Video Transcript
A particle was moving in a straight line at 172.8 kilometers per hour. If it decelerated over 120 meters to come to rest, find the deceleration 𝑎 of the particle and the time 𝑡 to cover this distance.
In order to answer this question, we will use our equations of uniform acceleration, or 𝑠𝑢𝑣𝑎𝑡 equations. The 𝑠 stands for the displacement, the 𝑢 for the initial velocity, 𝑣 for the final velocity, 𝑎 for acceleration, and 𝑡 for time.
We are told that the initial velocity is 172.8 kilometers per hour. It decelerates over 120 meters to come to rest. Therefore, the displacement 𝑠 is 120 meters, and the final velocity 𝑣 is zero kilometers per hour. We want to calculate the time 𝑡 and deceleration of the particle. We do need to be careful here because by default the 𝑠𝑢𝑣𝑎𝑡 equations use 𝑎 to represent acceleration. But in this question, 𝑎 has been defined to represent the deceleration. If we let the acceleration 𝑎 in the 𝑠𝑢𝑣𝑎𝑡 equation be equal to the negative of the deceleration 𝑎 from the question, then the answer we get will tell us the deceleration of the particle.
We notice here that our units are different. Our displacement is in meters, whereas the velocity is in kilometers per hour. We recall that there are 1,000 meters in one kilometer. There are 60 seconds in one minute and 60 minutes in one hour. As 60 multipled by 60 is 3,600. There are 3,600 seconds in one hour. Multiplying 172 by eight gives us 172,800. Therefore, our initial velocity is 172,800 meters per hour. We can then divide this by 3,600 to give us 48 meters per second. The initial velocity, 172.8 kilometers per hour, is the same as 48 meters per second.
As the particle was coming to rest, the final velocity will be zero meters per second. We can now substitute our values into the 𝑠𝑢𝑣𝑎𝑡 equations. In order to calculate the value of 𝑎, the deceleration, we will use 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values, we have zero squared is equal to 48 squared plus two multiplied by negative 𝑎 multiplied by 120. Zero squared is zero, and 48 squared is 2,304. Multiplying two by negative 𝑎 and 120 gives us negative 240𝑎. We can then add this to both sides of our equation to give us 240𝑎 is equal to 2,304. Dividing both sides by 240 gives us 𝑎 is equal to 9.6. The deceleration of the particle 𝑎 is equal to 9.6 meters per second squared.
As we now know all our values apart from 𝑡, we can use any of the equations of motion to calculate 𝑡. As a general rule, we wouldn’t use the value that we have calculated in the first part of the question. Therefore, we will use 𝑠 is equal to 𝑢 plus 𝑣 over two multiplied by 𝑡. The reason for this is that even if the first part of the question was incorrect, we could still get full marks on the second part.
Substituting in our values gives us 120 is equal to 48 plus zero divided by two multiplied by 𝑡. The right-hand side simplifies to 24𝑡. We can then divide both sides of this equation by 24, giving us 𝑡 is equal to five. The time taken for the particle to come to rest is five seconds.
