Lesson Explainer: Kinematic Equations Mathematics

In this explainer, we will learn how to apply the laws of the uniform acceleration motion of a particle in a straight line.

A particle in uniform motion changes its displacement in the time that it moves. The displacement of a particle, Δ𝑠, is the product of the time for which it moves, Δ𝑡, and its velocity, 𝑣. This can be expressed as Δ𝑠=𝑣Δ𝑡.

The initial displacement of the particle may have been zero or nonzero. If the particle has some initial displacement 𝑠i, then the final displacement of the particle, 𝑠f, is given by 𝑠=𝑠+𝑣Δ𝑡.fi

These expressions are equivalent as Δ𝑠=𝑠𝑠.fi

The change in time, Δ𝑡, can be expressed as Δ𝑡=𝑡𝑡,fi where 𝑡i is the initial time and 𝑡f is the final time.

This expression can be rearranged to define the velocity of a particle.

Definition: Velocity of a Particle

The velocity, 𝑣, of a particle is given by 𝑣=Δ𝑠Δ𝑡=𝑠𝑠𝑡𝑡,fifi where 𝑠i is the displacement of the particle at an instant 𝑡i and 𝑠f is the displacement of the particle at an instant 𝑡f.

The velocity of a particle that accelerates uniformly changes uniformly as the particle moves. The change of the velocity of the particle is the product of the time for which it moves and its acceleration. This can be expressed as 𝑣=𝑢+𝑎Δ𝑡, where 𝑢 is the initial velocity of the particle, 𝑣 is the final velocity of the particle, 𝑎 is the acceleration of the particle, and Δ𝑡 is the time interval over which the particle moves.

The change in time, Δ𝑡, can be expressed as Δ𝑡=𝑡𝑡,fi where 𝑡 is the initial time and 𝑡f is the final time.

This allows us to the write the expression 𝑣=𝑢+𝑎(𝑡𝑡).fi

This expression can be rearranged to define the acceleration of a particle.

Definition: Acceleration of a Particle

The acceleration, 𝑎, of a particle is given by 𝑎=Δ𝑣Δ𝑡=𝑣𝑢𝑡𝑡,fi where 𝑢 is the velocity of the particle at a time 𝑡i and 𝑣 is the velocity of the particle at 𝑡f.

If we assume that 𝑡i is zero, we see that 𝑡𝑡=𝑡0=𝑡.fiff

We can denote 𝑡𝑡fi as 𝑡. Doing this allows us to express acceleration as 𝑎=𝑣𝑢𝑡, and to express velocity as 𝑣=𝑢+𝑎𝑡.

Let us now look at an example in which the velocity of a uniformly accelerating particle is determined.

Example 1: Finding the Final Velocity of a Uniformly Accelerating Particle

If a particle started moving in a straight line with an initial velocity of 25.1 cm/s and a uniform acceleration of 2.4 cm/s2, determine its velocity after 9 seconds.

Answer

The velocity of the particle after it accelerates can be determined by the formula 𝑣=𝑢+𝑎𝑡.

We can substitute the values given for 𝑢, 𝑎, and 𝑡 into the formula.

We then find that 𝑣=25.1+2.4(9)=46.7/.cms

The displacement of a particle is the product of its velocity and the time for which it moves. The velocity of a particle depends on its acceleration and the time for which it accelerates, so the displacement of a particle that is initially at rest can be expressed in terms of its acceleration and the time that it accelerates for.

For a particle with a constant velocity, its displacement can be expressed as 𝑠=𝑠+𝑣𝑡.fi

If we assume that 𝑠i is zero and denote 𝑠f as 𝑠, this becomes 𝑠=𝑣𝑡.

For a uniformly accelerating particle that is initially at rest and has a final velocity 𝑣, the mean of the initial and final velocities is given by 𝑣=𝑣2.mean

The displacement of the particle in a time interval 𝑡 is then given by 𝑠=𝑣2𝑡.

For a particle initially at rest, it is the case that 𝑣=𝑎𝑡.

Substituting this expression for 𝑣 into the expression for 𝑠, we obtain 𝑠=𝑎𝑡2𝑡,𝑠=12𝑎𝑡.

For a uniformly accelerating particle that initially has a velocity 𝑢 and has a final velocity 𝑣, the mean of the initial and final velocities is given by 𝑣=𝑣+𝑢2.mean

The displacement of the particle in a time interval 𝑡 is given by 𝑠=𝑣+𝑢2𝑡.

This expression can also be shown using a graph, as in the following figure.

The graph shows that the area under the blue line consists of the sum of the area of a rectangle, 𝑠, that is given by 𝑠=𝑢𝑡 and the area of a right triangle, 𝑠, that is given by 𝑠=(𝑣𝑢)2𝑡.

