The portal has been deactivated. Please contact your portal admin.

Question Video: Finding the Acceleration of a Body Accelerating Then Decelerating between Two Points given the Total Distance between Them Mathematics

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 π‘Ž. Then it continued to move at the velocity it had acquired for a further 65 seconds. Finally, it decelerated with a rate of 2π‘Ž until it came to rest. Given that the distance between the two stations was 8.9 km, find the magnitude of π‘Ž and the velocity 𝑣 at which it moved during the middle leg of the journey.

06:58

Video Transcript

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 π‘Ž. Then it continued to move at the velocity it had acquired for a further 65 seconds. Finally, it decelerated with a rate of two π‘Ž until it came to rest. Given that the distance between the two stations was 8.9 kilometers, find the magnitude of π‘Ž and the velocity 𝑣 at which it moved during the middle leg of the journey.

In order to answer this question, we will use our equations of motion or SUVAT equations. We will create equations for the three legs of the journey and then solve them to calculate any unknowns. Let’s begin by considering the first part of the journey. The train accelerates from rest, so our initial velocity is equal to zero meters per second. It accelerates for 80 seconds at a constant acceleration of π‘Ž meters per second squared. We will call the velocity it reaches at this point 𝑣 and the displacement from the start point or the distance covered 𝑠 one.

We know that 𝑣 is equal to 𝑒 plus π‘Žπ‘‘. Substituting in our values, we have 𝑣 is equal to zero plus π‘Ž multiplied by 80. This gives us the equation 𝑣 is equal to 80π‘Ž. Another one of our equations states that 𝑠 is equal to 𝑒 plus 𝑣 divided by two multiplied by 𝑑. 𝑠 one is, therefore, equal to zero plus 𝑣 divided by two multiplied by 80. This simplifies to 𝑠 one is equal to 40𝑣. We will call these equations one and two and move on to the second part of the journey.

In the second part of the journey, the train travels at this constant velocity 𝑣. This means its acceleration is equal to zero meters per second squared. The time it takes is 65 seconds. And we will call the distance traveled 𝑠 two. Once again, we will use the equation 𝑠 is equal to 𝑒 plus 𝑣 divided by two multiplied by 𝑑. This gives us 𝑠 two is equal to 𝑣 plus 𝑣 divided by two multiplied by 65. As 𝑣 plus 𝑣 is equal to two 𝑣, this simplifies to 𝑠 two is equal to 65𝑣. We will call this equation three and now move on to the final part of the journey.

In this part of the journey, the train decelerates to rest. As the deceleration is equal to two π‘Ž, our value of π‘Ž will be equal to negative two π‘Ž meters per second squared. The final velocity 𝑣 is equal to zero meters per second. The initial velocity in this part of the journey is equal to 𝑣, the speed it was traveling for the second part of the journey. We will call the distance traveled in this part 𝑠 three and the time it takes 𝑑. We will begin by once again using 𝑠 is equal to 𝑒 plus 𝑣 over two multiplied by 𝑑. 𝑠 three is, therefore, equal to 𝑣 plus zero divided by two multiplied by 𝑑. This simplifies to 𝑠 three is equal to 𝑣𝑑 divided by two.

We will also use the equation 𝑣 equals 𝑒 plus π‘Žπ‘‘. Substituting in our values here gives us zero is equal to 𝑣 plus negative two π‘Ž multiplied by 𝑑. This simplifies to zero is equal to 𝑣 minus two π‘Žπ‘‘. And adding two π‘Žπ‘‘ to both sides of this equation gives us 𝑣 is equal to two π‘Žπ‘‘. We now have two further equations that we’ll number four and five.

The final part of our question tells us that the distance between the two stations was 8.9 kilometers. To use the equations of motion, we need this to be in meters. As there are 1000 meters in one kilometer, 8.9 kilometers is equal to 8900 meters. The distances 𝑠 one, 𝑠 two, and 𝑠 three must, therefore, sum to 8900. Using equations two, three, and four, this can be rewritten as 40𝑣 plus 65𝑣 plus 𝑣𝑑 over two is equal to 8900. 105𝑣 plus 𝑣𝑑 over two is equal to 8900. If we can work out the value of the time 𝑑, we can then calculate the velocity 𝑣.

Let’s consider equation one and equation five. These tell us that 𝑣 is equal to 80π‘Ž and two π‘Žπ‘‘, which means that 80π‘Ž must be equal to two π‘Žπ‘‘. At this stage, we know that π‘Ž is not equal to zero. Therefore, we can divide through by π‘Ž. We can then divide both sides of this equation by two, giving us 𝑑 is equal to 40. The time taken for the third part of the journey is 40 seconds. We can now substitute 𝑑 equals 40 to calculate 𝑣. 40 divided by two is equal to 20, and 105 plus 20 is 125. Therefore, 125𝑣 is equal to 8900. Dividing both sides of this equation by 125 gives us 𝑣 is equal to 71.2. The velocity 𝑣 is equal to 71.2 meters per second.

As we now know the value of 𝑣, we can use equation one to calculate the acceleration π‘Ž. 71.2 is equal to 80π‘Ž. Dividing both sides of this equation by 80 gives us π‘Ž is equal to 0.89. The magnitude of acceleration π‘Ž is equal to 0.89 meters per second squared. And the velocity 𝑣 is 71.2 meters per second.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.