1000 m / 1 km) * (1 h / 3600 s) = 5/18.
v = 40 * (5/18) m s⁻¹ = 200/18 m s⁻¹ = 100/9 m s⁻¹
Time (t) = 10 minutes
To convert minutes to seconds, we multiply by 60.
t = 10 * 60 seconds = 600 seconds
Now, we can use the kinematic equation: v = u + at
We have:
v = 100/9 m s⁻¹
u = 0 m s⁻¹
t = 600 s
Substitute the values into the equation:
100/9 = 0 + a * 600
Now, solve for acceleration (a):
a = (100/9) / 600
a = 100 / (9 * 600)
a = 100 / 5400
a = 1 / 54 m s⁻²
The acceleration is approximately 0.0185 m s⁻².
If we want to express the acceleration in km min⁻², we can do the following:
Initial velocity (u) = 0 km min⁻¹
Final velocity (v) = 40 km min⁻¹ (since the time is in minutes)
Time (t) = 10 minutes
Using v = u + at:
40 = 0 + a * 10
a = 40 / 10
a = 4 km min⁻²
Question:
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.
Concept in a Minute:
The speed of an object moving in a circular path is determined by the distance it travels (circumference of the circle) and the time it takes to complete one revolution. For an artificial satellite orbiting the Earth, this involves calculating the orbital circumference and dividing it by the orbital period.
Explanation:
The satellite moves in a circular orbit. The distance the satellite travels in one revolution is the circumference of this orbit.
The formula for the circumference of a circle is $C = 2 \pi r$, where $r$ is the radius of the circle.
In this case, the radius of the orbit is given as $r = 42250$ km.
So, the circumference of the orbit is $C = 2 \times \pi \times 42250$ km.
The time it takes for the satellite to complete one revolution is its orbital period, which is given as $T = 24$ hours.
The speed of the satellite ($v$) can be calculated by dividing the distance traveled (circumference) by the time taken (period):
Speed ($v$) = Distance / Time
$v = C / T$
$v = (2 \pi r) / T$
Now, substitute the given values:
$r = 42250$ km
$T = 24$ hours
$v = (2 \times \pi \times 42250 \text{ km}) / 24 \text{ hours}$
Using $\pi \approx 3.14159$:
$v = (2 \times 3.14159 \times 42250) / 24$ km/hour
$v = (265465.45) / 24$ km/hour
$v \approx 11061.06$ km/hour
The speed of the satellite is approximately 11061.06 km per hour.
If we want to convert this to km per second, we divide by the number of seconds in an hour (3600):
$v \approx 11061.06 / 3600$ km/second
$v \approx 3.0725$ km/second
Question:
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?
Concept in a Minute:
This question involves uniformly accelerated motion. We will use the equations of motion for constant acceleration. The relevant equations are:
1. v = u + at (final velocity = initial velocity + acceleration * time)
2. v^2 = u^2 + 2as (final velocity squared = initial velocity squared + 2 * acceleration * displacement)
3. s = ut + 0.5at^2 (displacement = initial velocity * time + 0.5 * acceleration * time squared)
Explanation:
The problem states that a ball is gently dropped, which implies its initial velocity (u) is 0 m/s.
The height from which it is dropped is the displacement (s), which is 20 m.
The rate at which its velocity increases is the acceleration (a), which is 10 m/s².
To find the velocity with which it strikes the ground (final velocity, v), we can use the equation v² = u² + 2as because we have u, a, and s.
v² = (0 m/s)² + 2 * (10 m/s²) * (20 m)
v² = 0 + 400 m²/s²
v = sqrt(400 m²/s²)
v = 20 m/s
To find the time (t) after which it will strike the ground, we can use the equation v = u + at or s = ut + 0.5at². Using v = u + at, since we have now calculated v:
20 m/s = 0 m/s + (10 m/s²) * t
20 m/s = (10 m/s²) * t
t = (20 m/s) / (10 m/s²)
t = 2 s
Alternatively, using s = ut + 0.5at²:
20 m = (0 m/s) * t + 0.5 * (10 m/s²) * t²
20 m = 0 + (5 m/s²) * t²
20 m = (5 m/s²) * t²
t² = (20 m) / (5 m/s²)
t² = 4 s²
t = sqrt(4 s²)
t = 2 s
Therefore, the ball will strike the ground with a velocity of 20 m/s after 2 seconds.
