Lesson Explainer: Applications on Motion with Uniform Acceleration Mathematics

In this explainer, we will learn how to solve problems involving motion of a particle with uniform acceleration through one or more sections of its path.

We may recall that the acceleration of a body is the rate at which the velocity of the body changes.

Let us first define average acceleration.

Definition: Average Acceleration

If in a time interval Δ𝑡=𝑡𝑡 the velocity of a body changes from an initial velocity, 𝑢, to a final velocity, 𝑣, the average acceleration of the body in the time interval is given by 𝑎=𝑣𝑢𝑡𝑡.

If a body accelerates uniformly, then the value of its acceleration is constant throughout the time interval over which it accelerates. This means that the instantaneous acceleration of the body is equal to its average acceleration.

Correspondingly, the displacement of a body while uniformly accelerating in a time interval is the mean of the displacement of the body at its initial and final velocities in that time interval.

We can then define the displacement of a uniformly accelerating body.

Definition: The Displacement of a Uniformly Accelerating Body

For a body that accelerates uniformly to change its velocity from an initial value, 𝑢, to a final value, 𝑣, in a time interval, Δ𝑡, the displacement of the body can be expressed as 𝑠=𝑣(Δ𝑡)+𝑢(Δ𝑡)2𝑠=(𝑣+𝑢)Δ𝑡2.

If the acceleration of a body that moves along a straight-line path is known, various kinematic relations of the motion of the body can be determined.

The relationship between the displacement of a uniformly accelerating body and its initial and final velocities can be expressed in a form that does not involve time. It has been previously stated that 𝑠=(𝑣+𝑢)Δ𝑡2, and that 𝑎=𝑣𝑢Δ𝑡; hence, Δ𝑡=𝑣𝑢𝑎.

From this, we can see that 2𝑠=(𝑣+𝑢)Δ𝑡=(𝑣+𝑢)𝑣𝑢𝑎,2𝑎𝑠=(𝑣+𝑢)(𝑣𝑢),2𝑎𝑠=𝑣𝑢.

The terms obtained are usually expressed as the kinematic formula that is defined below.

Definition: The Relation of Initial and Final Velocities of a Uniformly Accelerating Body to its Displacement

Consider a body that accelerates uniformly to change its velocity from an initial value, 𝑢, to a final value, 𝑣, over a displacement 𝑠. The initial and final velocities of the body are related to the displacement and acceleration of the body as shown by the formula 𝑣=𝑢+2𝑎𝑠.

It is important to note that the time interval over which the body moves is not a term in the expression.

Let us look at an example of a body that has uniform acceleration during part of its motion.

Example 1: Finding the Distance Covered by a Cyclist Moving with Uniform Acceleration Then at a Uniform Velocity

A cyclist, riding down a hill from rest, was accelerating at a rate of 0.5 m/s2. By the time he reached the bottom of the hill, he was traveling at 1.5 m/s. He continued traveling at this speed for another 9.5 seconds. Determine the total distance 𝑠 that the cyclist covered.

Answer

The motion of the cyclist can be separated into two parts: a part where the cyclist has uniform acceleration and a part where they have uniform velocity and hence have zero acceleration.

The part of the motion with zero acceleration is simpler, so it will be considered first although it is the second part of the motion of the cyclist to occur.

When the cyclist has reached the bottom of the hill, the velocity of the cyclist is stated to be 1.5 m/s. The cyclist moves at this speed for 9.5 seconds. The displacement of the cyclist in this time is given by 𝑠=1.5(9.5)=14.25.m

While the cyclist travels down the hill, the velocity of the cyclist changes from zero to 1.5 m/s. The distance that the cyclist would travel as their velocity increases could be immediately determined if the time interval in which the velocity changed was given, but this is not stated. The length of the time interval can be determined by using the formula 𝑎=𝑣𝑢Δ𝑡, and making Δ𝑡 the subject of the formula, to give Δ𝑡=𝑣𝑢𝑎.

