The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Vectors in Kinematics Mathematics

In this explainer, we will learn how to solve kinematic problems using the vector equations of motion.

We start by recalling that a vector is a mathematical object that has both magnitude (size) and direction; this is in contrast to a scalar, which has a size but no direction. For example, we can have scalar quantities such as a distance of 4 m or a speed of 5 m/s and vector quantities such as a displacement of 4 m east or a velocity of 5 m/s at 30∘ above the horizontal. Geometrically, we can represent a vector as a directed line segment, where the length of the line segment denotes the magnitude and the orientation of the arrow shows the direction.

Throughout this explainer, we will be working with vectors in the two-dimensional plane. In general, the actual location of a vector within the plane is unimportant, because a vector represents a translation rather than an object with a fixed position. If, however, the initial point is specified (commonly, this point is the origin), then this β€œanchors” the vector in the plane as a translation starting from that particular point.

In the two-dimensional plane, we can describe a vector in terms of its change in position (displacement) relative to the π‘₯- and 𝑦-axes. Recall that we use ⃑𝑖 and ⃑𝑗 to denote the vectors of length 1 (known as the β€œunit vectors”) in the positive π‘₯- and 𝑦-directions, respectively; so ⃑𝑖=ο€Ό10 and ⃑𝑗=ο€Ό01.

Using this notation, we can write any two-dimensional vector as 𝑝⃑𝑖+π‘žβƒ‘π‘—=ο€Όπ‘π‘žοˆ, where 𝑝 and π‘ž are numbers. For example, the diagram below shows the vector 3⃑𝑖+5⃑𝑗=ο€Ό35. In this context, ⃑𝑖-⃑𝑗 form and column vector form can be used interchangeably. We might also meet the equivalent form (𝑝,π‘ž), which in this case would be (3,5).

Two-dimensional vectors are especially useful when we want to describe the motion of objects in a plane. We model the objects as particles. Then, the displacement, velocity, and acceleration (all of which have both size and direction) are modeled as vector quantities, while time (which has a size but no direction) is a scalar quantity. Note that the standard units of displacement are metres (m), the standard units of velocity are metres per second (m/s), and the standard units of acceleration are metres per second squared (m/s2).

Having set up our model in this way, we can then analyze the motion of objects in a plane by applying vector forms of the standard equations of motion (otherwise known as the kinematic equations or SUVAT equations).

We start with the vector equation for the displacement of a particle moving in the plane with constant velocity. Note that this is the same as saying that the particle is moving in a straight line at a constant speed.

Equation: Displacement of a Particle Moving with Constant Velocity

If a particle starts from the point with position vector βƒ‘π‘ŸοŠ¦ and moves with constant velocity ⃑𝑣, then at time 𝑑 its displacement from its initial position is ⃑𝑣𝑑 and its position vector βƒ‘π‘Ÿ is given by the equation βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+⃑𝑣𝑑.

Note that, in general, the displacement of a particle from its initial position is given by ⃑𝑠=βƒ‘π‘Ÿβˆ’βƒ‘π‘ŸοŠ¦. From the above equation, we have βƒ‘π‘Ÿβˆ’βƒ‘π‘Ÿ=βƒ‘π‘£π‘‘οŠ¦, which is why the displacement element of the above equation is given by ⃑𝑠=⃑𝑣𝑑. This displacement can be thought of as the vector version of the standard kinematic equation 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ in the case where the velocity is constant (so the acceleration is zero).

We can apply this equation in several ways, depending on the information we are given in a question. If we know a particle’s starting position βƒ‘π‘ŸοŠ¦ and velocity ⃑𝑣 together with a value of the time 𝑑, then we can calculate βƒ‘π‘Ÿ, the position of the particle at time 𝑑. We can also work backward from the equation to find the time 𝑑 at which the particle is at a specified position.

