Question Video: Determining the Distance Covered by a Particle and Its Velocity given the Time and Acceleration Expression | Nagwa Question Video: Determining the Distance Covered by a Particle and Its Velocity given the Time and Acceleration Expression | Nagwa

Question Video: Determining the Distance Covered by a Particle and Its Velocity given the Time and Acceleration Expression Mathematics • Third Year of Secondary School

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The acceleration of a particle moving in a straight line, at time š‘” seconds, is given by š‘Ž = (39 āˆ’ 3š‘”) cm/sĀ², 0 ā‰¤ š‘” ā‰¤ 13. At time š‘” = 0 the particle is at rest, and when š‘” > 13 it moves with uniform velocity š‘£. Determine the velocity š‘£ and the distance, š‘‘, covered by the particle in the first 23 s of motion.

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Video Transcript

The acceleration of a particle moving in a straight line at time š‘” seconds is given by š‘Ž is equal to 39 minus three š‘” centimeters per second squared, where š‘” is greater than or equal to zero and less than or equal to 13. At time š‘” equals zero, the particle is at rest. And when š‘” is greater than 13, it moves with uniform velocity š‘£. Determine the velocity š‘£ and the distance š‘‘ covered by the particle in the first 23 seconds of motion.

In this question, we are given an expression in terms of time š‘” for the acceleration of a particle. And we are told that the particle accelerates for the first 13 seconds of its motion. It then travels with uniform velocity š‘£, which we need to calculate. We are also asked to calculate the distance š‘‘ covered by the particle in the first 23 seconds of motion. We begin by recalling that we can find an expression for the velocity in terms of š‘” by integrating our expression for the acceleration with respect to š‘”.

In this question, the velocity is equal to the integral of 39 minus three š‘” with respect to š‘”. We can integrate this expression term by term. Integrating the constant 39 with respect to š‘” gives us 39š‘”. And integrating three š‘” gives us three š‘” squared over two. Remembering to add our constant of integration š¶, we have 39š‘” minus three š‘” squared over two plus š¶. We can find the value of the constant using the fact that at time š‘” equals zero, the particle is at rest.

This means that š‘£ is equal to zero when š‘” is equal to zero. And substituting in these values, we have zero is equal to 39 multiplied by zero minus three multiplied by zero squared over two plus š¶. This means that the constant š¶ is also equal to zero. Our expression for the velocity is therefore equal to 39š‘” minus three š‘” squared over two. We want to calculate the velocity after 13 seconds. This is equal to 39 multiplied by 13 minus three multiplied by 13 squared over two. This simplifies to 507 minus 507 over two, which in turn is equal to 507 over two or 253.5. After 13 seconds, the particle stops accelerating and travels with uniform velocity. This means that our value of š‘£ is 253.5 centimeters per second.

The second part of our question asks us to calculate the distance covered by the particle in the first 23 seconds. After clearing some space, we recall that we can find an expression for the displacement š‘  by integrating our expression for the velocity with respect to š‘”. This means that in the first 13 seconds, the displacement will be equal to the integral of 39š‘” minus three š‘” squared over two with respect to š‘”. By adding the limits of zero and 13, we have a definite integral that will help us calculate the distance traveled in the first 13 seconds. Once again, we can integrate term by term. Integrating 39š‘” gives us 39š‘” squared over two. And integrating three š‘” squared over two gives us three š‘” cubed over six.

Next, we need to substitute our upper and lower limits and then find the difference between these two values. We also notice that the second term simplifies to š‘” cubed over two. Substituting in our values, we have the expression shown. The second part is simply equal to zero. And typing the remainder of our expression into our calculator gives us 2197. As this is just the distance covered in the first 13 seconds, we will call this š‘  sub one. And this is equal to 2197 centimeters.

After clearing some space, we can now consider what happens in the next 10 seconds up to a total of 23 seconds. During this period of time, the particle is traveling with uniform velocity. This means that we can use the equation distance is equal to velocity multiplied by time. If we let the distance traveled be š‘  sub two, we have š‘  sub two is equal to 253.5 multiplied by 10, which is equal to 2535. The particle travels a distance of 2535 centimeters from š‘” equals 13 to š‘” equals 23.

We can now calculate the total distance š‘‘ by adding š‘  sub one and š‘  sub two. We need to add 2197 and 2535. This is equal to 4732. The distance traveled by the particle in the first 23 seconds of motion is 4732 centimeters.

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