Using the values stated in the question, we have Δ𝑡=1.500.5=3.s

The displacement of the uniformly accelerating cyclist in these 3 seconds is the mean of the displacement of the cyclist in 3 seconds at their initial and final velocities, given by 𝑠=(𝑣+𝑢)Δ𝑡2𝑠=(1.5+0)32=2.25.m

The total displacement of the cyclist is, therefore, given by 𝑠=𝑠+𝑠=14.25+2.25=16.5.m

This example can also be solved by using the formula 𝑣=𝑢+2𝑎𝑠, to obtain 𝑠. The equation must be rearranged to make 𝑠 the subject. As 𝑢 is zero, this is done as follows: 𝑣=2𝑎𝑠𝑠=𝑣2𝑎.

Substituting known values, we see that 𝑠=1.52(0.5)=1.5=2.25,m which is equal to the value of 𝑠 previously determined.

Let us look at an example in which the relationship between the displacement of a uniformly accelerating body and its initial and final velocities is expressed in a form that does not involve the time interval over which the body accelerates.

Example 2: Finding the Acceleration of a Body Accelerating Then Decelerating between Two Points when given the Total Distance between Them

A train, starting from rest, began moving in a straight line between two stations. For the first 80 seconds, it moved with a constant acceleration 𝑎. Then it continued to move at the velocity it had acquired for a further 65 seconds. Finally, it decelerated with a rate of 2𝑎 until it came to rest. Given that the distance between the two stations was 8.9 km, find the magnitude of 𝑎 and the velocity 𝑣 at which it moved during the middle leg of the journey.

Answer

The motion of the train consists of three parts. In the first part, the train accelerates uniformly from rest; in the second part, the train does not accelerate; and in the third part, the train decelerates uniformly to rest.

In the first part of its motion, the train starts from rest and accelerates uniformly for 80 seconds. The acceleration of the train is given by 𝑎=𝑣𝑢Δ𝑡, and, therefore, after accelerating, the velocity of the train is given by 𝑣=𝑢+𝑎(Δ𝑡).

The train starts from rest, so 𝑣 is just 𝑣=𝑎(Δ𝑡)=80𝑎.

The displacement of the train as it accelerates can be determined by rearranging the formula 𝑣=𝑢+2𝑎𝑠 to make 𝑠 the subject. As 𝑢 is zero, this is done as follows: 𝑣=2𝑎𝑠𝑠=𝑣2𝑎=(80𝑎)2𝑎=3200𝑎=𝑠, where 𝑠 is the displacement of the train for the first part of its motion.

In the second part of its motion, the train stops accelerating. It has been shown that the velocity of the train in the second part of its motion is 80𝑎. The displacement of the train in the second part of its motion, 𝑠, is given by 𝑠=65(80𝑎)=5200𝑎.

In the third part of its motion, the displacement of the train is again determined using the formula 𝑣=𝑢+2𝑎𝑠.

The final velocity of the train is zero and the initial velocity of the train is 80𝑎. The acceleration of the train doubles its magnitude and acts in the opposite direction to the velocity of the train, and so the acceleration of the train in the third part of its motion has a negative sign. From this, 𝑠 can be determined as follows: 0=(80𝑎)+2(2𝑎)𝑠(80𝑎)=(4𝑎)𝑠(80𝑎)(4𝑎)=𝑠=1600𝑎.

The displacement of the train over its entire motion is 8.9 km. The displacements 𝑠, 𝑠, and 𝑠 are not in kilometres but rather in metres. The 8.9 km can be converted to 8‎ ‎900 m which is equal to the sum of the displacements of the train over the parts of its motion: 𝑠+𝑠+𝑠=3200𝑎+5200𝑎+1600𝑎𝑠+𝑠+𝑠=10000𝑎𝑠+𝑠+𝑠=8900.

Finally, 𝑎 can be determined by making it the subject of the equation: 𝑎=890010000=0.89/.ms

It has already been shown that the velocity of the train in the second part of its motion is 80𝑎, so the velocity 𝑣 at which it moved for the middle leg of the journey is given by 𝑣=80(0.89)=71.2/.ms

The motion of a body can be considered where the part of the motion of the body for which time, displacement, velocity, or acceleration are to be determined starts when the body is not at rest. Let us consider such an example.

Example 3: Displacement of a Moving Body That Comes to Rest

A body, moving in a straight line, covered 60 cm in 6 seconds while accelerating uniformly. Maintaining its velocity, it covered a further 52 cm in 5 seconds. Finally, it started decelerating at a rate double the rate of its former acceleration until it came to rest. Find the total distance covered by the body.

