Question Video: Finding the Displacement, Velocity, and Speed of a Particle | Nagwa Question Video: Finding the Displacement, Velocity, and Speed of a Particle | Nagwa

Question Video: Finding the Displacement, Velocity, and Speed of a Particle Mathematics • Third Year of Secondary School

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A particle moves in a straight line with respect to a stationary point with position vector 𝐫 = (−2𝑡² + 5𝑡 + 3)𝐢, where 𝑡 ≥ 0 and is measured in seconds. 𝐢 is a unit vector parallel to the straight line, and 𝐫 is measured in meters. Find the magnitude of the displacement after 2s. Find the total distance covered by the particle after 2s. Determine the magnitude of the average velocity vector of the particle between 𝑡 = 0 s and 𝑡 = 2 s. Determine the average speed of the particle between 𝑡 = 0 s and 𝑡 = 2 s.

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Video Transcript

A particle moves in a straight line with respect to a stationary point with position vector 𝐫 equal to negative two 𝑡 squared plus five 𝑡 plus three 𝐢, where 𝑡 is greater than or equal to zero and is measured in seconds. 𝐢 is a unit vector parallel to the straight line and 𝐫 is measured in meters. Find the magnitude of the displacement after two seconds. Find the total distance covered by the particle after two seconds. Determine the magnitude of the average velocity vector of the particle between 𝑡 equals zero seconds and 𝑡 equals two seconds. And determine the average speed of the particle between 𝑡 equals zero seconds and 𝑡 equals two seconds.

There are four parts to this question. We need to calculate the displacement, the total distance covered, the average velocity, and the average speed. And we are told that the particle moves with position vector 𝐫 is equal to negative two 𝑡 squared plus five 𝑡 plus three 𝐢. We begin by finding the magnitude of the displacement after two seconds. We can calculate this by subtracting the position vector at 𝑡 equals zero seconds from the position vector at 𝑡 equals two seconds. The position vector at 𝑡 equals two seconds is equal to negative two multiplied by two squared plus five multiplied by two plus three 𝐢. This simplifies to five 𝐢. The initial position vector when 𝑡 equals zero is equal to negative two multiplied by zero squared plus five multiplied by zero plus three 𝐢. This is equal to three 𝐢.

So our expression for the displacement after two seconds is five 𝐢 minus three 𝐢. This is equal to two 𝐢. And we need to calculate the magnitude of the displacement, which is therefore equal to two meters.

The second part of the question asks us to calculate the total distance covered by the particle after two seconds. In this question, the position vector is always colinear to the unit vector 𝐢, which means that the particle has a rectilinear motion. This means that the distance traveled between two points is given by the magnitude of the displacement vector if there is no change in direction. As the direction of motion is given by the velocity, we need to find the velocity function and then study its sign between 𝑡 equals zero and 𝑡 equals two seconds to know if there has been a change of direction between these two times.

The velocity vector is the first derivative of the position vector. Differentiating negative two 𝑡 squared plus five 𝑡 plus three with respect to 𝑡 gives us negative four 𝑡 plus five, so the velocity vector is equal to negative four 𝑡 plus five 𝐢. This means that the component of the velocity vector along the motion axis is given by the linear equation 𝐯 is equal to negative four 𝑡 plus five.

Setting 𝑣 equals zero, we can find the time at which it changes sign. This occurs at one and a quarter, or 1.25, seconds. Since the line 𝑣 is equal to negative four 𝑡 plus five has negative slope, we know that the velocity is positive when 𝑡 is greater than zero and less than 1.25. And the velocity is negative when 𝑡 is greater than 1.25 and less than two. This means that the particle moves in the positive direction and then changes direction at 𝑡 equals 1.25 seconds and then moves in the negative direction. We will call the distance covered in the first part of the journey 𝑑 sub one and the distance covered in the second part of the journey 𝑑 sub two.

The distance covered between 𝑡 equals zero and 𝑡 equals 1.25 seconds is given by the magnitude of the displacement vector between the position at 𝑡 equals zero and that at 𝑡 equals 1.25 seconds. Substituting 𝑡 equals 1.25 and 𝑡 equals zero gives us the following equation. And this simplifies to the absolute value of 3.125𝐢. 𝑑 sub one is therefore equal to 3.125 meters. We can then repeat this process for 𝑑 sub two as shown. This is equal to the absolute value of 1.125𝐢 such that 𝑑 sub two is equal to 1.125 meters. We are now in a position to calculate the total distance. We need to add 3.125 and 1.125. This is equal to 4.25 meters. The total distance covered by the particle after two seconds is 4.25 meters.

The third part of this question asks us to calculate the magnitude of the average velocity vector of the particle between 𝑡 equals zero and 𝑡 equals two seconds. We recall that the average velocity is the average rate of change of position with respect to time and satisfies the following equation. Since the final displacement vector at 𝑡 equals two seconds is equal to two 𝐢, the average velocity vector is equal to two 𝐢 divided by two, and this is equal to 𝐢 meters per second. The magnitude of this vector is equal to one, which means that the answer to the third part of this question is one meter per second.

The final part of this question asks us to calculate the average speed of the particle between 𝑡 equals zero and 𝑡 equals two seconds. We know that the average speed is equal to the total distance covered divided by the total time taken. In this question, we need to divide 4.25 meters by two seconds. Dividing 4.25 by two gives us 2.125, and we can therefore conclude that the average speed of the particle from 𝑡 equals zero to 𝑡 equals two seconds is 2.125 meters per second.

We now have answers to all four parts of this question. They are two meters, 4.25 meters, one meter per second, and 2.125 meters per second.

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