In the previous section we saw that:
■ The area enclosed between a velocity-time graph and the x axis will give the displacement of the object
• Based on that information, we will now derive the following:
Kinematic equations for uniformly accelerated motion
• That is., when objects move with uniform acceleration, certain equations can be applied to find various quantities like displacement, velocity, time duration etc.,
• We will derive those equations now
1. In a previous section, we derived Eq.3.1: v = v0 + at
• We know that it is the equation of the velocity-time graph for an object moving with uniform acceleration.
• Such a graph is shown below:
2. From the graph, the following points are clear:
• When the stop watch was turned on, the velocity of the object was 'v0'.
• When the stop watch showed time as 't' seconds, the velocity of the object was 'v'
3. Knowing these facts, we can calculate the 'area of the shaded portion'.
• For easy calculation, we split the area into a rectangle and a triangle
• Area of the rectangle = v0 × t
• Area of the triangle = (1⁄2 × base × altitude) = (1⁄2 × t × (v-v0))
• So total area = [v0 × t + (1⁄2 × t × (v-v0))]
4. Now consider Eq.3.1 that we derived earlier: v = v0 + at.
• Rearranging this, we get: (v-v0) = at
• Substituting this in (3), we get:
Total area = [v0 × t + (1⁄2 × t × (at))]
⟹ Total area = v0t + 1⁄2 × at2.
5. But this total area is the displacement 'x' in the duration between the following instances:
• Instance when the stop watch showed t1 = 0 s
• Instance when the stop watch showed t2 = t s
Obviously, the duration = (t2 - t1) = (t - 0) = t s
• So we can write:
Eq.3.2: x = v0t + 1⁄2 × at2
6. Now consider the result in (3) again:
Total area = [v0 × t + (1⁄2 × t × (v-v0))]
• But 'total area' is 'x'. So we can write:
x = [v0 × t + (1⁄2 × t × (v-v0))]
• This can be rearranged as shown below:
■ So we can write:
Displacement = average velocity × time
7. Now we can write a comparison between two items:
(i) Motion with uniform velocity
• There is only one value for velocity
• Multiplying this velocity by time will give the displacement x
(ii) Motion with uniform acceleration
• Velocity changes continuously with time
• Multiplying 'average velocity' by time will give the displacement x
8. Now consider again Eq.3.1 that we derived earlier: v = v0 + at.
• Rearranging it, we get: t = (v-v0)⁄a
• Substituting this in (6), we get: x = [(v+v0)⁄2 × (v-v0)⁄a] = [(v2-v02)⁄2a]
Rearranging, we get:
Eq.3.3: v2 = v02 + 2ax
So we have derived 3 equations. Let us write them in order:
Eq.3.1: v = v0 + at
Eq.3.2: x = v0t + 1⁄2 × at2
Eq.3.3: v2 = v02 + 2ax
• These 3 equations connect the five quantities: v0, v, a, t and x
• They are the kinematic equations of rectilinear motion with uniform acceleration
• Note that, they are applicable only when the object moves with uniform acceleration
We will see some solved examples:
Click here for Solved examples 3.2, 3.3, 3.4 and 3.5
• Since it is 'simply released', it's initial velocity v0 will be zero
2. The object will fall towards the ground. It will be under a constant acceleration
• This acceleration is the 'acceleration due to gravity'
• It is denoted by the symbol 'g' and has a value of 9.8 ms-2
3. We assume that, the object is under the same acceleration of 9.8 ms-2 during it's entire fall
■ But this is true only if the 'height of fall' is small compared to the 'radius of the earth'
Why is it so?
• Ans: The acceleration due to gravity is caused due to the pull by the earth towards it's centre.
• This pull will be different if the object is at a large distance from the earth
• So g cannot be given the same value of 9.8 ms-2, for the entire fall if the 'height of fall' is very large
• For the problems in this chapter, we can take g = 9.8 ms-2
We will see some solved examples:
Click here for Solved examples 3.6, 3.7, 3.8 and 3.9 .
• It will come to rest only after moving a certain distance. So we can write the following points
1. Consider the two instances:
(i) Instant at which brakes are applied
(ii) Instant at which the vehicle comes to rest
2. There will be a time duration between these two instances
■ The vehicle will move a certain distance in this duration
3. It is important to find this distance for safety purposes. Let us try:
• Let the vehicle be moving with a velocity of v0
• Let the 'application of brakes' produce a negative acceleration of '-a'
• Here acceleration is taken as negative because it is in a direction opposite to the direction of motion
4. Let the vehicle move a distance of 'ds' in the duration mentioned in (5) above
Then 'ds' is called the stopping distance
• Applying Eq.3.3 we get: v2 = v02 + 2ax.