We can denote the velocity at 𝑡2 as 𝑣. The following figure shows that the area of the blue-shaded right triangle of side length 𝑣𝑢2 equals the area of the white-shaded right triangle of side length 𝑣𝑢2.

The area below the blue line is, therefore, equal to that of the rectangle shown in the following figure.

This area is equal to the displacement of the particle at the instant 𝑡, given by 𝑠=𝑣+𝑢2𝑡.

For a particle with an initial velocity 𝑢, it is the case that 𝑣=𝑢+𝑎𝑡.

Substituting this expression for 𝑣 into the expression for 𝑠, we obtain 𝑠=𝑢+𝑎𝑡+𝑢2𝑡,𝑠=2𝑢+𝑎𝑡2𝑡,𝑠=𝑢𝑡+12𝑎𝑡.

Let us now look at an example in which the displacement of an accelerating body is determined.

Example 2: Finding the Distance Traveled by a Uniformly Accelerating Particle

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.

Answer

The ball is moving in a straight line, accelerating uniformly. The displacement of a uniformly accelerating body moving in a straight line is given by the formula 𝑠=𝑢𝑡+12𝑎𝑡, where 𝑢 is the initial velocity of the body and 𝑎 is the acceleration of the body.

The ball in the question is stated to be decelerating uniformly. A decelerating body accelerates in the opposite direction to the direction of its velocity when it starts to accelerate. The sign of the acceleration therefore has the opposite sign to that of the initial velocity.

Substituting the values given in the question, we have 𝑠=16.3(2)+12(3)2𝑠=32.6+12(3)(4)=32.66=26.6.m

Let us look at another such example.

Example 3: Calculating the Initial and Final Velocity of a Uniformly Accelerating Particle

A particle, moving in a straight line, was accelerating at a rate of 22 cm/s2 in the same direction as its initial velocity. If the magnitude of its displacement 10 seconds after it started moving was 29 m, calculate the magnitude of its initial velocity 𝑣 and its velocity 𝑣 at the end of this period.

Answer

The displacement of a uniformly accelerating body is given by the formula 𝑠=𝑢𝑡+12𝑎𝑡, where 𝑢 is the initial velocity of the body and 𝑎 is the acceleration of the body. In this question, 𝑢 is denoted by 𝑣.

We are given the displacement of the body, its acceleration, and the time that it accelerates for. Substituting these values into the formula, we have 29=10𝑣+120.2210.

The displacement is given in metres, so the acceleration is converted from 22 cm/s2 to 0.22 m/s2.

This expression can be rearranged to make 𝑣 the subject: 𝑣=290.221010=1.8/.ms

The final velocity, 𝑣, of the body is given by 𝑣=𝑣+𝑎𝑡.

Substituting known values, we find that 𝑣=1.8+0.22(10)=4/.ms

If the time that a body moves for is not known, but the displacement and the initial velocity of the body are known, then the final velocity can be determined. Equivalently, if the displacement and the final velocity of the body are known, then the initial velocity can be determined.

The relation between the initial and final velocity where the time is unknown involves the kinematic formulas 𝑣=𝑢+𝑎𝑡 and 𝑠=𝑢𝑡+12𝑎𝑡.

The formula 𝑣=𝑢+𝑎𝑡 can be rearranged to express 𝑡 in terms of velocity and acceleration: 𝑡=𝑣𝑢𝑎.

This expression for 𝑡 can be substituted into 𝑠=𝑢𝑡+12𝑎𝑡.

This gives us 𝑠=𝑢𝑣𝑢𝑎+12𝑎𝑣𝑢𝑎.

This expression can be rearranged as follows: 𝑠=𝑢𝑣𝑢𝑎+12𝑎𝑣𝑢𝑎𝑣𝑢𝑎𝑠=𝑢𝑣𝑢𝑎+12𝑎𝑣+𝑢2𝑢𝑣𝑎𝑠=𝑢𝑣𝑢𝑎+𝑣+𝑢2𝑢𝑣2𝑎𝑠=2𝑢𝑣2𝑢2𝑎+𝑣+𝑢2𝑢𝑣2𝑎𝑠=2𝑢𝑣2𝑢+𝑣+𝑢2𝑢𝑣2𝑎,2𝑎𝑠=2𝑢𝑣2𝑢+𝑣+𝑢2𝑢𝑣2𝑎𝑠=𝑣𝑢𝑣=𝑢+2𝑎𝑠.

Let us now look at an example of how to model the motion of a particle that occurs in an unknown time interval.

Example 4: Finding the Final Velocity of a Uniformly Accelerating Particle

A particle was moving in a straight line with a constant acceleration of 2 cm/s2. Given that its initial velocity was 60 cm/s, find the velocity of the body to the nearest centimetre per second when it was 15 m from the starting point.

