# Question Video: Finding the Initial and Final Velocity of a Body Moving with Uniform Acceleration Given Its Displacement and Time Mathematics

A particle moving in a straight line, was accelerating at a rate of 22 cm/s² in the same direction as its initial velocity. If the magnitude of its displacement 10 seconds after it started moving was 29 m, calculate the magnitude of its initial velocity 𝑣₀ and its velocity 𝑣 at the end of this period.

03:48

### Video Transcript

A particle moving in a straight line was accelerating at a rate of 22 centimeters per square second in the same direction as its initial velocity. If the magnitude of its displacement 10 seconds after it started moving was 29 meters, calculate the magnitude of its initial velocity 𝑣 naught and its velocity 𝑣 at the end of this period.

We’re given that the particle is accelerating at a constant rate of 22 centimeters per square second. This means to answer this question, we’re going to need to use the kinematic equations. These are, of course, equations of constant acceleration. For a starting velocity 𝑣 naught, a velocity 𝑣 after 𝑡 time units, an acceleration 𝑎, and a displacement Δ𝑥, they are as shown.

What we do is begin by listing everything we know about our motion. We’ve already said we know that the acceleration is constant, and it’s 22 centimeters per square second. It’s in the same direction as its initial velocity. Now, we don’t know its initial velocity, but by assuming that they’re in the same direction, we can take both acceleration and 𝑣 naught to be positive. We’re also told that the magnitude of the displacement 10 seconds after it started moving was 29 meters. Remember, displacement can have a direction. So, by considering just the magnitude, we’re thinking about the distance; that’s 29 meters. Time 𝑡 is 10 seconds.

Now, the question actually asks us to calculate the magnitude of the initial velocity and its velocity at the end of the period. Let’s begin by calculating its initial velocity 𝑣 naught. In this case, we’re not interested in 𝑣, so we go through our equations and eliminate those containing 𝑣. Those are one, three, and four. Our next step would normally be to substitute everything we know about the motion of our particle into that second equation. We do have a little bit of a problem though. We notice that the units for our acceleration and our displacement are different. We need them to be the same. So, we multiply displacement by 100 and we find it’s actually equal to 2900 centimeters.

Then, substituting everything we know into this formula, and we get 2900 equals 10𝑣 naught plus a half times 22 times 10 squared. A half times 22 times 10 squared is 1100. So, we subtract 1100 from both sides, and we find that 1800 is equal to 10 times 𝑣 naught. Our final step is to divide through by 10. 1800 divided by 10 is 180. Now, we’re working in centimeters. So, our velocity, our initial velocity 𝑣 naught, is 180 centimeters per second. We might choose to give our answer in meters per second by dividing through by 100. And when we do, we find that 𝑣 naught is 1.8 meters per second.

We’re not quite finished. We’re still looking to calculate its velocity 𝑣 at the end of the period. Now that we know 𝑣 naught, we can actually use any of our equations. So, let’s use the first one. We substitute everything we know about the motion of our particle into this formula, continuing to work in centimeters and centimeters per second. When we do, we get 𝑣 is 180 plus 22 times 10. 22 times 10 is 220. And 180 plus 220 is 400. We’re still, of course, working in centimeters per second. To give our answer in meters per second, we’ll divide through by 100. And when we do, we find that the velocity 𝑣 at the end of the motion is four meters per second.