Question:
What is the quantity which is measured by the area occupied below the velocity-time graph?
Concept in a Minute:
The relationship between velocity, time, and displacement. Specifically, understanding that velocity is the rate of change of displacement with respect to time, and how this relationship can be represented graphically. The fundamental idea of integration, which is essentially finding the area under a curve, is also crucial.
Explanation:
The velocity-time graph plots velocity on the y-axis and time on the x-axis. The area occupied below this graph represents the accumulation of velocity over a period of time.
Let’s consider a small time interval, Δt. During this small interval, we can assume that the velocity, v, is approximately constant. The displacement, Δx, that occurs during this time interval can be approximated as:
Δx ≈ v × Δt
If we consider the velocity-time graph, this product (v × Δt) corresponds to the area of a small rectangle under the curve, with height v and width Δt.
When we sum up these small displacements over the entire duration represented by the graph, we get the total displacement. Mathematically, this summation of infinitely small quantities is represented by integration. Therefore, the area under the velocity-time graph is the integral of velocity with respect to time.
∫ v dt = Displacement
Thus, the quantity measured by the area occupied below the velocity-time graph is displacement.
Question:
A train is travelling at a speed of 90 km h−1. 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.
Concept in a Minute:
This question involves the application of uniformly accelerated motion. The key concepts are the kinematic equations of motion, specifically those relating initial velocity, final velocity, acceleration, and displacement. We will need to convert units to be consistent.
Explanation:
We are given the initial speed of the train, the uniform acceleration (which is negative because it’s deceleration), and we need to find the distance travelled before the train comes to rest.
Step 1: Identify the given information.
Initial speed (u) = 90 km/h
Acceleration (a) = -0.5 m/s²
Final speed (v) = 0 m/s (since the train is brought to rest)
Step 2: Convert the initial speed from km/h to m/s.
To convert km/h to m/s, multiply by (5/18).
u = 90 * (5/18) m/s
Step 3: Calculate the value of u in m/s.
u = 5 * 5 m/s
u = 25 m/s
Step 4: Choose the appropriate kinematic equation.
We have u, v, and a, and we need to find the displacement (s). The kinematic equation that relates these quantities is:
v² = u² + 2as
Step 5: Rearrange the equation to solve for s.
2as = v² – u²
s = (v² – u²) / (2a)
Step 6: Substitute the known values into the equation.
s = (0² – 25²) / (2 * -0.5)
Step 7: Calculate the displacement (s).
s = (-625) / (-1)
s = 625 meters
Answer:
The train will go 625 meters before it is brought to rest.
Question:
Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1. On his return trip along the same route, there is less traffic and the average speed is 30 km h−1. What is the average speed for Abdul’s trip?
Concept in a Minute:
Average speed is defined as total distance traveled divided by the total time taken. For a journey with different speeds over equal distances, the average speed is not simply the arithmetic mean of the speeds. Instead, it involves the harmonic mean concept.
Explanation:
Let the distance between Abdul’s home and school be D km.
Trip to school:
Average speed = 20 km/h
Time taken (t1) = Distance / Speed = D / 20 hours
Return trip:
Average speed = 30 km/h
Time taken (t2) = Distance / Speed = D / 30 hours
Total distance traveled = Distance to school + Distance back home = D + D = 2D km
Total time taken = Time to school + Time back home = t1 + t2 = (D/20) + (D/30) hours
To add the fractions for total time, find a common denominator, which is 60:
Total time taken = (3D/60) + (2D/60) = 5D/60 hours
Now, calculate the average speed for the entire trip:
Average speed = Total distance / Total time taken
Average speed = 2D / (5D/60)
To divide by a fraction, multiply by its reciprocal:
Average speed = 2D * (60 / 5D)
Average speed = (2D * 60) / 5D
Cancel out D from the numerator and denominator:
Average speed = (2 * 60) / 5
Average speed = 120 / 5
Average speed = 24 km/h
Therefore, the average speed for Abdul’s trip is 24 km/h.