Answer

As the time for which the particle moved is not known, the velocity of the particle is determined using the formula 𝑣=𝑢+2𝑎𝑠, where 𝑣 is the final velocity, 𝑢 is the initial velocity, 𝑎 is the acceleration, and 𝑠 is the displacement.

The velocity and acceleration are given in centimetres per second and centimetres per second squared, respectively, so the displacement is converted from 15 metres to 1‎ ‎500 centimetres. Substituting the known values into the formula, we obtain 𝑣=60+2(2)(1500)𝑣=9600.

To the nearest centimetre per second, 𝑣 is 98 cm/s.

Let us now look at an example of the motion of a particle that requires analyzing its motion in two separate time intervals.

Example 5: Using Kinematic Equations to Solve a Multistep Problem

A body was uniformly accelerating in a straight line such that it covered 72 m in the first 3 seconds and 52 m in the next 4 seconds. Find its acceleration 𝑎 and its initial velocity 𝑣.

Answer

The distance covered by the particle in the first 3 seconds is greater than in the subsequent 4 seconds; hence, the particle is decelerating from an unknown initial velocity.

The mean velocity of a uniformly accelerating particle in a time interval is given by 𝑣=𝑣+𝑢2,mean where 𝑢 is the initial velocity and 𝑣 is the final velocity. The mean velocity is also given by 𝑣=𝑠𝑡,mean where 𝑠 is the displacement and 𝑡 is the time interval length. We have, therefore, that 𝑠𝑡=𝑣+𝑢2.

This expression can be rearranged to give 𝑣+𝑢=2𝑠𝑡.

In the first 3 seconds that the particle moves, we see that 𝑣+𝑢=2(72)3=48.

In the next 4 seconds that the particle moves, we see that 𝑣+𝑢=2(52)4=26.

The initial velocity 𝑢 equals the final velocity 𝑣; hence, we see that 𝑣+𝑢=𝑣+𝑣, and 𝑣+𝑣=𝑣+48𝑢.

We have, therefore, that 𝑣+48𝑢=26𝑣𝑢=22.

The final velocity, after 7 seconds of acceleration, is 22 m/s less than the initial velocity. The acceleration of the particle in the direction of its initial velocity is therefore given by 𝑎=227/.ms

The initial velocity, 𝑢, can now be determined using the formula 𝑠=𝑢𝑡+12𝑎𝑡.

Substituting known values, we obtain 72+52=7𝑢+122277124=7𝑢2214(49)124=7𝑢77𝑢=124+777=2017/.ms

Another way in which this question can be solved is by using simultaneous equations.

We can use the formula 𝑠=𝑢𝑡+12𝑎𝑡 for the first 3 seconds of motion, obtaining 72=3𝑢+𝑎2372=3𝑢+92𝑎.

We can use the same formula for the subsequent 4 seconds of motion, obtaining 124=7𝑢+𝑎27124=7𝑢+492𝑎.

Both of these equations contain two unknowns. To eliminate one of the unknowns, we can multiply one of the equations by a factor to make the coefficient for that unknown equal to the coefficient of the unknown in the other equation.

We multiply 72=3𝑢+92𝑎 by 73, obtaining 168=7𝑢+212𝑎.

We can now subtract the equation 124=7𝑢+492𝑎 from the equation 168=7𝑢+212𝑎.

This gives us 44=212492𝑎𝑎=44𝑎=4414𝑎=227/.ms

This allows 𝑢 to be found by substitution, the same way as in the first method of solving the question.

Let us now summarize what we have learned in these examples.

Key Points

  • The velocity, 𝑣, of a particle is given by 𝑣=𝑠𝑠𝑡𝑡,fifi where 𝑠i is the displacement of the particle at a time 𝑡i and 𝑠f is the displacement of the particle at 𝑡f.
  • The acceleration, 𝑎, of a particle is given by 𝑎=𝑣𝑢𝑡𝑡,fi where 𝑢 is the velocity of the particle at a time 𝑡i and 𝑣 is the velocity of the particle at 𝑡f.
  • The displacement of a particle can be expressed in terms of acceleration and time by the formula 𝑠=𝑢𝑡+12𝑎𝑡, where 𝑠 is the displacement of the particle, 𝑎 is the acceleration of the particle, 𝑢 is the initial velocity of the particle, and 𝑡 is the time for which the particle accelerates.
  • The velocity of a particle before and after acceleration can be expressed in terms of acceleration and displacement by the formula 𝑣=𝑢+2𝑎𝑠, where 𝑢 is the initial velocity of the particle, 𝑣 is the final velocity of the particle, 𝑎 is the acceleration of the particle, and 𝑠 is the displacement of the particle.
  • If the acceleration of a body is in the opposite direction to the direction of its initial velocity, then the acceleration and initial velocity have opposite signs.

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