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In this video, we will learn how to solve problems involving motion of a particle with uniform acceleration through one or more sections of its path.
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We will begin by recalling the equations of motion and how we can use them to calculate unknowns.
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The equations of motion, also known as SUVAT equations, are used when acceleration π is constant.
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They are known as SUVAT equations because they contain the following variables: π for displacement, π’ is the initial velocity, π£ is the final velocity or velocity at time π‘, π is the acceleration, and π‘ is the time.
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The initial velocity π’ is also sometimes written as π£ sub zero.
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For the purposes of this video, we will use the letter π’.
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However, these letters are interchangeable.
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Before using the equations, it is important that our measurements are in the correct units.
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For displacement, we use meters, for the velocities meters per second, acceleration is measured in meters per second squared, and time in seconds.
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Depending on which variables weβre dealing with, we use one of the five equations.
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π£ is equal to π’ plus ππ‘.
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π£ squared is equal to π’ squared plus two ππ .
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π is equal to π’π‘ plus a half ππ‘ squared.
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π is equal to π£π‘ minus a half ππ‘ squared.
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And π is equal to π’ plus π£ divided by two all multiplied by π‘.
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Before we look at some specific questions, we will also look at how we can solve this type of problem using a velocityβtime graph.
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A velocityβtime graph has time in seconds going along the horizontal or π₯-axis and velocity in meters per second going up the π¦- or vertical axis.
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When dealing with constant acceleration, the velocityβtime graph will be made up of straight lines.
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These straight lines can have positive or negative slope or be horizontal.
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When the graph is horizontal, the acceleration is equal to zero meters per second squared.
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When the graph is sloping diagonally upwards or downwards, we can calculate the acceleration by dividing the change in velocity by the change in time.
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This is the same as calculating the slope or the gradient.
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If the slope is positive, the acceleration will be positive.
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And if the slope is negative, the acceleration will be negative.
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This is also known as a deceleration.
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The greater the slope of the graph, the greater the acceleration or deceleration.
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We can also think of the change of velocity over the change in time as being the rise over the run.
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We can also calculate the displacement of the object from the graph.
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It is the area trapped between the graph and the π₯-axis.
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We can sometimes calculate this in one go.
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In the diagram shown, we would calculate the area of the trapezoid or trapezium.
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Alternatively, we could split the area into triangles and rectangles and calculate each area separately.
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The displacement, and in this case, also the distance travelled, is equal to the sum of these areas.
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Whilst the velocityβtime graph can be used to solve some of the problems that follow, we will focus in this video on using the equations of motion.
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A cyclist riding down a hill from rest was accelerating at a rate of 0.5 meters per square second.
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By the time he reached the bottom of the hill, he was traveling at 1.5 meters per second.
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He continued traveling at this speed for another 9.5 seconds.
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Determine the total distance π that the cyclist covered.
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There are two parts to the cyclistβs journey that we need to consider, firstly when he was accelerating down the hill and secondly when he continued to travel at this constant speed.
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We know that when traveling at a constant speed, acceleration is equal to zero.
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In order to answer this question, we will use the equations of motion or SUVAT equations.
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Letβs begin by considering the first part of the journey.
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The cyclist accelerates from rest.
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Therefore, his initial velocity is zero meters per second.
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We are told that the acceleration is 0.5 meters per second squared.
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At the bottom of the hill, the cyclist has reached a speed of 1.5 meters per second.
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Therefore, his final velocity π£ for this part of the journey is 1.5 meters per second.
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We are trying to calculate the total distance traveled.
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Therefore, we will call the displacement or distance for the first part of the journey π one.
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We will now use the equation that π£ squared is equal to π’ squared plus two ππ .
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Substituting in our values gives us 1.5 squared is equal to zero squared plus two multiplied by 0.5 multiplied by π sub one.
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This simplifies to 2.25 is equal to one multiplied by π sub one.
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The distance traveled in the first part of the journey is, therefore, equal to 2.25 meters.
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Letβs now consider the second part of the journey.
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The cyclist is traveling at a constant speed.
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Therefore, the acceleration is zero meters per second squared.
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Both the initial and final velocities are equal to 1.5 meters per second.
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And we are told he cycles at this speed for 9.5 seconds.
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We will call the distance covered in this part of the journey π sub two.
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We know that the displacement, or in this case the distance π , is equal to π’ plus π£ divided by two multiplied by π‘.
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Substituting in our values, we have π two is equal to 1.5 plus 1.5 divided by two multiplied by 9.5.
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This simplifies to 1.5 multiplied by 9.5, which is equal to 14.25.
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The distance covered in the second part of the journey is 14.25 meters.
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We can now calculate the total distance covered by adding 2.25 and 14.25.
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This gives us an answer of 16.5 meters.
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The cyclist covers a total distance of 16.5 meters.
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In our next question, weβll consider the motion of a train between two stations.
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A train, starting from rest, began moving in a straight line between two stations.
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For the first 80 seconds, it moved with a constant acceleration π.
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Then it continued to move at the velocity it had acquired for a further 65 seconds.
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Finally, it decelerated with a rate of two π until it came to rest.
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Given that the distance between the two stations was 8.9 kilometers, find the magnitude of π and the velocity π£ at which it moved during the middle leg of the journey.
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In order to answer this question, we will use our equations of motion or SUVAT equations.
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We will create equations for the three legs of the journey and then solve them to calculate any unknowns.
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Letβs begin by considering the first part of the journey.
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The train accelerates from rest, so our initial velocity is equal to zero meters per second.
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It accelerates for 80 seconds at a constant acceleration of π meters per second squared.
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We will call the velocity it reaches at this point π£ and the displacement from the start point or the distance covered π one.