In general, we assume that ⃑𝑖 and ⃑𝑗 are the unit vectors due east and north respectively. Therefore, if we are told that a particle is due east of the origin, it will have position vector 𝑝⃑𝑖 with 𝑝>0, and if it is due west, it will have position vector 𝑝⃑𝑖 with 𝑝<0. If the particle is due north of the origin, it will have position vector π‘žβƒ‘π‘— with π‘ž>0, and if it is due south, it will have position vector π‘žβƒ‘π‘— with π‘ž<0. The same relationships hold for the velocity of the particle; so, for example, if the particle is traveling due east with a constant velocity, then its velocity will be given by 𝑐⃑𝑖 with 𝑐>0, and if it is traveling due north with a constant velocity, then its velocity will be given by 𝑑⃑𝑗 with 𝑑>0.

Let us now look at an example where we can use the above equation.

Example 1: Finding the Position Vector and Time Using Vector Kinematics

A particle starts from a point with position vector ο€Ί5⃑𝑖+8⃑𝑗 m and travels with a constant velocity of ο€Ί3βƒ‘π‘–βˆ’βƒ‘π‘—ο† m/s.

  1. Find the position vector of the particle 2 seconds later.
  2. Find the time at which the particle is due east of the origin.

Answer

Recall that if a particle starts from the point with position vector βƒ‘π‘ŸοŠ¦ and moves with constant velocity ⃑𝑣, then at time 𝑑 its displacement from its initial position is ⃑𝑣𝑑 and its position vector βƒ‘π‘Ÿ is given by the equation βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+⃑𝑣𝑑.

In this question, we are given the vectors for a particle’s starting point βƒ‘π‘Ÿ=ο€Ί5⃑𝑖+8βƒ‘π‘—ο†οŠ¦m and its constant velocity ⃑𝑣=ο€Ί3βƒ‘π‘–βˆ’βƒ‘π‘—ο†/ms.

Part 1

We need to find the position of the particle 2 seconds after the start. To do this, we substitute 𝑑=2 into the above equation, along with the given values for βƒ‘π‘ŸοŠ¦ and ⃑𝑣. Hence, using vector addition, we get βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+⃑𝑣𝑑=ο€Ί5⃑𝑖+8⃑𝑗+2ο€Ί3βƒ‘π‘–βˆ’βƒ‘π‘—ο†=ο€Ί5⃑𝑖+8⃑𝑗+ο€Ί6βƒ‘π‘–βˆ’2⃑𝑗=ο€Ί11⃑𝑖+6⃑𝑗.m

Part 2

We need to find the time at which the particle is due east of the origin. At this time, the position vector of the particle will have a ⃑𝑗-component of zero and so will be of the form 𝑝⃑𝑖 for some number 𝑝>0. Applying the equation again and then writing the ⃑𝑖- and ⃑𝑗-components separately, we get βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+⃑𝑣𝑑=ο€Ί5⃑𝑖+8⃑𝑗+ο€Ί3βƒ‘π‘–βˆ’βƒ‘π‘—ο†π‘‘=5⃑𝑖+8⃑𝑗+3π‘‘βƒ‘π‘–βˆ’π‘‘βƒ‘π‘—=(5+3𝑑)⃑𝑖+(8βˆ’π‘‘)⃑𝑗.

Setting the coefficient of ⃑𝑗 equal to 0 gives the equation 8βˆ’π‘‘=0, which we can solve for 𝑑. Thus, we get 8βˆ’π‘‘=08=𝑑.

So, we conclude that the particle is due east of the origin after 8 seconds.

Notice that although we were not asked to work out the actual position vector of the particle when it is due east of the origin, we could do this by substituting 𝑑=8 back into the equation we obtained for βƒ‘π‘Ÿ; so, βƒ‘π‘Ÿ=(5+3Γ—8)⃑𝑖=29⃑𝑖m.

Next, we can solve problems that involve particles traveling with constant acceleration by applying the following two vector equations.

Equation: Velocity and Displacement of a Particle Moving with Constant Acceleration

For a particle moving in the plane with constant acceleration, we have the equations ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘βƒ‘π‘ =⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘,and where ⃑𝑒 is the particle’s initial velocity, βƒ‘π‘Ž is the acceleration, ⃑𝑣 is the velocity at time 𝑑, and ⃑𝑠 is the displacement at time 𝑑.