⟹ 02 = v02 + 2(-a)ds =
⟹ v02 = 2ads
⟹ Eq.3.4: ds = [v02⁄2a]
5. The significance of ds is shown in the fig.3.51(a) below:
(i) The car moves with a initial velocity of v0
(ii) At the instant when it reaches P, brakes are applied
(iii) The car will come to rest only after travelling a distance of ds from P
(iv) If we measure ds from P, we will be able to mark the point Q at which the car comes to rest
(v) In fig.a, we see that, Q falls before the obstacle present on the road
(vi) So the car will not crash into the obstacle
• However, it may be noted that, if the obstacle moves towards the car, then the above ds will not be sufficient
• Vehicle manufactures and High way designers do extensive research and calculations taking into account all the factors, to ensure safety
• It is the responsibility of the drivers to keep the speed below accepted limits
6. Now let us see what happens if we increase the speed:
• If the car travels with double the velocity as in the above case, we can write:
Initial velocity = 2v0
• Substituting in Eq.3.4, we get: ds = [(2v0)2⁄2a] = [4v02⁄2a]
• This new value of ds is 4 times the previous value. That is., the stopping distance increases by 4 times
• This is shown in fig.3.51(b)
(i) The car moves with a initial velocity of 2v0
(ii) At the instant when it reaches P, brakes are applied
(iii) The car will come to rest only after travelling a distance of 4ds from P
(iv) If we measure 4ds from P, we will be able to mark the point Q at which the car comes to rest
(v) In fig.b, we see that, Q falls after the obstacle present on the road
(vi) So the car will crash into the obstacle
■ We can write: Smaller the velocity, smaller will be the 'required ds'
• For example, while driving a car, if an obstacle appears suddenly on the road, some time will pass before the driver slams the brakes. This 'elapsed time' is called reaction time.
• It is the time taken to observe, think and act
• Reaction time depends on the complexity of the situation and also on the individual
• We can effectively use the 'acceleration due to gravity' to measure the reaction time of any individual. Let us see how it is done:
1. Drop a small object with out any pre notice
2. The individual whose reaction time is to be determined, must catch the object before it hits the ground
3. Then we want the distance between the following two points:
(i) Point of release
(ii) Point of catch
• This distance is 'h', the height of fall
• If we use a meter scale as the falling object, we can easily measure 'h'
4. Then we apply Eq.3.2: x = v0t + 1⁄2 × at2.
• Substituting the values, we get:
h = 0 + 1⁄2 × gt2.
⟹ t2 = 2h⁄g
⟹ Eq.3.5: t = √[2h⁄g].
• This 't' is the time required by the object to fall from the 'point of release' to the 'point of catch'
• So this 't' is the reaction time.
■ The area enclosed between a velocity-time graph and the x axis will give the displacement of the object
• Based on that information, we will now derive the following:
Kinematic equations for uniformly accelerated motion
• That is., when objects move with uniform acceleration, certain equations can be applied to find various quantities like displacement, velocity, time duration etc.,
• We will derive those equations now
1. In a previous section, we derived Eq.3.1: v = v0 + at
• We know that it is the equation of the velocity-time graph for an object moving with uniform acceleration.
• Such a graph is shown below:
 |
| Fig.3.50 |
• When the stop watch was turned on, the velocity of the object was 'v0'.
• When the stop watch showed time as 't' seconds, the velocity of the object was 'v'
3. Knowing these facts, we can calculate the 'area of the shaded portion'.
• For easy calculation, we split the area into a rectangle and a triangle
• Area of the rectangle = v0 × t
• Area of the triangle = (1⁄2 × base × altitude) = (1⁄2 × t × (v-v0))
• So total area = [v0 × t + (1⁄2 × t × (v-v0))]
4. Now consider Eq.3.1 that we derived earlier: v = v0 + at.
• Rearranging this, we get: (v-v0) = at
• Substituting this in (3), we get:
Total area = [v0 × t + (1⁄2 × t × (at))]
⟹ Total area = v0t + 1⁄2 × at2.
5. But this total area is the displacement 'x' in the duration between the following instances:
• Instance when the stop watch showed t1 = 0 s
• Instance when the stop watch showed t2 = t s
Obviously, the duration = (t2 - t1) = (t - 0) = t s
• So we can write:
Eq.3.2: x = v0t + 1⁄2 × at2
6. Now consider the result in (3) again:
Total area = [v0 × t + (1⁄2 × t × (v-v0))]
• But 'total area' is 'x'. So we can write:
x = [v0 × t + (1⁄2 × t × (v-v0))]
• This can be rearranged as shown below:
■ So we can write:
