Question Video: Finding the Final Velocity of a Body Moving with Uniform Acceleration given Its Initial Velocity and the Covered Distance | Nagwa Question Video: Finding the Final Velocity of a Body Moving with Uniform Acceleration given Its Initial Velocity and the Covered Distance | Nagwa

Question Video: Finding the Final Velocity of a Body Moving with Uniform Acceleration given Its Initial Velocity and the Covered Distance Mathematics • Second Year of Secondary School

A particle was moving in a straight line with a constant acceleration of 2 cm/s². Given that its initial velocity was 60 cm/s, find the velocity of the body when it was 15 m from the starting point.

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Video Transcript

A particle was moving in a straight line with a constant acceleration of two centimeters per square second. Given that its initial velocity was 60 centimeters per second, find the velocity of the body when it was 15 meters from the starting point.

We have a constant acceleration, so we’re going to use our kinematic equations. For a starting velocity 𝑣 naught, an acceleration 𝑎, and a velocity 𝑣 after 𝑡 time units, the first equation is 𝑣 equals 𝑣 naught plus 𝑎𝑡. We introduce Δ𝑥 to represent the displacement of the objects. And we get Δ𝑥 equals 𝑣 naught 𝑡 plus a half 𝑎𝑡 squared. Our third equation Δ𝑥 is equal to a half 𝑣 naught plus 𝑣 times 𝑡. And then we have one further equation. That is 𝑣 squared equals 𝑣 naught squared plus two times 𝑎 times Δ𝑥.

Let’s list what we know about the motion of our particle. We have acceleration as two centimeters per square second, an initial velocity of 60 centimeters per second, and a displacement Δ𝑥 of 15 meters. Now, in fact, our acceleration and velocity are given in centimeters per second and centimeters per square second. So, we want the units to be the same for Δ𝑥. We multiply 15 by 100 and we see that that’s 1500 centimeters. We’re looking to find the velocity of the body when it’s this distance away from the starting point. Notice that that means we’re not interested in the time that this takes.

So, we go through to our four kinematic equations and we get rid of those that contain 𝑡. Well, that’s equation one, two, and three. That leaves us with just one equation, 𝑣 squared equals 𝑣 naught squared plus two times 𝑎 times Δ𝑥. We’re going to substitute everything we know about the motion of our particle into this equation. When we do, we get 𝑣 squared equals 60 squared plus two times two times 1500. 60 squared is 3600. And two times two times 1500 is 6000. So, 𝑣 squared is equal to 9600.

It follows that we can solve for 𝑣 by finding the square root of both sides of our equation. 𝑣 is therefore equal to the square root of 9600. That’s 97.97 and so on. Correct to the nearest whole number, that’s 98. And we can therefore say that the velocity of the body when it’s 15 meters from the starting point is 98 centimeters per second.

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