Note that the above equations are simply the vector versions of the standard kinematic equations 𝑣=𝑒+π‘Žπ‘‘ and 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨. Again, if we know a particle’s initial velocity ⃑𝑒 and constant acceleration βƒ‘π‘Ž together with a value of the time 𝑑, then we can use the first equation to calculate ⃑𝑣, the velocity of the particle at time 𝑑, and the second equation to calculate ⃑𝑠, the displacement of the particle at time 𝑑. We can also work backward from the equations to find a missing time, acceleration, or initial velocity, depending on the information we are given in a question.

Our next example will help us test our understanding of these two equations.

Example 2: Finding the Acceleration, Displacement, and Time Using Vector Kinematics

An ice skater skates on a flat ice rink. At time 𝑑=0, the skater is at a fixed point 𝑂 and is traveling with velocity ο€Ί3.6βƒ‘π‘–βˆ’0.8⃑𝑗 m/s. At time 𝑑=10, the skater is traveling with velocity ο€Ίβˆ’8.4⃑𝑖+3.2⃑𝑗 m/s. The ice skater has a position vector ⃑𝑠 at time 𝑑 seconds, relative to 𝑂.

By modeling the ice skater as a particle with constant acceleration, answer the following.

  1. Work out the acceleration of the ice skater.
  2. Work out an expression for ⃑𝑠 in terms of 𝑑.
  3. Work out the time at which the ice skater is directly northeast of 𝑂.

Answer

Recall that for a particle moving in the plane with constant acceleration, we have the equations ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘βƒ‘π‘ =⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘,and where ⃑𝑒 is the initial velocity, βƒ‘π‘Ž is the acceleration, ⃑𝑣 is the velocity at time 𝑑, and ⃑𝑠 is the displacement at time 𝑑.

Here, we are given the vectors for the ice skater’s initial velocity, ⃑𝑒=ο€Ί3.6βƒ‘π‘–βˆ’0.8⃑𝑗/ms, and their velocity after 10 seconds, ⃑𝑣=ο€Ίβˆ’8.4⃑𝑖+3.2⃑𝑗/ms.

Part 1

We need to find the ice skater’s acceleration; so, we substitute for ⃑𝑒, ⃑𝑣, and 𝑑 in the equation ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘ to get ο€Ίβˆ’8.4⃑𝑖+3.2⃑𝑗=ο€Ί3.6βƒ‘π‘–βˆ’0.8⃑𝑗+10βƒ‘π‘Ž.

To simplify this equation, we subtract 3.6⃑𝑖 from both sides, add 0.8⃑𝑗 to both sides, and then collect the ⃑𝑖- and ⃑𝑗-components separately on the left-hand side, which gives βˆ’8.4⃑𝑖+3.2βƒ‘π‘—βˆ’3.6⃑𝑖+0.8⃑𝑗=10βƒ‘π‘Žβˆ’8.4βƒ‘π‘–βˆ’3.6⃑𝑖+3.2⃑𝑗+0.8⃑𝑗=10βƒ‘π‘Žβˆ’12⃑𝑖+4⃑𝑗=10βƒ‘π‘Ž.

Lastly, we divide through by 10 to get βƒ‘π‘Ž=ο€Ίβˆ’1.2⃑𝑖+0.4⃑𝑗/ms.

Part 2

We need to find an expression for ⃑𝑠, the skater’s position vector relative to 𝑂 at time 𝑑 seconds. Observe that since the skater started at the fixed point 𝑂, finding an expression for ⃑𝑠 in terms of 𝑑 is equivalent to finding an expression for the skater’s displacement at time 𝑑 seconds. Therefore, we need the equation ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘οŠ¨. We then substitute the value of ⃑𝑒 given in the question and the value of βƒ‘π‘Ž that we obtained in part 1, as follows: ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘=ο€Ί3.6βƒ‘π‘–βˆ’0.8⃑𝑗𝑑+12ο€Ίβˆ’1.2⃑𝑖+0.4⃑𝑗𝑑=ο€Ί3.6βƒ‘π‘–βˆ’0.8⃑𝑗𝑑+ο€Ίβˆ’0.6⃑𝑖+0.2⃑𝑗𝑑.

