Video Transcript
An object has an initial velocity that decreases to 10 meters per second as the object accelerates in the opposite direction to its velocity. The object moves along a 60-meter-long straight line accelerating with a magnitude of 6.5 meters per second squared. What is the object’s initial velocity to the nearest meter per second?
Okay, so in this question, we’ve got an object that’s accelerating, and it has some unknown initial velocity. Let’s label this as 𝑢. The question is asking us to find the object’s initial velocity. And so what we’ve got to do here is to work out the value of this quantity 𝑢. We’re told that the object’s velocity decreases until it ends up at a final value of 10 meters per second. We’ll label this final velocity as 𝑣. We’re also told that the object ends up at this final velocity after accelerating over a distance of 60 meters. Let’s label this distance as 𝑠.
The question tells us that the object accelerates with a magnitude of 6.5 meters per second squared and that this acceleration happens in the opposite direction to the object’s velocity. So, if the object’s velocity is in this direction pointing to the right, that means its acceleration is in this direction to the left. Since the velocity of 10 meters per second that we’re given is a positive value, then this means that the direction of the object’s velocity is being taken as the positive direction. Since the acceleration is in the opposite direction to this, this means that the acceleration is in the negative direction. So, the object accelerates with a magnitude of 6.5 meters per second squared in the negative direction.
This means that the object has an acceleration, which we’ve labeled as 𝑎, of negative 6.5 meters per second squared. Now, perhaps the idea of a negative acceleration might seem kind of strange because usually we tend to think of acceleration as meaning speeding up. However, acceleration just means any change in the object’s velocity over time, and this can involve speeding up or slowing down. A negative acceleration, also known as a deceleration, is one that acts to slow down the object.
With all this in mind, let’s draw a quick sketch showing what happens to the object. Let’s suppose that this orange dot represents our object. We know that it starts out moving at some initial velocity that we’ve labeled as 𝑢, and we’ll suppose that this initial velocity is directed to the right. After traveling along a straight line for a distance of 60 meters, which we labeled as 𝑠, our object ends up over here, with a final velocity of 𝑣 equal to 10 meters per second. And this final velocity will also be directed to the right.
During this motion, we know that the object experiences an acceleration 𝑎 of negative 6.5 meters per second squared. To answer this question, we need to find a way to work out the object’s initial velocity by using its final velocity, the distance it travels, and its acceleration. We can recall that these four quantities are linked by one of the kinematic equations. Specifically, that equation is this one here, which says that the square of the final velocity 𝑣 is equal to the square of the initial velocity 𝑢 plus two times the acceleration 𝑎 times the distance 𝑠 over which this acceleration occurs. This equation can only be used when the object is moving along a straight line and the acceleration is constant.
The question tells us explicitly that the object moves along a straight line, so this first condition is met. We also have a value for the object’s acceleration of negative 6.5 meters per second squared. This value is fixed; it’s not changing in time. And so the acceleration of the object is constant, and our second condition is also met. Since both these conditions are met, this means we’re good to go ahead and use this equation to solve our problem. We want to use the equation to find the value of the object’s initial velocity 𝑢. This means we need to rearrange the equation to make 𝑢 the subject.
We start from our equation that says 𝑣 squared is equal to 𝑢 squared plus two times 𝑎 times 𝑠. Then, the first thing that we do is to subtract two times 𝑎 times 𝑠 from each side of the equation. Then, on the right-hand side, we’ve got the two terms plus two times 𝑎 times 𝑠 and minus two times 𝑎 times 𝑠, and these two terms cancel each other out. If we swap the left- and right-hand sides of the equation over, we then have that 𝑢 squared is equal to 𝑣 squared minus two times 𝑎 times 𝑠. Finally, we take the square root of both sides. And since the square root of 𝑢 squared is simply 𝑢, then this gives us that 𝑢 is equal to the square root of 𝑣 squared minus two times 𝑎 times 𝑠.
Now that we’ve got this expression for the initial velocity 𝑢, all that remains is to substitute these values into the right-hand side. Let’s clear ourselves a bit more space so that we can do this. When we substitute in our values, we get that 𝑢 is equal to the square root of the square of 10 meters per second, that’s our final velocity 𝑣, minus two times negative 6.5 meters per second squared, that’s the acceleration 𝑎, times 60 meters, the distance 𝑠. The square of 10 meters per second works out as 100 meters squared per second squared. And two times negative 6.5 meters per second squared times 60 meters gives us negative 780 meters squared per second squared.
Since we’re subtracting a negative number, then the two negative signs negate each other. So we have that 𝑢 is equal to the square root of 100 meters squared per second squared plus 780 meters squared per second squared. We can then add together these two terms to find that 𝑢 is equal to the square root of 880 meters squared per second squared. Evaluating the square root gives a result of 29.66 meters per second, where the ellipses indicate that there are further decimal places. This result is the initial velocity of the object, which is what we were asked to find.
The question asks for this initial velocity to the nearest meter per second. Rounding to the nearest meter per second gives us our answer that the object’s initial velocity is 30 meters per second.