Answer

The motion of the body consists of three parts. In the first part, the body accelerates uniformly, in the second part, the body does not accelerate, and in the third part, the body accelerates uniformly to rest.

The initial velocity of the body is not given, but it is stated that in the second part of the motion of the body a displacement of 52 cm occurs in a time of 5 seconds as the body maintains its velocity. This tells us that the velocity in the second part of the motion of the body is given by 𝑣=525=10.4/.cms

The velocity of the body in the second part of its motion must equal the velocity at the end of the first part of its motion. From this, we can determine the initial velocity of the body in the first part of its motion.

In the first part of the motion of the body, the body is displaced 60 cm in 6 seconds. Applying the formula 𝑠=(𝑣+𝑢)Δ𝑡2 and making 𝑢 the subject gives us 𝑢=2𝑠Δ𝑡𝑣.

By substituting known values, we obtain 𝑢=120610.4=9.6/.cms

We can now find the acceleration in the first part of the journey by applying 𝑣=𝑢+2𝑎𝑠.

Making 𝑎 the subject gives us 𝑎=𝑣𝑢2𝑠.

By substituting known values, we obtain 𝑎=10.49.6120=215/.cms

For the third part of the motion of the body, we again apply the formula 𝑣=𝑢+2𝑎𝑠.

Making 𝑠 the subject gives us 𝑠=𝑣𝑢2𝑎.

The initial velocity of the body in the third part of its motion equals the velocity of the body in the second part of its motion, and the final velocity of the body in the third part of its motion is zero. The acceleration of the body in the third part of its motion is in the opposite direction to 𝑢, so has a negative sign, and has double the magnitude of the acceleration in the first part of the motion of the body.

Substituting the values of 𝑢, 𝑣, and 𝑎, we obtain 𝑠=010.422=202.8,cm where 𝑠 is the displacement of the body in the third part of its motion.

The displacement of the body in the first part of its motion is stated to be 60 cm, and the displacement of the body in the second part of its motion is stated to be 52 cm.

The displacement of the body over the three parts of its motion is given by 𝑠=60+52+202.8=314.8.cm

Let us now look at an example where the motions of two bodies that have equal accelerations are compared.

Example 4: Studying the Motion of Bullets Horizontally Fired at Two Different Wooden Blocks

A bullet was fired horizontally at a wooden block. It entered the block at 80 m/s and penetrated 32 cm into the block before it stopped. Assuming that its acceleration 𝑎 was uniform, find 𝑎 in km/s2. If, under similar conditions, another bullet was fired at the wooden block that was 14 cm thick, determine the velocity 𝑣 at which the bullet exited the wooden block.

Answer

For the first bullet, its final velocity is zero. The initial velocity and displacement are given so the acceleration can be determined using the formula 𝑣=𝑢+2𝑎𝑠, making 𝑎 the subject to obtain 𝑎=𝑣𝑢2𝑠.

The velocity is in metres per second, and the displacement is in centimetres. To obtain an acceleration in metres per second squared, we convert the 32 cm to 0.32 m. By substituting known values, we obtain 𝑎=0802(0.32)=10000/.ms

This value is quite large and can be more conveniently expressed as 10 km/s2. The negative value is consistent with the acceleration of the bullet being in the direction opposite to its initial velocity.

For the second bullet, the same value of 𝑎 is used, but the value of 𝑠 is changed. The 14 cm displacement is also converted to a displacement in metres, of 0.14 m. The same formula is used as in the case of the first bullet, but now with 𝑣 as the subject: 𝑣=𝑢+2𝑎𝑠𝑣=𝑢+2𝑎𝑠.

By substituting known values, we obtain 𝑣=80+2(10000)(0.14)=60/.ms

Let us now look at an example where the motions of two bodies that have different accelerations are combined.

Example 5: Finding the Time Taken for a Moving Object to Catch Up with Another

A speeding car, moving at 96 km/h, passed by a police car. 12 seconds later, the police car started pursuing it. Accelerating uniformly, the police car covered a distance of 134 m until its velocity was 114 km/h. Maintaining this speed, it continued until it caught up with the speeding car. Find the time it took for the police car to catch the other car starting from the point the police car began moving.

Answer

This question can be more easily answered by keeping separate track of the displacement and velocity of each car at various times. The following table tracks these changes.