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We know that π£ is equal to π’ plus ππ‘.
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Substituting in our values, we have π£ is equal to zero plus π multiplied by 80.
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This gives us the equation π£ is equal to 80π.
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Another one of our equations states that π is equal to π’ plus π£ divided by two multiplied by π‘.
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π one is, therefore, equal to zero plus π£ divided by two multiplied by 80.
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This simplifies to π one is equal to 40π£.
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We will call these equations one and two and move on to the second part of the journey.
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In the second part of the journey, the train travels at this constant velocity π£.
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This means its acceleration is equal to zero meters per second squared.
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The time it takes is 65 seconds.
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And we will call the distance traveled π two.
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Once again, we will use the equation π is equal to π’ plus π£ divided by two multiplied by π‘.
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This gives us π two is equal to π£ plus π£ divided by two multiplied by 65.
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As π£ plus π£ is equal to two π£, this simplifies to π two is equal to 65π£.
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We will call this equation three and now move on to the final part of the journey.
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In this part of the journey, the train decelerates to rest.
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As the deceleration is equal to two π, our value of π will be equal to negative two π meters per second squared.
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The final velocity π£ is equal to zero meters per second.
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The initial velocity in this part of the journey is equal to π£, the speed it was traveling for the second part of the journey.
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We will call the distance traveled in this part π three and the time it takes π‘.
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We will begin by once again using π is equal to π’ plus π£ over two multiplied by π‘.
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π three is, therefore, equal to π£ plus zero divided by two multiplied by π‘.
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This simplifies to π three is equal to π£π‘ divided by two.
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We will also use the equation π£ equals π’ plus ππ‘.
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Substituting in our values here gives us zero is equal to π£ plus negative two π multiplied by π‘.
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This simplifies to zero is equal to π£ minus two ππ‘.
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And adding two ππ‘ to both sides of this equation gives us π£ is equal to two ππ‘.
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We now have two further equations that weβll number four and five.
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The final part of our question tells us that the distance between the two stations was 8.9 kilometers.
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To use the equations of motion, we need this to be in meters.
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As there are 1000 meters in one kilometer, 8.9 kilometers is equal to 8900 meters.
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The distances π one, π two, and π three must, therefore, sum to 8900.
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Using equations two, three, and four, this can be rewritten as 40π£ plus 65π£ plus π£π‘ over two is equal to 8900.
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105π£ plus π£π‘ over two is equal to 8900.
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If we can work out the value of the time π‘, we can then calculate the velocity π£.
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Letβs consider equation one and equation five.
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These tell us that π£ is equal to 80π and two ππ‘, which means that 80π must be equal to two ππ‘.
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At this stage, we know that π is not equal to zero.
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Therefore, we can divide through by π.
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We can then divide both sides of this equation by two, giving us π‘ is equal to 40.
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The time taken for the third part of the journey is 40 seconds.
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We can now substitute π‘ equals 40 to calculate π£.
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40 divided by two is equal to 20, and 105 plus 20 is 125.
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Therefore, 125π£ is equal to 8900.
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Dividing both sides of this equation by 125 gives us π£ is equal to 71.2.
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The velocity π£ is equal to 71.2 meters per second.
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As we now know the value of π£, we can use equation one to calculate the acceleration π.
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71.2 is equal to 80π.
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Dividing both sides of this equation by 80 gives us π is equal to 0.89.
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The magnitude of acceleration π is equal to 0.89 meters per second squared.
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And the velocity π£ is 71.2 meters per second.
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In our final question, we will study the motion of two bullets fired at two different wooden blocks.
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A bullet was fired horizontally at a wooden block.
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It entered the block at 80 meters per second and penetrated 32 centimeters into the block before it stopped.
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Assuming that its acceleration π was uniform, find π.
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If under similar conditions, another bullet was fired at a wooden block that was 14 centimeters thick, determine the velocity at which the bullet exited the wooden block.
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In our first scenario, we are told that a bullet is fired with velocity 80 meters per second.
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It penetrates 32 centimeters into the block.
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And as there are 100 centimeters in one meter, this is equal to 0.32 meters.
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To calculate the value of π, we will use the equations of uniform acceleration or SUVAT equations.
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We know that π , the displacement or distance, is equal to 0.32 meters, the initial velocity is 80 meters per second, the final velocity is zero meters per second, and we are trying to calculate the acceleration π.
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We will do this using the equation π£ squared is equal to π’ squared plus two ππ .
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Substituting in our values gives us zero squared is equal to 80 squared plus two π multiplied by 0.32.
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This simplifies to zero is equal to 6400 plus 0.64π.
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We can then subtract 6400 from both sides and then divide by 0.64.
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This gives us a value of π equal to negative 10000.
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The acceleration of the bullet is negative 10000 meters per second squared.
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As there are 1000 meters in a kilometer, this can also be written as negative 10 kilometers per second squared.
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In our second scenario, the bullet travels straight through a wooden block of thickness 14 centimeters or 0.14 meters.
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Using the same values of π’ and π, we can now calculate π£, the velocity, at which the bullet exits the wooden block.
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We know that π£ squared is equal to π’ squared plus two ππ .
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Substituting in our values, we can calculate π£ squared.
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This is equal to 3600.
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Square rooting both sides and knowing that π£ must be positive, we get π£ is equal to 60.
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The bullet exits the block at a velocity of 60 meters per second.
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We will now summarize the key points from this video.
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The equations of motion can be used when the acceleration of an object is constant.
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Before using any of the five equations, it is important weβre in the correct units of measurement.
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Our standard units are meters, seconds, meters per second, and meters per second squared.