Video Transcript
An object has an initial velocity of 12 meters per second. The object accelerates at 2.5 meters per second squared in the same direction of its velocity for a time of 1.5 seconds. What is the displacement of the object during this time? Answer to one decimal place.
In this question, we’ve got an object that’s accelerating. And let’s suppose that this blue circle here represents that object. We’re told that it has an initial velocity equal to 12 meters per second, and we’ll label this initial velocity as 𝑢. We’re not told anything about the direction of its motion. But for the sake of this diagram, let’s suppose that its initial velocity is directed to the right. Whatever direction the object is moving in, we’re told that it accelerates in the same direction as its velocity. So as we’ve drawn things in our diagram, that means that the acceleration is directed to the right. Let’s call this acceleration 𝑎, and we’re told in the question that it has a value of 2.5 meters per second squared.
After accelerating for a time of 1.5 seconds, which we’ve labeled as 𝑡, the object will be displaced by some amount from the position it was in before it started accelerating. Since the object accelerates in the same direction it was initially moving in, then its displacement must be in the same direction again. So in our diagram, that’s to the right. We’ve labeled this unknown displacement as 𝑠, and this is what the question is asking us to work out.
To find the value of this displacement 𝑠, we can recall one of the kinematic equations, which says that 𝑠 is equal to 𝑢 times 𝑡 plus a half times 𝑎 times 𝑡 squared. In this equation, the 𝑠 on the left-hand side is the displacement of the object. Then, on the right-hand side, 𝑢 is the initial velocity that the object was moving with, 𝑎 is the object’s acceleration, and 𝑡 is the time that the object accelerates for. Now, we should point out that this kinematic equation only applies for motion where the acceleration is constant. Luckily for us, the acceleration here, 2.5 meters per second squared to the right, is indeed constant.
Also conveniently, the labels that we’ve given the quantities in this question all match up with those in the equation. 𝑢 is the initial velocity of the object, which we know is equal to 12 meters per second. 𝑎 is the object’s acceleration, which is 2.5 meters per second squared. And 𝑡 is the time it accelerates for, which we know is 1.5 seconds. We can take those values and substitute them into this equation in order to calculate 𝑠, the object’s displacement.
When we do that, we end up with this expression here. The first term on the right-hand side is 12 meters per second multiplied by 1.5 seconds. The units of seconds and per second cancel out. And this first term works out as 18 meters. Then, we add to that this second term which is equal to a half times 2.5 meters per second squared times the square of 1.5 seconds. When we take the square of this time in units of seconds, that will give us a quantity in units of second squared. Those second squared will then cancel out the per second squared in the acceleration. And so, we’ll be left with just units of meters.
Evaluating the second term gives a result of 2.8125 meters and then adding together these two terms, we get a result for 𝑠 of 20.8125 meters. Notice, however, that the question wants our answer to one decimal place. Rounding to this level of precision gives us a result for 𝑠 of 20.8 meters. We have found then that, to one decimal place, the displacement of the object during the time that it accelerates is equal to 20.8 meters.