Question:
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?
Concept in a Minute:
This question involves the concept of uniformly accelerated motion. We will use the first equation of motion, which relates final velocity, initial velocity, acceleration, and time.
Explanation:
The problem describes a trolley moving down an inclined plane with a constant acceleration. We are asked to find its velocity after a certain time, starting from rest.
Given:
Acceleration (a) = 2 cm s⁻²
Time (t) = 3 s
Assuming the trolley starts from rest, its initial velocity (u) = 0 m/s.
We need to find the final velocity (v).
The first equation of motion is:
v = u + at
Substituting the given values into the equation:
v = 0 m/s + (2 cm s⁻²) × (3 s)
v = 6 cm s⁻¹
Therefore, the velocity of the trolley 3 seconds after the start will be 6 cm s⁻¹.
Detailed Steps:
1. Identify the given quantities: acceleration (a) and time (t).
2. Identify what needs to be found: final velocity (v).
3. Assume the initial velocity (u) based on the phrase “after the start,” which implies starting from rest.
4. Recall the first equation of uniformly accelerated motion: v = u + at.
5. Substitute the known values of u, a, and t into the equation.
6. Perform the calculation to find v.
7. State the final answer with the correct units.
Question:
State which of the following situation is possible and give an example:
An object moving with an acceleration but with uniform speed.
Concept in a Minute:
Acceleration is the rate of change of velocity. Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. Uniform speed means the magnitude of velocity is constant. Therefore, acceleration can exist even if the speed is uniform, provided the direction of motion is changing.
Explanation:
Yes, this situation is possible. An object moving with an acceleration but with uniform speed occurs when the object’s direction of motion is continuously changing, even though its speed remains constant. This is because acceleration is a vector quantity, and a change in direction constitutes a change in velocity, thus resulting in acceleration.
Example:
Consider a car moving in a circular path at a constant speed. The speed of the car (magnitude of its velocity) is uniform. However, the direction of the car’s velocity is continuously changing as it moves along the circle. This change in direction means the car is accelerating. This acceleration is directed towards the center of the circle and is known as centripetal acceleration. The magnitude of this acceleration is given by $a_c = v^2/r$, where $v$ is the uniform speed and $r$ is the radius of the circular path.
Question:
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 × 108m s−1.
Concept in a Minute:
The question requires understanding the relationship between distance, speed, and time. The fundamental formula is: Distance = Speed × Time. We are given the speed of the signal (speed of light) and the time it took to reach the ground station. We need to calculate the distance.
Explanation:
The signal from the spaceship travels at the speed of light.
Speed of signal = 3 × 108 m s−1.
The time taken for the signal to reach the ground station is given as five minutes.
Time = 5 minutes.
To use the formula Distance = Speed × Time, the units of time must be consistent. Since the speed is given in meters per second (m s−1), we need to convert the time from minutes to seconds.
1 minute = 60 seconds.
So, 5 minutes = 5 × 60 seconds = 300 seconds.
Now, we can calculate the distance:
Distance = Speed × Time
Distance = (3 × 108 m s−1) × (300 s)
Distance = 3 × 108 × 300 meters
Distance = 3 × 108 × 3 × 102 meters
Distance = 9 × 10(8+2) meters
Distance = 9 × 1010 meters.
Therefore, the distance of the spaceship from the ground station was 9 × 1010 meters.
Question:
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Concept in a Minute:
Distance is the total path length covered by an object. Displacement is the shortest straight-line distance between the initial and final positions of an object. Displacement is a vector quantity, meaning it has both magnitude and direction, while distance is a scalar quantity.
Explanation:
Yes, an object can have moved through a distance and still have zero displacement. This occurs when the object returns to its starting point.
Example:
Consider a person walking around a circular track. If the person starts at a point on the track and completes one full lap, returning to the exact same starting point, the distance they have covered is equal to the circumference of the track. However, their initial position and final position are the same. Therefore, the displacement is zero.