Our final step is to collect the ⃑𝑖- and ⃑𝑗-components separately on the right-hand side, from which we deduce that the required expression for ⃑𝑠 in terms of 𝑑 is ⃑𝑠=ο€Ίο€Ή3.6π‘‘βˆ’0.6𝑑⃑𝑖+ο€Ή0.2π‘‘βˆ’0.8𝑑⃑𝑗.m

Part 3

We need to find the time at which the ice skater is directly northeast of 𝑂. It helps to draw a sketch of the situation, looking down from above at the vector representing the skater’s position at that particular time.

We can see that when the skater is directly northeast of 𝑂, their position vector is at 45∘ to both north and east, meaning that the ⃑𝑖- and ⃑𝑗-components must be equal. Therefore, we take the expression for ⃑𝑠 that we obtained in part 2, set the two components equal, and solve for 𝑑: 3.6π‘‘βˆ’0.6𝑑=0.2π‘‘βˆ’0.8𝑑0=0.2𝑑+0.6π‘‘βˆ’0.8π‘‘βˆ’3.6𝑑0=0.8π‘‘βˆ’4.4𝑑0=0.4𝑑(2π‘‘βˆ’11).

The above factored quadratic equation implies that either 0.4𝑑=0 or 2π‘‘βˆ’11=0; so, 𝑑=0 or 𝑑=5.5. We can rule out the case 𝑑=0 because this time corresponds to when the skater is at 𝑂. Thus, we conclude that the ice skater is directly northeast of 𝑂 after 5.5 seconds.

One advantage of representing a mathematical quantity in unit vector form is that it is straightforward to calculate the magnitude of the vector, which is a scalar, by applying the Pythagorean theorem. In particular, if we have a particle traveling with a constant velocity, which is the same as saying that it is traveling in a straight line at a constant speed, then the speed of the particle is the magnitude of the velocity vector. An example of calculating the speed from the velocity vector is shown below.

We can also use the same principle to calculate the distance between two points, provided we know their respective position vectors. We begin by calculating the displacement from the initial point to the end point, which we obtain by subtracting the position vector of the initial point from the position vector of the end point. Then, we apply the Pythagorean theorem to work out the magnitude of this displacement vector, which gives the corresponding distance between the two points. We show an example in the diagram below.

Here, we have point 𝐴 with position vector 2⃑𝑖+⃑𝑗 and point 𝐡 with position vector 4⃑𝑖+5⃑𝑗. The displacement from 𝐴 to 𝐡 is given by ο€Ί4⃑𝑖+5βƒ‘π‘—ο†βˆ’ο€Ί2⃑𝑖+⃑𝑗=ο€Ί2⃑𝑖+4⃑𝑗.m

Then, applying the Pythagorean theorem to work out the magnitude of this displacement vector, we get √2+4=√20=4.472….οŠͺ

So, the distance between 𝐴 and 𝐡 is 4.47 m to 2 decimal places.

Note that in the particular case where we have a particle traveling between two points at a constant velocity ⃑𝑣 for a time period of 𝑑, we can calculate the displacement by rearranging the equation βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+βƒ‘π‘£π‘‘οŠ¦ into the form ⃑𝑣𝑑=βƒ‘π‘Ÿβˆ’βƒ‘π‘ŸοŠ¦. The vector ⃑𝑣𝑑 represents the displacement from the initial position βƒ‘π‘ŸοŠ¦ to the end position βƒ‘π‘Ÿ. So, we can find it by substituting for βƒ‘π‘ŸοŠ¦ and βƒ‘π‘Ÿ. Recall again that a particle traveling at a constant velocity must be traveling in a straight line at a constant speed. Therefore, the magnitude of this displacement vector, as well as being the distance between the two points, will be the distance traveled by the particle.

