Question Video: Finding the Net Displacement of an Accelerating Object | Nagwa Question Video: Finding the Net Displacement of an Accelerating Object | Nagwa

# Question Video: Finding the Net Displacement of an Accelerating Object Physics • First Year of Secondary School

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An object has an initial velocity of 32 m/s. The object accelerates at 12 m/s² in the opposite direction to its velocity for a time of 5.5 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 32 meters per second. The object accelerates at 12 meters per second squared in the opposite direction to its velocity for a time of 5.5 seconds. What is the net displacement of the object in the direction of its initial velocity during this time?

In this question, we’re being asked about the motion of an object. Let’s represent this object by this blue circle here. The question tells us that it has an initial velocity of 32 meter per second, and we will label this initial velocity as 𝑢. We can recall that velocity is a vector quantity, which means that it has both a magnitude and a direction. This value of 32 meters per second is the magnitude of the object’s initial velocity. We’re not told what direction this initial velocity is in, but for the sake of our sketch, let’s suppose that it’s directed to the right.

What we do know is that whatever the direction of the initial velocity, the acceleration acts in the opposite direction to this. In our sketch then, since we’ve got the initial velocity directed to the right, then the acceleration must be to the left. We’re told that the acceleration has a value of 12 meters per second squared. This value is the magnitude of the acceleration and its direction is to the left. Since it has both a magnitude and a direction, then, just like velocity, acceleration is also a vector quantity. Now, instead of just thinking about left and right, it can be more helpful to define a positive and a negative direction.

Let’s choose the direction of the initial velocity, which is to the right, as the positive direction. If right is the positive direction, then this makes left, the opposite direction, negative. The object has an acceleration of 12 meters per second squared in the negative direction to the left. We could equally express this as an acceleration to the right equal to negative 12 meters per second squared. We can say then that the value of the acceleration in the direction of the object’s initial velocity is negative 12 meters per second squared. Let’s label this acceleration as 𝑎.

The fact that this acceleration value is negative means that it’s acting to reduce the velocity as measured in the original direction of motion of the object. We’re told that the object experiences this acceleration for a time of 5.5 seconds, which we’ll label as 𝑡. The question wants us to find the object’s net displacement in the direction of its initial velocity after this time of 5.5 seconds has passed. We’ve labeled this net displacement as 𝑠. If the value of 𝑠 is positive, so that’s a positive net displacement in the direction of the object’s initial velocity, then this means that the net displacement 𝑠 is in the same direction as the initial velocity 𝑢.

In our sketch, this direction is to the right. And since we’ve drawn the displacement vector pointing to the right, and therefore we’ve drawn the object displaced to the right from where it started, then we’ve kind of implicitly assumed that 𝑠 is positive in order to draw this sketch. If however when we calculate 𝑠, we get a negative value, then this means that the net displacement is to the left in the opposite direction to the initial velocity.

In order to work out the value of 𝑠, we can recall that there’s a kinematic equation which relates an object’s initial velocity, its acceleration, the time that it accelerates for, and its displacement after that time. That equation says that 𝑠 is equal to 𝑢 times 𝑡 plus a half times 𝑎 times 𝑡 squared. We can use this equation whenever we’ve got an object that’s moving in a straight line with an acceleration that has a constant value.

In this question, the acceleration of the object has a single numerical value of negative 12 meters per second squared. Since this value isn’t changing in time, then we can safely say that the acceleration is constant. Additionally, the acceleration acts along the same line as the object’s initial velocity. Since there’s no acceleration in any other direction than this left–right direction, then all of the object can do is to speed up or slow down along this line. We can safely say then that the motion will take place along a straight line. Since both of these two conditions are met, this means that we can go ahead and use this equation to find the value of 𝑠.

If we substitute in that the initial velocity 𝑢 is equal to 32 meters per second, the acceleration 𝑎 is negative 12 meters per second squared and the time 𝑡 has a value of 5.5 seconds. Then, we get this expression for the net displacement 𝑠. Since the values for 𝑢 and 𝑎 that we’ve substituted in are being measured in the direction of the object’s initial velocity, then the value for 𝑠 that we’ll calculate will be the net displacement in the direction of the object’s initial velocity.

Let’s now evaluate this expression beginning with the first term on the right-hand side. We’ve got a velocity in units of meters per second multiplied by a time in units of seconds. We can see then that the units of seconds and per second will cancel, just leaving units of meters. Multiplying 32 by 5.5, we find that this first term is equal to 176 meters. We then have to add this second term, which is equal to a half, multiplied by an acceleration in meters per second squared multiplied by the square of a time in units of seconds.

When we take the square of a quantity in units of seconds, then we’ll get a result in units of seconds squared. Then, the second squared and the per second squared will cancel out, leaving units of meters just like we had in the first term. We can then rewrite this second term as a half multiplied by negative 12 multiplied by the square of 5.5 in units of meters. When we do this multiplication, we get a result for the second term of negative 181.5 meters.

Overall then, we have that 𝑠 is equal to 176 meters plus negative 181.5 meters. We can write that more simply as 176 meters minus 181.5 meters. This works out as a net displacement in the direction of the object’s initial velocity equal to negative 5.5 meters. Since this value is negative, then this means that the object doesn’t actually end up displaced to the right, like we drew it in our sketch. Instead, it ends up displaced to the left in the opposite direction to the initial velocity. It’s worth taking a moment to think about how this can happen.

When the object is initially moving to the right, how can it end up displaced to the left? Let’s recall that we can represent vectors using arrows, where the direction of the arrow gives the direction of the vector and the length of the arrow gives the magnitude of the vector. Initially, the object’s velocity vector is directed to the right, and it starts out with a magnitude or length equal to 32 meters per second. However, for the following 5.5-second time interval, the object is accelerating to the left.

We can describe this as an acceleration vector that’s pointing to the left with a constant length or magnitude equal to 12 meters per second squared. Since the acceleration of an object is equal to the rate of change of that object’s velocity with time, then as we get further and further through this 5.5-second time interval, then this oppositely directed acceleration vector causes the velocity vector to get shorter and shorter and shorter. At some point, the length of this velocity vector instantaneously gets reduced to zero. At this time, the object is instantaneously stationary as the acceleration has slowed it down and brought it to a stop.

Since the object is still being constantly accelerated to the left, then as soon as it has stopped moving to the right, the object will start moving to the left. We then have a velocity vector that’s pointing to the left, and the length of this vector grows as the acceleration continues. The key point is that although at the beginning of the time interval the object was moving to the right, by the end of the interval it’s moving in the opposite direction to the left. Throughout this time interval then, the object starts out moving to the right.

And then because of the acceleration to the left, its velocity vector to the right gets shorter and shorter as the object slows down, eventually coming to a complete stop. At this moment, the object stops moving to the right and immediately starts moving back to the left. And now because of the acceleration to the left, its velocity vector, which is leftward pointing, is growing in length as more time goes on. Eventually the object ends up displaced to the left from where it started, so that’s a negative displacement in the direction of its initial velocity.

Since we found that 𝑠 was equal to negative 5.5 meters, then that means that the object ends up displaced to the left by 5.5 meters from where it started. The question asks for the displacement of the object measured in the direction of its initial velocity. We have found that the net displacement of the object in this direction is equal to negative 5.5 meters.

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