Another example is a car moving on a straight road from point A to point B, and then returning to point A. The distance covered would be twice the distance between A and B. But since the car ends up at its starting point (A), its displacement is zero.
Question:
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?
Concept in a Minute:
The question involves uniform acceleration in a straight line. The key concept is to use the equations of motion for uniformly accelerated linear motion to find the distance traveled. Specifically, we need to identify the appropriate equation that relates initial velocity, acceleration, time, and displacement (distance in this case, as the motion is in a straight line).
Explanation:
The motorboat starts from rest, which means its initial velocity (u) is 0 m s⁻¹.
The boat accelerates at a constant rate (a) of 3.0 m s⁻².
The time (t) for which it accelerates is 8.0 s.
We need to find the distance (s) the boat travels.
We can use the second equation of motion:
s = ut + (1/2)at²
Substitute the given values into the equation:
s = (0 m s⁻¹)(8.0 s) + (1/2)(3.0 m s⁻²)(8.0 s)²
s = 0 + (1/2)(3.0 m s⁻²)(64.0 s²)
s = (1/2)(3.0 * 64.0) m
s = (1/2)(192.0) m
s = 96.0 m
Therefore, the boat travels 96.0 meters during this time.
Question:
When will you say a body is at non-uniform acceleration?
Concept in a Minute:
Uniform acceleration means the velocity of the object changes by the same amount in every equal interval of time. Non-uniform acceleration means the velocity of the object changes by different amounts in equal intervals of time.
Explanation:
A body is said to be at non-uniform acceleration when its velocity changes by varying amounts in equal intervals of time. This means that the rate at which the velocity is changing is not constant. In other words, the acceleration itself is changing with time. For example, if an object’s velocity increases from 0 m/s to 2 m/s in the first second, and then from 2 m/s to 5 m/s in the second second, and from 5 m/s to 10 m/s in the third second, it is experiencing non-uniform acceleration because the change in velocity in each second is different (2 m/s, 3 m/s, 5 m/s respectively). Mathematically, this implies that the acceleration is not a constant value.
Question:
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Concept in a Minute:
Understanding the relationship between distance and time.
Recognizing that uniform motion means constant velocity and non-uniform motion means changing velocity.
Visualizing these relationships on a graph.
Explanation:
For uniform motion, the object covers equal distances in equal intervals of time. This means its velocity is constant. On a distance-time graph, this is represented by a straight line. The slope of this straight line indicates the velocity of the object.
For non-uniform motion, the object covers unequal distances in equal intervals of time. This means its velocity is changing. On a distance-time graph, this is represented by a curved line. The slope of the tangent to the curve at any point indicates the instantaneous velocity of the object at that moment. If the curve is becoming steeper, the speed is increasing; if it is becoming less steep, the speed is decreasing.
Question:
State which of the following situation is possible and give an example:
An object with a constant acceleration but with zero velocity.
Concept in a Minute:
Acceleration is the rate of change of velocity. Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. Constant acceleration means the velocity is changing at a steady rate. Zero velocity means the object is instantaneously at rest.
Explanation:
Yes, this situation is possible.
Consider an object thrown vertically upwards. At the highest point of its trajectory, just for an instant, its velocity is zero. However, the acceleration due to gravity is still acting on it, which is constant and directed downwards. This constant downward acceleration will cause the object to start moving downwards, thus changing its velocity from zero.
Example:
An object thrown vertically upwards reaches its maximum height. At this peak, its instantaneous velocity is zero, but the acceleration due to gravity (approximately 9.8 m/s² downwards) is still acting on it.
Question:
State which of the following situation is possible and give an example:
An object moving in a certain direction with an acceleration in the perpendicular direction.
Concept in a Minute:
This question deals with the concept of projectile motion, which is a specific case of motion with constant acceleration in a direction perpendicular to the initial velocity. It requires understanding that velocity and acceleration are vectors and can be independent of each other.
Explanation:
Yes, this situation is possible. This describes projectile motion.
