Question Video: Finding the Distance Covered by a Particle Moving with Uniform Acceleration till It Reaches a Given Velocity | Nagwa Question Video: Finding the Distance Covered by a Particle Moving with Uniform Acceleration till It Reaches a Given Velocity | Nagwa

# Question Video: Finding the Distance Covered by a Particle Moving with Uniform Acceleration till It Reaches a Given Velocity Mathematics • Second Year of Secondary School

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Given that a particle started moving from rest with a constant acceleration of 3.5 m/s² until its velocity became 378 km/h, find the distance it covered.

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### Video Transcript

Given that a particle started moving from rest with a constant acceleration of 3.5 metres per square second until its velocity became 378 kilometres per hour, find the distance it covered.

And the first thing we should notice is that the particle is moving with a constant acceleration. This tells us we’re going to need to use the equations of constant acceleration. Now, these are often called the SUVAT equations, where SUVAT is an acronym from each of the variables. 𝑠 is displacement, 𝑢 is initial velocity, 𝑣 is final velocity, 𝑎 is acceleration, and 𝑡 is time.

So, let’s see what we know about our particle. Firstly, it started moving from rest, so its initial velocity 𝑢 must have been zero. It has a constant acceleration of 3.5 metres per square second. And we’re told its final velocity is 378 kilometres per hour. Now, we do have a little bit of a problem; our units for acceleration and velocity are different. So, we’re going to convert the final velocity from kilometres per hour into metres per second.

Now, we can do this in stages. We can begin by converting from kilometres per hour into metres per hour by multiplying by 1000. And that’s because there are 1000 metres in a kilometre. So, we see that the particle has a final velocity of 378000 metres per hour. Then, to convert from hours into seconds, we divide by 3600. And that’s because if we’re converting from hours to seconds normally, we will multiply it by 60 and then multiply it by 60 again. This time though this is per hours and per seconds, so we do the opposite. We divide by 60 and then divide by 60 again, or divide by 3600. By the way, we end up with a final velocity of 105 metres per second.

We’re looking to find the distances covered, so we’re looking for 𝑠. And we’re not interested in 𝑡 at all. Now, the only equation in our list that does not include 𝑡 is this fourth one, 𝑣 squared equals 𝑢 squared plus two 𝑎 𝑠. So, let’s substitute what we have into this equation. We have 𝑣 squared, so that’s 105 squared equals 𝑢 squared, that’s zero squared plus two 𝑎𝑠, so two times 3.5 times 𝑠.

105 squared is 11025, whereas two times 3.5 is seven. So, we have 11025 equals seven 𝑠. And we can solve for 𝑠 by dividing through by seven. And when we do, we find that 𝑠 is equal to 1575. But what are the units? Well, our units for acceleration and for velocity were in metres per second squared and metres per second. So, our units for distance must be metres. And we can say then that the particle covered 1575 metres in its journey.

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