We will find some of the above ideas useful when tackling the next example.

Example 3: Finding a Missing Time Using Vector Kinematics

A radio-controlled car starts from a point with a position vector ο€Ίβˆ’6⃑𝑖+⃑𝑗 m relative to a fixed origin. It travels in a straight line toward the point with the position vector ο€Ί9⃑𝑖+9⃑𝑗 m with a constant speed of 5 m/s. Work out the time taken for the car to travel to the point with the position vector ο€Ί9⃑𝑖+9⃑𝑗 m.

Answer

Recall that to work out the distance between two points from their position vectors, we first calculate the displacement from the initial point to the end point by subtracting the position vector of the initial point from the position vector of the end point. Then, we apply the Pythagorean theorem to work out the magnitude of this displacement vector, which is the corresponding distance between the two points.

In the particular case where a particle travels between two points at a constant velocity ⃑𝑣 for a time period of 𝑑, we can find the displacement vector ⃑𝑣𝑑 by rearranging the equation βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+βƒ‘π‘£π‘‘οŠ¦ into the form ⃑𝑣𝑑=βƒ‘π‘Ÿβˆ’βƒ‘π‘ŸοŠ¦ and substituting for the initial position vector βƒ‘π‘ŸοŠ¦ and the end position vector βƒ‘π‘Ÿ. The magnitude of this displacement vector, as well as being the distance between the two points, will be the distance traveled by the particle.

In this question, we are told that a radio-controlled car travels in a straight line with a constant speed of 5 m/s. This is the same as saying that it has a constant velocity, which we can label as ⃑𝑣. The car travels from the point with position vector βƒ‘π‘Ÿ=ο€Ίβˆ’6⃑𝑖+βƒ‘π‘—ο†οŠ¦m to the point with position vector βƒ‘π‘Ÿ=ο€Ί9⃑𝑖+9⃑𝑗m. Therefore, the displacement from the initial point to the end point is given by the vector ⃑𝑣𝑑=βƒ‘π‘Ÿβˆ’βƒ‘π‘Ÿ=ο€Ί9⃑𝑖+9βƒ‘π‘—ο†βˆ’ο€Ίβˆ’6⃑𝑖+⃑𝑗=9⃑𝑖+9⃑𝑗+6βƒ‘π‘–βˆ’βƒ‘π‘—=15⃑𝑖+8⃑𝑗.

We sketch this vector below.

As the car moves in a straight line, the distance it travels is the magnitude of this displacement vector. So, we apply the Pythagorean theorem to get √15+8=√289=17.m

Finally, since we now know that the car travels 17 m in a straight line at a constant speed of 5 m/s, we can calculate the time taken by rearranging the equation speeddistancetime= into the form timedistancespeed=. This gives the time taken as 175=3.4seconds.

Remember that to represent vectors, we can use ⃑𝑖-⃑𝑗 form and column vector form interchangeably. Let us now look at an example expressed in column vector form.

Example 4: Finding the Velocity and Distance Using Vector Kinematics

An airplane travels with a constant acceleration of ο€Ό1.5βˆ’1 m/s2. When 𝑑=0, the velocity of the airplane is ο€Όβˆ’4030 m/s.

  1. Find the velocity of the airplane after 10 s.
  2. Find the distance of the airplane from its starting point after 10 s to the nearest metre.

Answer

Recall that for a particle moving in the plane with a constant acceleration, we have the equations ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘βƒ‘π‘ =⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘,and where ⃑𝑒 is the initial velocity, βƒ‘π‘Ž is the acceleration, ⃑𝑣 is the velocity at time 𝑑, and ⃑𝑠 is the displacement at time 𝑑.

In this question, we are given the column vectors for an airplane’s initial velocity ⃑𝑒=ο€Όβˆ’4030/ms and its constant acceleration βƒ‘π‘Ž=ο€Ό1.5βˆ’1/ms.

