 Question Video: Acceleration over Distance and Time | Nagwa Question Video: Acceleration over Distance and Time | Nagwa

# Question Video: Acceleration over Distance and Time Physics

An object has an initial velocity of 10 m/s. The object accelerates at 1.5 m/s² in the opposite direction to its velocity for a time of 2.5 s. What is the net displacement of the object in the direction of its initial velocity during this time?

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### Video Transcript

An object has an initial velocity of 10 meters per second. The object accelerates at 1.5 meters per second squared in the opposite direction to its velocity for a time of 2.5 seconds. What is the net displacement of the object in the direction of its initial velocity during this time?

Okay so we’ve been told that we’ve got an object that is initially moving at 10 meters per second. So let’s say it’s moving towards the right initially, and let’s say that its initial velocity is called 𝑢. We can then write down over here that 𝑢 is equal to 10 meters per second. Then we’ve been told in the question that the object accelerates at 1.5 meters per second squared in the opposite direction to its velocity. So let’s say its acceleration, which we’ll call 𝑎, is 1.5 meters per second squared in the opposite direction to 𝑢.

Now we need to be very careful here. We’re going to say that 𝑎 is negative 1.5 meters per second squared. This is because the acceleration is in the opposite direction to 𝑢. But then this means that 𝑎 cannot simply be the magnitude or size of the acceleration. Now normally when we draw diagram like this, the arrow deals with the direction and the letter whatever it is, in this case 𝑎, would normally just show the size of the acceleration. However, this time we are including the negative sign in the acceleration value because we’re going to need it later from when we conduct our calculation.

Anyway, so we’re told that this object is accelerating in the opposite direction to its velocity for a time of 2.5 seconds. So even though the object is initially moving to the right and accelerating to the left as we’ve drawn it, the object is still gonna move towards the right although getting slower and slower. This happens for a time 𝑡, which we will call 2.5 seconds. What we’ve been asked to do is to find the displacement of the object in the direction of its initial velocity, in other words the distance between this point and this point.

Let’s call that displacement 𝑠, and we’ll say that we’re trying to find out the value of 𝑠, so 𝑠 is equal to question mark. Now the way that we can do this is to recall a particular SUVAT equation this SUVAT equation is 𝑠 is equal to 𝑢𝑡 plus half 𝑎𝑡 squared, where of course 𝑠 is the displacement of the object, 𝑢 is its initial velocity, 𝑡 is the time taken for which the object is accelerating at a rate 𝑎, and of course we’ve already seen what 𝑡 means and half is just a constant. So all we need to do at this point is to plug in the values of 𝑢, 𝑡, and 𝑎. But before we do that, we can see that we’ve been given all of these values in their standard units. This is a good thing.

This means that our final answer of 𝑠 is going to be in its standard unit, which is the meter, because remember 𝑠 represents a distance or displacement, and the standard unit of that is the meter. So when we plug in the values, we get 𝑠 is equal to 𝑢𝑡 plus half 𝑎𝑡 squared. And this is where it’s important that we kept the negative sign of the acceleration. Because the object is accelerating in the opposite direction to the direction it’s moving in, this will give us a different smaller value for 𝑠 than if the object was accelerating at the same rate but in the same direction as its motion. But this is not the case, and so we need to account for this by including the negative sign.

But anyway, now that we’ve successfully considered this, we need to evaluate the right-hand side of this equation. Doing so gives us a value of 20.3125. And as we said earlier, the unit is going to be meters. But then there is a sneaky final part of the question: it says answer to one decimal place. So we need to round our answer to one decimal place. Now this is the first decimal place, but it’s the next value that tells us what’s going to happen to this digit. Now the second decimal place is a one, and one is less than five. So our first decimal place is going to stay the same; it’s not going to round up. And therefore, to one decimal place, the displacement of the object in the direction of its initial velocity is 20.3 meters. And that is our final answer.