A car that was initially moving at a steady speed travels a distance of 45 meters while accelerating in a straight line at 1.5 meters per second squared for 15 seconds. What was the car’s initial velocity in the direction of its acceleration?
Okay, so in this scenario, we’ve got a car. Now let’s say that this block that we’re drawing is representing our car. We’ve been told that this car was initially moving at a steady speed. So let’s arbitrarily say that, initially, it was moving in this direction at a steady speed, which we will call 𝑢. But then, we’ve been told that the car accelerates, in other words, changes its velocity in a straight line at 1.5 meters per second squared. So let’s then say that the car starts accelerating in this direction with an acceleration, which we will call 𝑎. And that happens to be 1.5 meters per second squared.
Now we’ve been told that while this car accelerates at 𝑎, it travels a distance of 45 meters. In other words then, the car travels that distance represented by the blue arrow. And we’ll call this distance 𝑠. We know that 𝑠 is 45 meters in a straight line. And, additionally, we know that the acceleration of the car took a time, which we’ll call 𝑡, of 15 seconds. Now what we’ve been asked to do is to find the car’s initial velocity in the direction of its acceleration. And because we’ve said the acceleration of the car is in this direction towards the right, we therefore have to say that anything moving in this direction towards the right is moving in the positive direction. And, consequently, anything moving towards the left is moving in the negative direction.
Now, at this point, it’s worth noting that in this diagram, we’ve arbitrarily drawn the acceleration towards the right. And we’ve also arbitrarily drawn the initial velocity towards the right. We could have chosen to draw both of them to the left as well, but that’s not relevant. The important point, though, is that there is no guarantee that 𝑢 and 𝑎 are in the same direction. In other words, in this diagram, we’ve drawn both acting towards the right. So what we’re trying to say here is not that we could have drawn them both towards the left, but rather that if we’ve chosen right to be the direction of the acceleration, then you may even be towards the left.
In other words, the acceleration might be in the opposite direction to the car’s initial velocity. However, this is not a problem. If that ends up being the case, then whatever value we’ll calculate for 𝑢 will end up being a negative value. And this will tell us that the acceleration and the initial velocity were in the opposite direction to each other.
So with all that being said, how are we going to actually go about calculating the car’s initial velocity? Well, to do this, we need to recall a relationship between the car’s initial velocity, the acceleration of the car, the distance traveled in a straight line by the car, and the time taken for that acceleration to occur. The relationship that we’re looking for is one of the kinematic equations. Specifically, this relationship, which tells us that the distance moved in a straight line by an object is equal to the initial velocity of the object multiplied by the time over which the object accelerates plus half multiplied by the acceleration of the object multiplied by the time of acceleration squared. And it’s important to remember that this equation only works if the acceleration is constant, which in this case it is. We’ve been told that the acceleration is 1.5 meters per second squared. And that is a constant value.
Now, at this point, we can see that we know the values of 𝑠, 𝑡, and 𝑎. And the only unknown in this equation is 𝑢. So we can rearrange to solve for 𝑢, which actually happens to be the initial velocity of the car. And that’s exactly what we’re trying to find. So to do this, we can start by subtracting half 𝑎 𝑡 squared from both sides of the equation. When we do this on the left, we’re left with 𝑠 minus half 𝑎𝑡 squared. And on the right, we have 𝑢𝑡 plus half 𝑎𝑡 squared minus half 𝑎𝑡 squared. Therefore, these two terms cancel out and we’re just left with 𝑢𝑡 on the right. Then to rearrange to solve for 𝑢, we can divide both sides of the equation by the time 𝑡. This way, the 𝑡 on the right cancels. And at this point, our equation tells us that 𝑠 minus one-half 𝑎𝑡 squared all divided by 𝑡 is equal to 𝑢.
So all that’s left for us to do now is to substitute in the values. We can say that 𝑢 is equal to 𝑠 which is 45 meters minus one-half 𝑎 which is 1.5 meters per second squared 𝑡 squared, which is 15 seconds whole squared. And we divide this whole thing by 15 seconds, which is the time 𝑡. So now we’ve plugged in all the values. Let’s quickly consider the unit.
In the numerator of our fraction here, we can see that units get very messy. We’ve got meters over here. But then in this set of parentheses being multiplied together, we’ve got meters per second squared and seconds whole squared. So if we just think about this final term here, we can see that the units are going to be meters per second squared from this term multiplied by seconds squared because remember, we’ve got seconds in this pair of parentheses. But we also have to remember to square that unit. And so, meters per second squared multiplied by second squared means that the second squared cancel. And the overall unit here is going to be just meters. And hence, we’ve got 45 meters here in the numerator minus whatever this value is in meters as well. And hence, the unit in the numerator itself is going to be meters.
And so, we can safely say that we’ve got meters here in the numerator. And we’re dividing this by the unit of seconds because remember, in the denominator, all we’ve got is 15 seconds which is the amount of time taken for the acceleration of the car to occur. And this means that the overall unit of a fraction is meters per second, which is perfect because on the left-hand side we’re calculating the initial velocity of the car which will be measured in meters per second. So with all the units being worried about already, all we need to do is to calculate the numerical value. So let’s start with the numerator.
We’ve got 45 meters minus half multiplied by 1.5 multiplied by 15 squared because remember, we need to square the 15 as well. And so what that leaves us within the numerator is negative 123.75 meters. Now, it’s quite interesting that we’ve got a negative value. We’ll come back to this in a second. But for now, let’s finish evaluating our fraction. We’ve got negative 123.75 meters divided by 15 seconds and that ends up being negative 8.25 meters per second. In other words then, we found the initial velocity at which the car was moving, which happens to be negative 8.25 meters per second.
So what does it mean for this value to be negative? Well, as we said earlier, when we were drawing our diagram, we made the assumption that the initial velocity and the acceleration were both acting in the same direction. And we did all our mathematics, assuming that that was true. However, as it turns out, we’ve got a negative value for our initial velocity. And because we said earlier that the direction of acceleration was positive because that’s what the question told us to do. Remember, we have to find the car’s initial velocity in the direction of its acceleration. This means that the maths is telling us that the car’s initial velocity was actually in the opposite direction to its acceleration.
In other words then, if we’re saying that the car’s 1.5 meters per second squared acceleration was towards the right, then this means that actually the car’s initial velocity was towards the left. The car was initially moving towards the left. And then, an acceleration towards the right of 1.5 meters per second squared for time of 15 seconds resulted in it having a displacement of 45 meters.
But anyway, so as the answer to our question, we can say that the car’s initial velocity in the direction of its acceleration was negative 8.25 meters per second.