Part 1

We need to find the velocity of the airplane after 10 seconds. To do this, we substitute 𝑑=10 into the first of the above equations, together with the given values for ⃑𝑒 and βƒ‘π‘Ž. Thus, using scalar multiplication and vector addition, we get ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘=ο€Όβˆ’4030+10ο€Ό1.5βˆ’1=ο€Όβˆ’40+1530βˆ’10=ο€Όβˆ’2520/.ms

Notice that we could also represent this velocity in terms of the standard unit vectors as ο€Ίβˆ’25⃑𝑖+20⃑𝑗 m/s.

Part 2

We need to find the distance of the airplane from its starting point after 10 seconds. Recall that the distance is the magnitude of the displacement. Therefore, our strategy will be to first calculate the displacement of the airplane from its starting point after 10 seconds by using the equation ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘οŠ¨. Then, we can apply the Pythagorean theorem on our resulting displacement vector to calculate the corresponding distance, rounding our answer to the nearest metre.

Substituting for ⃑𝑒, βƒ‘π‘Ž, and 𝑑, we get ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘=10ο€Όβˆ’4030+12ο€Ό1.5βˆ’110=ο€Όβˆ’400300+ο€Ό75βˆ’50=ο€Όβˆ’325250.m

Notice that we could also represent this displacement in terms of the standard unit vectors as ο€Ίβˆ’325⃑𝑖+250⃑𝑗 m.

Even though the ⃑𝑖-component of the displacement vector is negative, we can still calculate the magnitude of this vector. This should become apparent if we draw a diagram of the situation.

The negative sign of the ⃑𝑖-component simply tells us that its direction is negative (due west). This means we can apply the Pythagorean theorem as usual, which gives a magnitude of √325+250=√168125=410.030….

Rounding our answer to the nearest meter, we conclude that the distance of the airplane from its starting point after 10 seconds is 410 m.

In our final example, we explore the idea of two particles colliding. Observe that this is equivalent to saying that at a certain time, the particles will have the same position vector; we can use this fact to help us tackle these types of problems.

Example 5: Finding the Time of Collision Using Vector Kinematics

Samar and Nader are skating on a large flat ice rink. At 𝑑=0, Samar is at a fixed point 𝑂 and is traveling with a velocity of ο€Ίβˆ’1.8⃑𝑖+0.4⃑𝑗 m/s. She has a constant acceleration of ο€Ί1.2βƒ‘π‘–βˆ’0.2⃑𝑗 m/s2. Nader travels so that he has the position vector βƒ‘π‘Ÿ=(1.5π‘‘βˆ’4.2)βƒ‘π‘—οŒ²m relative to 𝑂 at time 𝑑.

At time π‘‘οŒΌ, Samar and Nader collide. Find π‘‘οŒΌ.

Answer

Recall that for a particle moving in the plane with constant acceleration, we have the equation ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘, where ⃑𝑒 is the initial velocity, βƒ‘π‘Ž is the acceleration, and ⃑𝑠 is the displacement at time 𝑑.

Here, we are given Samar’s initial velocity ⃑𝑒=ο€Ίβˆ’1.8⃑𝑖+0.4⃑𝑗/ms and constant acceleration βƒ‘π‘Ž=ο€Ί1.2βƒ‘π‘–βˆ’0.2⃑𝑗/ms. We are also given an equation for Nader’s position vector βƒ‘π‘ŸοŒ² as a function of 𝑑. Since we are told that Samar and Nader collide at time π‘‘οŒΌ, then, at this time, they will be at the same position.

Note that as Samar is at 𝑂 at 𝑑=0, then her position vector at time 𝑑, which we can label as βƒ‘π‘ŸοŒ©, will be the same as her displacement vector, ⃑𝑠. Hence, our strategy will be to use the equation ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘οŠ¨ to derive an equation for Samar’s position vector as a function of 𝑑. Once we have done this, we can compare it with βƒ‘π‘ŸοŒ² and work out the value of π‘‘οŒΌ.

