Video: Finding the Velocity and Displacement of a Particle Moving with Uniform Acceleration

A particle moves along the π‘₯-axis in the direction of π‘₯ increasing. It starts at π‘₯ = 37 cm with an initial velocity of 47 cm/s and moves with uniform acceleration of 51 cm/sΒ² in the same direction as its motion. Determine its velocity and its displacement from the origin after 6 seconds.

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Video Transcript

A particle moves along the π‘₯-axis in the direction of π‘₯ increasing. It starts at π‘₯ equals 37 centimetres with an initial velocity of 47 centimetres per second and moves with uniform acceleration of 51 centimetres per second squared in the same direction as its motion. Determine its velocity and its displacement from the origin after six seconds.

Well, this type of question is something that we that we call a π‘ π‘’π‘£π‘Žπ‘‘ question or question that uses the π‘ π‘’π‘£π‘Žπ‘‘ equations. But what do we mean by π‘ π‘’π‘£π‘Žπ‘‘? Well, π‘ π‘’π‘£π‘Žπ‘‘ relates the variables that we’re gonna be looking at, which are displacement, which is 𝑠; initial velocity, 𝑒; final velocity, 𝑣; acceleration, π‘Ž; and time, 𝑑. And see that equations that we’re gonna look at all relate to problems where we have a constant acceleration. So the acceleration does not change. There are, in fact, a variety of π‘ π‘’π‘£π‘Žπ‘‘ equations. But the five that we tend to use are these five here. And we can find different ones at variations thereof. So we have 𝑣 equals 𝑒 plus π‘Žπ‘‘, 𝑠 equals 𝑒𝑑 plus half π‘Žπ‘‘ squared, 𝑠 equals a half multiplied by 𝑒 plus 𝑣 multiplied by 𝑑, 𝑣 squared equals 𝑒 squared plus two π‘Žπ‘ , and 𝑠 equals 𝑣𝑑 minus a half π‘Žπ‘‘ squared.

Now, the first thing we want to do and the first thing I ever did with this type of question is identify what my 𝑠, 𝑒, 𝑣, π‘Ž, and 𝑑 are or whichever ones we can. Well, in our scenario, the 𝑠 and the 𝑣 we don’t know cause they’re what we’re looking to find. Our initial velocity 𝑒 is 47 because it’s 47 centimetres per second. Our acceleration is 51 because it’s 51 centimetres per second squared. And our 𝑑 is equal to six cause our time is six cause it’s six seconds. And it’s worth stating that I also checked our units so we could see that everything is in centimetres or seconds or centimetres per second, centimetres per second squared. So we don’t have to do anything with changing any of the units.

Well, first of all, what I’m gonna do is I’m gonna use equation one. And we’ll use equation one to find out what 𝑣, our final velocity, is. So if I multiply in our values, what we gonna get is 𝑣 is equal to 𝑒, which is 47, plus π‘Ž, which is 51, multiplied by our 𝑑, which is six, which is gonna be equal to 47 plus 306, which is gonna give us a velocity of 353 centimetres per second. Great, so we found 𝑣. So now, we’ve got a choice of equations we could use to find out our 𝑠, so our displacement. The equation I’m gonna use is, in fact, equation two. And I’m gonna use this just because it doesn’t rely on a value we’ve just found.

So if we use equation two what we gonna get is 𝑠 is equal to 𝑒𝑑, which is 47 multiplied by six, plus a half π‘Žπ‘‘ squared, so a half multiplied by 51 multiplied by six squared. So if we calculate this, we’re gonna get 282 plus 918 which is gonna give us a final answer for the displacement of 1200 centimetres. Okay, great, so we’ve found the velocity and we’ve found the displacement. What we can do now is just quickly check the displacement because, as I said, you could have used another equation to do this. And what we’re gonna now use is use equation three, which does, in fact, use 𝑣 our velocity cause it’s a good way of double-checking whether our 𝑠 and our 𝑣 are correct.

So if we use the third equation, we’re gonna have 𝑠 is equal to a half multiplied by ⁠— there we got 47 plus 353, then multiplied by six which is gonna give us 200 multiplied by six, which works out as 1200 centimetres, which confirms yes, we’re correct and our values for 𝑠 and 𝑣 are correct.

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