Question Video: Finding the Total Distance Covered by a Body Moving with Uniform Acceleration Mathematics

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 to the rate of its former acceleration until it came to rest. Find the total distance covered by the body.

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Video Transcript

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

We can immediately see from the question that there are three parts to the body’s journey. Firstly, it accelerates, then it travels at a constant velocity, and finally decelerates to rest. In order to answer the question, we’ll need to use the equations of motion or SUVAT equations. 𝑠 represents the displacement with a standard unit of meters. 𝑒 and 𝑣 are the initial and final velocities measured in meters per second. π‘Ž is the acceleration which has units of meters per second squared. And 𝑑 is the time with a standard unit of seconds. We notice that the distances in this question are given in centimeters. We will therefore consider the velocities in centimeters per second and the acceleration in centimeters per second squared.

We will now consider the information we are given in the question for each part of the journey. For the first part of the journey, we have a displacement of 60 centimeters and a time of six seconds. We will let the initial and final velocities be 𝑣 sub zero and 𝑣 sub one centimeters per second, respectively. And we will let the acceleration be π‘Ž. In the second part of the journey, we are also given the displacement and time. These are 52 centimeters and five seconds. As the body is traveling at a constant speed, its acceleration is zero centimeters per second squared. The velocity it is traveling at is the same as the final velocity of the first part, 𝑣 sub one centimeters per second.

Finally, we have the third part of the journey where the body decelerates to rest; therefore, its final velocity is zero centimeters per second. The initial velocity of this part of the journey is still 𝑣 sub one centimeters per second. We will let the unknown displacement in time be 𝑠 sub three and 𝑑 sub three, respectively. We know that the deceleration is double the rate of the former acceleration. This means that the deceleration is equal to two π‘Ž centimeters per second squared. And therefore, the acceleration is equal to negative two π‘Ž. Our aim in this question is to find the total distance covered by the body. And as we are told the distance covered in the first and second parts of the journey, it would seem sensible that we just need to calculate the distance covered in the third part.

Unfortunately, we did not have enough information to do this as there are too many unknowns. We will need to calculate the values of 𝑣 sub one and π‘Ž first. Let’s begin by considering part one. One of our equations of motion states that 𝑠 is equal to 𝑣𝑑 minus a half π‘Žπ‘‘ squared. Substituting in our values, we have 60 is equal to 𝑣 sub one multiplied by six minus a half multiplied by π‘Ž multiplied by six squared. The right-hand side simplifies to six 𝑣 sub one minus 18π‘Ž. We can then divide through by six such that 10 is equal to 𝑣 sub one minus three π‘Ž. This equation still has two unknowns, so we’ll need to find another equation to be able to solve this.

Let’s now consider part two and the same equation. This time we have 52 is equal to 𝑣 sub one multiplied by five minus a half multiplied by zero multiplied by five squared. This time the right-hand side simplifies to just five 𝑣 sub one. And dividing both sides of the equation by five, we have 𝑣 sub one is equal to 10.4. We can now replace this value in parts one, two, and three. Substituting 𝑣 sub one for 10.4, we have the equation 10 is equal to 10.4 minus three π‘Ž. By adding three π‘Ž and subtracting 10 from both sides, we have three π‘Ž is equal to 0.4. We can then divide both sides of this equation by three such that π‘Ž is equal to two fifteenths. The acceleration of the body in the first part of its journey is two fifteenths centimeters per second squared.

We can now calculate the acceleration in the third part of the journey. Negative two multiplied by two fifteenths is negative four fifteenths. We now have values of 𝑒, 𝑣, and π‘Ž in part three. And we can now use the equation 𝑣 squared is equal to 𝑒 squared plus two π‘Žπ‘  to calculate the value of 𝑠 sub three. Substituting in the values, we have zero squared is equal to 10.4 squared plus two multiplied by negative four over 15 multiplied by 𝑠 sub three. The left-hand side is equal to zero and the right-hand side simplifies to 108.16 minus eight fifteenths 𝑠 sub three. This means that eight fifteenths 𝑠 sub three is equal to 108.16. And dividing both sides by eight fifteenths gives us 𝑠 sub three is equal to 202.8. The distance covered by the body in the third part of the journey is 202.8 centimeters.

We now have the distance covered in each part of the journey and can use these to find the total distance. We need to find the sum of 60, 52, and 202.8. This is equal to 314.8. The total distance covered by the body is 314.8 centimeters.

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