Video: Finding the Displacement of a Body Moving with Uniform Acceleration in a Given Time given Its Initial Velocity

A small ball started moving horizontally at 16.3 m/s. It moved in a straight line with a uniform deceleration of 3 m/s². Determine the distance the ball covered in the first 2 seconds.

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Video Transcript

A small ball started moving horizontally at 16.3 meters per second. It moved in a straight line with a uniform deceleration of three meters per square second. Determine the distance the ball covered in the first two seconds.

The ball has a uniform deceleration. In other words, it has a uniform negative acceleration. That’s an indication to us that we’re going to use the kinematic equations, the equations of uniform acceleration. And so, the first thing that we do is list those four equations. For an initial velocity 𝑣 naught, an acceleration of 𝑎, a velocity of 𝑣 at time 𝑡 time units, and a displacement of Δ𝑥, they are as shown.

Our next step is to eliminate all but one of these equations, and we do so by listing everything we know about the motion of our object. We’re told that the small ball starts moving at 16.3 meters per second, so that’s its initial velocity. It has a uniform deceleration of three meters per square second. In other words, it’s slowing down. So, we say its acceleration is negative three. Time 𝑡 is two seconds, and we’re looking to find the distance that the ball covers.

We recall that distance is simply the magnitude of the displacement. So, distance here is Δ𝑥. That’s change in 𝑥. And that means we’re not interested in 𝑣, the final velocity of the object. So, we go through our equations and we eliminate every equation that contains 𝑣. That’s one, three, and four, leaving us, of course, with just equation two.

Our next job is to substitute what we know about our ball into this equation. Δ𝑥 is what we’re trying to calculate. Then, 𝑣 naught 𝑡 is 16.3 times two. A half 𝑎𝑡 squared becomes a half times negative three times two squared. Then, 16.3 times two is 32.6. And a half times negative three times two squared is negative six. So, Δ𝑥 is 32.6 minus six, which is 26.6, and we’re measuring that in meters. So, the ball covers a distance of 26.6 meters in the first two seconds of its motion.

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