Displacement = average velocity × time
7. Now we can write a comparison between two items:
(i) Motion with uniform velocity
• There is only one value for velocity
• Multiplying this velocity by time will give the displacement x
(ii) Motion with uniform acceleration
• Velocity changes continuously with time
• Multiplying 'average velocity' by time will give the displacement x
8. Now consider again Eq.3.1 that we derived earlier: v = v0 + at.
• Rearranging it, we get: t = (v-v0)⁄a
• Substituting this in (6), we get: x = [(v+v0)⁄2 × (v-v0)⁄a] = [(v2-v02)⁄2a]
Rearranging, we get:
Eq.3.3: v2 = v02 + 2ax
So we have derived 3 equations. Let us write them in order:
Eq.3.1: v = v0 + at
Eq.3.2: x = v0t + 1⁄2 × at2
Eq.3.3: v2 = v02 + 2ax
• These 3 equations connect the five quantities: v0, v, a, t and x
• They are the kinematic equations of rectilinear motion with uniform acceleration
• Note that, they are applicable only when the object moves with uniform acceleration
We will see some solved examples:
Click here for Solved examples 3.2, 3.3, 3.4 and 3.5
Motion of object under free fall
1. Consider an object released from the top of a tall building or a cliff• Since it is 'simply released', it's initial velocity v0 will be zero
2. The object will fall towards the ground. It will be under a constant acceleration
• This acceleration is the 'acceleration due to gravity'
• It is denoted by the symbol 'g' and has a value of 9.8 ms-2
3. We assume that, the object is under the same acceleration of 9.8 ms-2 during it's entire fall
■ But this is true only if the 'height of fall' is small compared to the 'radius of the earth'
Why is it so?
• Ans: The acceleration due to gravity is caused due to the pull by the earth towards it's centre.
• This pull will be different if the object is at a large distance from the earth
• So g cannot be given the same value of 9.8 ms-2, for the entire fall if the 'height of fall' is very large
• For the problems in this chapter, we can take g = 9.8 ms-2
We will see some solved examples:
Click here for Solved examples 3.6, 3.7, 3.8 and 3.9 .
Stopping distance of vehicles
• When brakes are applied to a moving vehicle, it will not come to rest immediately.• It will come to rest only after moving a certain distance. So we can write the following points
1. Consider the two instances:
(i) Instant at which brakes are applied
(ii) Instant at which the vehicle comes to rest
2. There will be a time duration between these two instances
■ The vehicle will move a certain distance in this duration
3. It is important to find this distance for safety purposes. Let us try:
• Let the vehicle be moving with a velocity of v0
• Let the 'application of brakes' produce a negative acceleration of '-a'
• Here acceleration is taken as negative because it is in a direction opposite to the direction of motion
4. Let the vehicle move a distance of 'ds' in the duration mentioned in (5) above
Then 'ds' is called the stopping distance
• Applying Eq.3.3 we get: v2 = v02 + 2ax.
⟹ 02 = v02 + 2(-a)ds =
⟹ v02 = 2ads
⟹ Eq.3.4: ds = [v02⁄2a]
5. The significance of ds is shown in the fig.3.51(a) below:
 |
| Fig.3.51 |
(ii) At the instant when it reaches P, brakes are applied
(iii) The car will come to rest only after travelling a distance of ds from P
(iv) If we measure ds from P, we will be able to mark the point Q at which the car comes to rest
(v) In fig.a, we see that, Q falls before the obstacle present on the road
(vi) So the car will not crash into the obstacle
• However, it may be noted that, if the obstacle moves towards the car, then the above ds will not be sufficient
• Vehicle manufactures and High way designers do extensive research and calculations taking into account all the factors, to ensure safety
• It is the responsibility of the drivers to keep the speed below accepted limits
6. Now let us see what happens if we increase the speed:
• If the car travels with double the velocity as in the above case, we can write:
Initial velocity = 2v0
• Substituting in Eq.3.4, we get: ds = [(2v0)2⁄2a] = [4v02⁄2a]
• This new value of ds is 4 times the previous value. That is., the stopping distance increases by 4 times
• This is shown in fig.3.51(b)
(i) The car moves with a initial velocity of 2v0
(ii) At the instant when it reaches P, brakes are applied
(iii) The car will come to rest only after travelling a distance of 4ds from P
(iv) If we measure 4ds from P, we will be able to mark the point Q at which the car comes to rest
(v) In fig.b, we see that, Q falls after the obstacle present on the road
(vi) So the car will crash into the obstacle
■ We can write: Smaller the velocity, smaller will be the 'required ds'
Reaction time
• If some incident happens suddenly with out any pre notice, we will need some time to respond to that incident.• For example, while driving a car, if an obstacle appears suddenly on the road, some time will pass before the driver slams the brakes. This 'elapsed time' is called reaction time.
• It is the time taken to observe, think and act
• Reaction time depends on the complexity of the situation and also on the individual
• We can effectively use the 'acceleration due to gravity' to measure the reaction time of any individual. Let us see how it is done:
1. Drop a small object with out any pre notice
2. The individual whose reaction time is to be determined, must catch the object before it hits the ground
3. Then we want the distance between the following two points:
(i) Point of release
(ii) Point of catch
• This distance is 'h', the height of fall
• If we use a meter scale as the falling object, we can easily measure 'h'
4. Then we apply Eq.3.2: x = v0t + 1⁄2 × at2.
• Substituting the values, we get:
h = 0 + 1⁄2 × gt2.
⟹ t2 = 2h⁄g
⟹ Eq.3.5: t = √[2h⁄g].
• This 't' is the time required by the object to fall from the 'point of release' to the 'point of catch'
• So this 't' is the reaction time.
In the next section, we will see relative velocity.
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