To derive an equation for Samar’s position vector βƒ‘π‘ŸοŒ©, we take the equation ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘οŠ¨ and substitute for ⃑𝑠, ⃑𝑒, and βƒ‘π‘Ž, which gives βƒ‘π‘Ÿ=ο€Ίβˆ’1.8⃑𝑖+0.4⃑𝑗𝑑+12ο€Ί1.2βƒ‘π‘–βˆ’0.2⃑𝑗𝑑=ο€Ίβˆ’1.8⃑𝑖+0.4⃑𝑗𝑑+ο€Ί0.6βƒ‘π‘–βˆ’0.1⃑𝑗𝑑=ο€Ήβˆ’1.8𝑑+0.6𝑑⃑𝑖+ο€Ή0.4π‘‘βˆ’0.1𝑑⃑𝑗.

So, βƒ‘π‘Ÿ=ο€Ί0.6𝑑(π‘‘βˆ’3)⃑𝑖+0.1𝑑(4βˆ’π‘‘)βƒ‘π‘—ο†οŒ©m.

Now, note that as Nader’s position vector is given by the equation βƒ‘π‘Ÿ=(1.5π‘‘βˆ’4.2)βƒ‘π‘—οŒ²m, this tells us that the ⃑𝑖-component of his position vector is zero. This is equivalent to saying that at all times, Nader is traveling along the line running north–south through 𝑂.

Consequently, we can find π‘‘οŒΌ, the time at which Samar and Nader collide, by setting the ⃑𝑖-component of Samar’s position vector equal to zero and reading off the corresponding value of 𝑑. Therefore, we must have 0.6𝑑(π‘‘βˆ’3)=0, which implies 𝑑=0 or 𝑑=3. We can rule out the case 𝑑=0 because this is the time when Samar is at 𝑂 at the start. So, we are left with 𝑑=3.

To check that Samar and Nader do indeed collide after 3 seconds, we can substitute this value of 𝑑 into their respective position vector equations and check that we get the same ⃑𝑗-component. For Nader, we have βƒ‘π‘Ÿ=(1.5π‘‘βˆ’4.2)⃑𝑗=(1.5Γ—3βˆ’4.2)⃑𝑗=(4.5βˆ’4.2)⃑𝑗=0.3⃑𝑗.m

For Samar, we have βƒ‘π‘Ÿ=0.1𝑑(4βˆ’π‘‘)⃑𝑗=(0.1Γ—3)(4βˆ’3)⃑𝑗=0.3⃑𝑗.m

Since the two ⃑𝑗-components match, giving a position of 0.3 m due north of 𝑂, we conclude that Samar and Nader collide after 3 seconds. So, 𝑑=3. The sketch below shows their respective positions over the first 3 seconds.

At 𝑑=0, Samar is at 𝑂 and Nader is at the point with position vector (1.5Γ—0βˆ’4.2)⃑𝑗=βˆ’4.2⃑𝑗m. Samar then skates along the curved route shown, while Nader skates due north in a straight line. At 𝑑=𝑑=3, they collide at the point with position vector 0.3⃑𝑗 m.

Let us finish by recapping some key concepts from this explainer.

Key Points

  • If a particle starts from the point with position vector βƒ‘π‘ŸοŠ¦ and moves with constant velocity ⃑𝑣, then, at time 𝑑, its displacement from its initial position is ⃑𝑣𝑑 and its position vector βƒ‘π‘Ÿ is given by the equation βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+⃑𝑣𝑑.
  • For a particle moving in the plane with constant acceleration, we have the equations ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘βƒ‘π‘ =⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘,and where ⃑𝑒 is the particle’s initial velocity, βƒ‘π‘Ž is the acceleration, ⃑𝑣 is the velocity at time 𝑑, and ⃑𝑠 is the displacement at time 𝑑.
  • If a particle travels at a constant velocity, then its speed is the magnitude of its velocity vector.
  • The distance between two points is the magnitude of the displacement vector between the two points. If a particle travels from one point to another at a constant velocity, then the distance it travels is the magnitude of the displacement vector between the two points.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.