# Video: Finding the Distance Covered by a Particle given the Expression of Its Velocity with Time by Using Integration

A particle’s velocity in metres per second as a function of time is 𝑣(𝑡) = 5𝑡 − 3. What distance does it travel between 𝑡 = 0 and 𝑡 = 1?

06:18

### Video Transcript

A particle’s velocity in metres per second as a function of time is 𝑣 of 𝑡 equals five 𝑡 minus three. What distance does it travel between 𝑡 equals zero and 𝑡 equals one?

So, I’ve shown our function on a quick sketch here. And, we can see we’ve got two points. We’ve got the point at zero, negative three, and another point at one, two. And, we want to find the distance it travels between 𝑡 equals zero and 𝑡 equals one. So, I’ve shown that here, so the — this section of our function that we want to find. We’ll have to think, well, how we solve this problem? We could look at the definite integral between one and zero of our function 𝑣 of 𝑡. However, we’re trying to find the distance it has traveled between time equals zero and time equals one. But, this definite integral would give us the net displacement. So, what could we do to solve this problem? How could we get the net distance?

Well, to find the net distance, what we want to do is think about our function into subintervals. And, I’ve circled these here in blue. Well, to know how to deal within these two subintervals, we’ll have to find a certain point. And, I’ve marked that point here in pink. So, let’s find that point. What we can see from the sketch that at this point, our function is equal to zero. So, 𝑣 of 𝑡 is equal to zero. So now, we can set our function equal to zero. So, we have five 𝑡 minus three equals zero. So then, if we add three to each side of the equation, we’re gonna get five 𝑡 equals three. And then, when we divide it by five, we get 𝑡 is equal to three over five. So, that’s great. We’ve now found the point that we need to divide our function in two subintervals over 𝑡 equals zero and 𝑡 equals one.

So therefore, our subintervals are gonna be zero, three over five, and three over five, one. And, it’s worth noting that over the first interval, our function is going to be negative. And over the second interval, it’s going to be positive. So now, thinking back to what we did earlier, we can say that the distance, which is what we’re looking for, is gonna be the definite integral between the limits one and zero of the modulus or absolute value of our function 𝑣 of 𝑡. So, that’s the difference we’re wanting to find it as the absolute value or modulus of our function.

Well, we can do that using our subintervals cause we can say that that is equal to the definite integral between the limits of three over five and zero of negative 𝑣 of 𝑡, and that’s because, as we said, the first subinterval is negative, plus the definite integral between one and three over five of 𝑣 of 𝑡. So, we’re gonna use that now to find the distance traveled between 𝑡 equals zero and 𝑡 equals one. So, this is gonna give us, in our scenario, the definite integral between the limits one and zero of the modulus, where the absolute value of 𝑣 of 𝑡 is equal to the definite integral between three over five and zero of three minus five 𝑡 plus the definite interval between one and three over five of five 𝑡 minus three. And, we had three minus five 𝑡 because this is negative 𝑣 of 𝑡.

Okay, great, so what we actually have here is the net change term because we’re trying to find the net change or net distance traveled between 𝑡 equals zero and 𝑡 equals one. So, now to calculate the value, what we’re gonna have to deal with is definite integrals. And to remind ourselves how to do that, what we do with a definite integral, if we’re trying to find the definite integral of 𝑓 of 𝑥 between the limits 𝑏 and 𝑎. This is equal to the integral of 𝑓 of 𝑥 would be substituted in for 𝑥 minus the integral of 𝑓 of 𝑥 with 𝑎 substituted in for 𝑥.

So now, when we do the first stage, which is integrating, we get three 𝑡 minus five 𝑡 squared over two between the limits three over five and zero plus five 𝑡 squared over two minus three 𝑡 between the limits one and three over five. And just to remind us how we integrated, let’s think about five 𝑡 where we have five 𝑡. What we do is we’ve raised the exponent by one. So, we add one to the exponent. So, it’s five 𝑡 to the power of one plus one, which is five 𝑡 squared. And then, we divide by the new exponent.

So, this is gonna be equal to three multiplied by three over five minus five multiplied by three over five squared over two then minus zero. Because if we substitute in 𝑡 equals zero, it just gives us zero. Plus, then, we’ve got five multiplied by one squared over two minus three multiplied by one minus five multiplied by three over five squared over two minus three multiplied by three over five. So, when we calculate this, we’re gonna get nine over five minus nine over 10 plus, and, we’ve got negative a half minus negative nine over 10. And, we’ve got that because if we’ve got five multiplied by three over five squared over two minus three multiplied by three over five, we get nine over 10 minus 18 over 10, which is negative nine over 10.

Okay, great, so we’re gonna simplify this remembering that if you’re subtracting negative, it turns positive. But before we do that, we need to make sure that everything has the same denominator. So, when we do that, we’re gonna get 18 over 10 minus nine over 10 minus five over 10 plus nine over 10. So, this is gonna give us 13 over 10. So therefore, we can say that if a particle’s velocity in metres per second as a function of time is 𝑣 of 𝑡 equals five 𝑡 minus three. Then, the distance it travels between 𝑡 equals zero and 𝑡 equals one is gonna be 13 over 10 metres or 1.3 metres.