Example:
Consider a ball thrown horizontally from the top of a table.
The ball’s initial velocity is horizontal.
The acceleration acting on the ball is due to gravity, which is always downwards (perpendicular to the initial horizontal velocity).
As the ball moves horizontally, gravity continuously accelerates it downwards, causing its path to curve. The horizontal component of velocity remains constant (ignoring air resistance), while the vertical component of velocity increases due to the downward acceleration.
Question:
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Concept in a Minute:
Speed-time graph: This graph plots speed on the vertical axis and time on the horizontal axis.
Meaning of a straight line parallel to the time axis: This signifies that the value on the vertical axis (speed) remains constant as the value on the horizontal axis (time) changes.
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 not changing with time. In other words, the speed is constant. When an object moves with a constant speed, its velocity may or may not be constant. However, the question specifically talks about speed. Therefore, we can conclude that the object is moving with a uniform speed. Uniform speed implies that the object covers equal distances in equal intervals of time.
Final Answer: The object is moving with uniform speed.
Question:
When will you say a body is at uniform acceleration?
Concept in a Minute:
Uniform acceleration means that the velocity of an object changes by the same amount in equal time intervals. This implies that the acceleration itself is constant.
Explanation:
A body is said to be in uniform acceleration when its velocity changes at a constant rate. In simpler terms, for every equal interval of time, the change in velocity of the body is the same. This means the acceleration, which is the rate of change of velocity, is constant. For example, if an object starts from rest and its velocity increases by 2 m/s every second, it is undergoing uniform acceleration. Mathematically, if the velocity of the body at time t is v(t), then for uniform acceleration, the derivative of velocity with respect to time, dv/dt, which represents acceleration, is a constant value.
Question:
A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start?
Concept in a Minute:
This question involves the kinematics of uniformly accelerated motion. The key concept is the use of equations of motion to relate displacement, initial velocity, acceleration, and time. Specifically, we will use the equation that directly links these variables when acceleration is constant.
Explanation:
We are given:
Uniform acceleration (a) = 4 m s⁻²
Time (t) = 10 s
The car starts from rest, which means its initial velocity (u) = 0 m s⁻¹.
We need to find the distance (s) covered.
The relevant equation of motion is:
s = ut + ½at²
Substitute the given values into the equation:
s = (0 m s⁻¹)(10 s) + ½(4 m s⁻²)(10 s)²
s = 0 + ½(4 m s⁻²)(100 s²)
s = 2 m s⁻² * 100 s²
s = 200 m
Therefore, the racing car will cover a distance of 200 meters in 10 seconds after the start.
Question:
What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Concept in a Minute:
Distance-time graph: A graph that plots distance traveled against time.
Speed: The rate at which an object changes its position.
Straight line parallel to the time axis: Indicates a constant value for the quantity plotted on the vertical axis.
Explanation:
In a distance-time graph, the vertical axis represents distance and the horizontal axis represents time. If the graph is a straight line parallel to the time axis, it means that the distance from the starting point remains constant over time. This implies that the object is not moving. Therefore, the object is at rest. Its speed is zero.
Question:
What does the odometer of an automobile measure?
Concept in a Minute:
The question asks about the function of an odometer. An odometer is a device that measures the distance traveled by a vehicle. This is a direct recall of a common automotive instrument.
Explanation:
The odometer of an automobile is an instrument used to measure and record the distance traveled by the vehicle. It is typically found on the dashboard of a car. As the vehicle moves, the odometer’s internal mechanism counts the rotations of the wheels and translates this into a reading of the total distance covered. This distance is usually displayed in miles or kilometers.
Question:
What does the path of an object look like when it is in uniform motion?
Concept in a Minute:
Uniform motion means moving at a constant speed in a straight line. This implies that the object’s velocity is constant, meaning both its speed and direction are unchanging.
Explanation:
When an object is in uniform motion, its path is a straight line. This is because its velocity is constant, which means it is not changing its speed and is not changing its direction. Any change in speed or direction would result in a curved path or a change in the straight line. Therefore, a straight line accurately represents the path of an object undergoing uniform motion.
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