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 𝑎. Then it continued to move at
the velocity it had acquired for a further 65 seconds. Finally, it decelerated with a
rate of two 𝑎 until it came to rest. Given that the distance between
the two stations was 8.9 kilometers, find the magnitude of 𝑎 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 𝑎 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 𝑢
plus 𝑎𝑡. Substituting in our values, we
have 𝑣 is equal to zero plus 𝑎 multiplied by 80. This gives us the equation 𝑣
is equal to 80𝑎. Another one of our equations
states that 𝑠 is equal to 𝑢 plus 𝑣 divided by two multiplied by 𝑡. 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 𝑢 plus 𝑣 divided by two multiplied by 𝑡. 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 𝑎, our value of 𝑎 will be equal to negative two 𝑎 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 𝑡. We will begin by once again
using 𝑠 is equal to 𝑢 plus 𝑣 over two multiplied by 𝑡. 𝑠 three is, therefore, equal
to 𝑣 plus zero divided by two multiplied by 𝑡. This simplifies to 𝑠 three is
equal to 𝑣𝑡 divided by two.
We will also use the equation
𝑣 equals 𝑢 plus 𝑎𝑡. Substituting in our values here
gives us zero is equal to 𝑣 plus negative two 𝑎 multiplied by 𝑡. This simplifies to zero is
equal to 𝑣 minus two 𝑎𝑡. And adding two 𝑎𝑡 to both
sides of this equation gives us 𝑣 is equal to two 𝑎𝑡. 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 𝑣𝑡 over two is equal to
8900. 105𝑣 plus 𝑣𝑡 over two is
equal to 8900. If we can work out the value of
the time 𝑡, we can then calculate the velocity 𝑣.
Let’s consider equation one and
equation five. These tell us that 𝑣 is equal
to 80𝑎 and two 𝑎𝑡, which means that 80𝑎 must be equal to two 𝑎𝑡. At this stage, we know that 𝑎
is not equal to zero. Therefore, we can divide
through by 𝑎. We can then divide both sides
of this equation by two, giving us 𝑡 is equal to 40. The time taken for the third
part of the journey is 40 seconds. We can now substitute 𝑡 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 𝑎. 71.2 is equal to 80𝑎. Dividing both sides of this
equation by 80 gives us 𝑎 is equal to 0.89. The magnitude of acceleration
𝑎 is equal to 0.89 meters per second squared. And the velocity 𝑣 is 71.2
meters per second.