Speeding CarPolice Car
𝑡()s𝑣(/)ms𝑠()m𝑣(/)ms𝑠()m

Initially, the police car is at rest and has zero displacement while the speeding car has zero displacement and a velocity of 96 km/h.

Times in the question are given in seconds and distances are given in meters, so it is convenient to express velocities in metres per second (m/s).

The initial velocity of the speeding car is given by 96(1000)60(60)=803/.ms

We add this information to the table.

Speeding CarPolice Car
𝑡()s𝑣(/)ms𝑠()m𝑣(/)ms𝑠()m
0803000

For the next 12 seconds, the speeding car travels at this velocity and the police car remains at rest. The displacement of the speeding car after 12 seconds is given by 𝑠=12803=320.m

We add this information to the table.

Speeding CarPolice Car
𝑡()s𝑣(/)ms𝑠()m𝑣(/)ms𝑠()m
0803000
1280332000

At 𝑡=12seconds, the police car begins accelerating. After it accelerates, it has a velocity of 114 km/h. This can be converted to metres per second in the same way as was done for the speeding car: 114(1000)60(60)=953/.ms

The time for which the police car accelerates is not stated, but the displacement of the police car over the acceleration is stated to be 134 m. Knowing this, we can determine the time interval for the acceleration by applying the formula 𝑠=(𝑣+𝑢)Δ𝑡2.

Knowing that 𝑢 is zero, we have 𝑠=𝑣Δ𝑡2.

Rearranging to make Δ𝑡 the subject, we obtain Δ𝑡=2𝑠𝑣=2(134)395=80495.s

The time at which the police car ceases accelerating is given by 12+80495=194495.s

In the time that the police car has accelerated, the speeding car has increased its displacement by 80380495=64320285=428819.m

The total displacement of the speeding car at this instant is given by 320+428819=1036819.m

We add this information to the table.

Speeding CarPolice Car
𝑡()s𝑣(/)ms𝑠()m𝑣(/)ms𝑠()m
0803000
1280332000
1944958031036819953134

After this time, both cars have constant velocity. The time interval in which the police car catches up with the speeding car once they both have uniform velocities is the time interval in which the displacements of the car become equal.

This time interval can be determined by dividing the difference in the displacements of the cars by the difference in the velocities of the cars.

The difference in the displacements of the cars is given by 1036819134=1036819254619=782219.m

The difference in the velocities of the cars is given by 953803=153=5/.ms

The time interval is given by 5=782295.s

This time interval is in addition to the time interval required for the police car to reach constant velocity, which we showed earlier to equal 80495 s.

Summing these time intervals, we obtain 782295+80495=862695=90.8.s

A different method by which this result could be obtained is to define an equation relating 𝑠 to 𝑡 for each of the cars.

For the speeding car, the equation is just the equation for constant motion at the speed of the speeding car: 𝑠=803𝑡.

For the police car, a similar equation can be written, but it must include the displacement of the speeding car from the police car after both cars were moving at their final velocities. This displacement is given by 1036819134=1036819254619=782219.m

The equation for the police car is then 𝑠=953𝑡782219.

The value of 𝑡 for which the police car catches up with the speeding car is where the values of 𝑠 for both cars are equal, which is where 803𝑡=953𝑡782219.

We can rearrange this as follows: 803𝑡953𝑡=782219153𝑡=782219153𝑡=7822195𝑡=782219𝑡=782295.

The time interval in which the police car accelerated must be added to this time, giving a total time of 782295+80495=862695=90.8.s

Let us now summarize what has been learned in these examples.

Key Points

  • If in a time interval Δ𝑡=𝑡𝑡, the velocity of a body changes from an initial velocity, 𝑢, to a final velocity, 𝑣, the average acceleration of the body in the time interval is given by 𝑎=𝑣𝑢𝑡𝑡.
  • If a body accelerates uniformly, then the value of its acceleration is constant throughout the time interval during which it accelerates. This means that the instantaneous acceleration of the body is equal to its average acceleration.
  • The displacement of a body while uniformly accelerating in a time interval is the mean of the displacement of the body at its initial and final velocities in that time interval: 𝑠=(𝑣+𝑢)Δ𝑡2.
  • The relationship between the displacement of a uniformly accelerating body and its initial and final velocities can be expressed in a form that does not involve time: 𝑣=𝑢+2𝑎𝑠.

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