A car decelerated from 144 kilometers per hour to 48 kilometers per hour. Given that while decelerating it covered 1200 meters, find the time 𝑡 it took. If the car maintained this rate of deceleration, how much further would it travel before it stops?
In order to answer this question, we will use the equations of motion or SUVAT equations. 𝑠 is the displacement, 𝑢, sometimes written as 𝑣 sub zero, is the initial velocity, 𝑣 is the final velocity, 𝑎 is the acceleration, and 𝑡 is the time. We are told that the initial speed of the car was 144 kilometers per hour, the final speed was 48 kilometers per hour, and it covered a distance of 1200 meters. We have a problem here with our units as we have both meters and kilometers.
We could convert the displacement into kilometers by using the fact there are 1000 meters in one kilometer. Alternatively, we could convert from kilometers per hour to meters per second. There are 60 seconds in one minute, and 60 minutes in one hour. As 60 multiplied by 60 is 3600, there are 3600 seconds in one hour. We can, therefore, convert from kilometers per hour to meters per second by multiplying by 1000 and then dividing by 3600. This is the same as dividing by 3.6. 144 divided by 3.6 is equal to 40. 48 divided by 3.6 is equal to 40 over three, or 13.3 recurring.
This means that our initial velocity is 40 meters per second and the final velocity is 40 over three meters per second. The first part of our question asks us to calculate the time 𝑡 that it took to decelerate between these two values. We will use the equation 𝑠 is equal to 𝑢 plus 𝑣 divided by two multiplied by 𝑡. Substituting in our values we get 1200 is equal to 40 plus 40 over three divided by two multiplied by 𝑡. The calculation in the brackets simplifies to 80 over three or eighty thirds so that we have 1200 is equal to 80 over three multiplied by 𝑡.
Multiplying both sides of this equation by three gives us 3600 is equal to 80𝑡. We can then divide both sides by 80 so that 𝑡 is equal to 45. The time taken for the car to decelerate from 144 kilometers per hour to 48 kilometers per hour is 45 seconds.
The second part of the question wants us to calculate how much further the car would travel before it stops. In order to do this, we firstly need to calculate the value of 𝑎. One way of doing this is using the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values here, we have 40 over three squared is equal to 40 squared plus two 𝑎 multiplied by 1200. This simplifies to 1600 over nine is equal to 1600 plus 2400𝑎. We can divide each of the terms by 100. Subtracting 16 from both sides of this equation gives us negative 128 over nine is equal to 24𝑎. We can then divide both sides by 24 so that 𝑎 is equal to negative 16 over 27.
The acceleration of the car is equal to negative 16 over 27 meters per square second. Therefore, the deceleration is 16 over 27 meters per square second. This is the magnitude or absolute value of 𝑎. We will now clear some space to work out the final part of this question.
We need to calculate the distance traveled from this point until the car stops. This means that the initial velocity is now 40 over three meters per second. The value of 𝑎 is negative 16 over 27 meters per second squared as the car maintains the rate of deceleration. The final velocity is equal to zero meters per second.
We will once again use the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values, we have zero squared is equal to 40 over three squared plus two multiplied by negative 16 over 27 multiplied by 𝑠. This simplifies to zero is equal to 1600 over nine minus 32 over 27 multiplied by 𝑠. The numerators are both divisible by nine. And the denominators are divisible by 32. Adding 𝑠 over three to both sides gives us 𝑠 over three is equal to 50. We can then multiply both sides by three, giving us a value of 𝑠 equal to 150.
The car travels 150 meters further before it stops.