# Video: Rate of Velocity Change

An object accelerates at 5 m/s² for 0.25 s. How much does its velocity increase?

03:25

### Video Transcript

An object accelerates at five metres per second squared for 0.25 seconds. How much does its velocity increase?

Okay, so in this question we’ve been told that we’ve got an object. Let’s say this blob is our object. And we’ve been told that it’s accelerating. Let’s say it’s accelerating to the right at five metres per second squared for a total time of 0.25 seconds. So let’s say that the object ends up here at the end of the 0.25 seconds. Now, we’ve been asked to find out how much its velocity increases by. In other words, we can say that, initially, the object had some velocity, which we’ll call 𝑣 one. And at the end of the 0.25 seconds, it had another velocity which we’ll call 𝑣 two.

We’ve been asked to find out the difference between 𝑣 two and 𝑣 one because we’ve been asked to find out how much the velocity of the object increases by. Or, in other words, we’ve been asked to find the change in velocity of the object because we use Δ to represent change in and 𝑣 to represent velocity, where of course this change in velocity is the same thing as the final velocity, 𝑣 two, minus the initial velocity, 𝑣 one. And we don’t know what this is. Now, it’s important to note that we don’t actually need to figure out the individual values of 𝑣 two and 𝑣 one. We just need to find the difference between them, Δ𝑣. And the way we can do this is to recall the definition of acceleration.

Acceleration, which we’ll call 𝑎, is defined as the rate of change of velocity which, in other words, is the change in velocity of an object divided by the change in time over which this velocity change is occurring. In other words, if we were to say that this object starts moving at, for example, three o’clock in the afternoon. Then 0.25 seconds later, so that’s three o’clock plus 0.25 seconds, is the final point at which we’ll consider the object. And so we’ve been asked to find out the change in velocity of the object over this 0.25 second time interval. In other words then, we can say that the change in time between the start and the finish is actually 0.25 seconds, which is what we’ve been told in the question. We’ve been told that the object is accelerating at five metres per second squared for 0.25 seconds.

So in this case, we know the acceleration and we know the time interval. And we’ve been asked to find out Δ𝑣, the change in velocity. So to do this, we need to rearrange the equation. We can do this by multiplying both sides of the equation by Δ𝑡, which means that Δ𝑡 cancels on the right-hand side. And so, we’ve only got Δ𝑣 left on the right-hand side and Δ𝑡 multiplied by 𝑎 on the left. Or, we can move the right-hand side to the left and the left-hand side to the right, which gives us Δ𝑣 is equal to Δ𝑡 times 𝑎. Or, we can write it as 𝑎 times Δ𝑡. And then, we can substitute in the values for 𝑎 and for Δ𝑡. When we do that, we get Δ𝑣 is equal to five metres per second squared multiplied by 0.25 seconds.

Considering the units very quickly, we see that we’ve got metres per second squared multiplied by seconds. In which case, a power of seconds in the numerator cancels with one of the powers of seconds in the denominator. And what we’re left with is metres divided by one power of seconds. Or, in other words, metres per second which is the unit of velocity. And this is good because we’re actually finding the change in velocity. So all that’s left to do now is to evaluate five times 0.25 which ends up being 1.25. And hence, we’ve found our final answer. The velocity of the object changes by or, more specifically, increases by 1.25 metres per second.

And the reason that we know the velocity is increasing is because we’ve been told that the object is accelerating at five metres per second squared. The fact that the acceleration is positive means that the acceleration is in the same direction as the velocity. Therefore, the object is only going to speed up. And hence, the velocity is increasing.