EQUATIONS OF MOTIONS

EQUATIONS OF MOTIONS

Equations of motions are formulae that re used in solving motion related problem.

There are for equations of motion. The four equations of motions are:

·                     V = u + at

·                     V² = U² + 2as

·                     S or h = ut + ½at²

·                     S = ((v+u)t) / 2

Definition of variables:

U is the initial velocity of the object. It is measured in meter per second (m/s).

V is the final velocity of the object. It is measured in meter per second (m/s).

A is the acceleration of the object. It is measured in meter per second square(m/s sqr.)

S is the distance travelled by the object. It is measured in metre, (m)

T is the time taken by the object to travel a given distance. It is measured in second (s).

DERIVATION OF THE EQUATIONS OF MOTION

Let us see how the four equations are derived.

DERIVATION OF EQUATION 1

Equation 1 is derived from the formula of acceleration. The derivation is as follows:

From the formula of acceleration,

                  Acceleration = change in velocity / time

                  Acceleration a = (final velocity v - initial velocity u) / time t

Using their symbols, the formula will become:

     

                  A = (v-u) / t

Cross multiply and make v the subject of the formula.

                  A*t = v-u

                  V = u + at which is the equation 1.

DERIVATION OF EQUATION 2

To obtain equation 2, we will use the formula for average velocity and substitute for v = u + at in the formula of average velocity as follows:

                  Average velocity Vaverage = (v+u)/2

At this point, you substitute for v.

Therefore,

                  Vaverage = (v+u)/2

                  Vaverage = (u+at+u)/2   

                  Vaverage = (2u+at)/2 …..       equation 1

At this point, you pen the bracket:

                  Vaverage = u+½at

Also, the formula for average velocity is,

                  Vaverage = distance s / time t

                  Vaverage = s/t  …..               equation 2

You equate the two equations as follows:

                  equation 1 = equation 2           

                  Vaverage  of equation 1 = Vaverage of equation 2

                  u+½at = s/t 

You make s the subject by cross multiplying t.

                  (u+½at)t = s

                  ut+½at*t = s

                  ut+½at² = s   

                  s = ut+½at²  … equation 2

DERIVATION OF EQUATION 3

Equation 3 can be obtained by squaring both sides of equation 1.

Therefore,

                  v² = (u+at)²

                  v² = (u+at)*(u+at)

                  v² = u²+2uat+a²t²

                  v² = u²+2uat+(at)²

At this point, you factorise 2a from 2uat+(at)².

Therefore, you will get,

                  v² = u²+2a(ut+½(at)² …….*

At this point, you have to recall that,

                  s = ut+½at² 

you will have to substitute s for ut+½at² in equation *

therefore,

                  v² = u²+2a(ut+½(at)² …….*

                  v² = u²+2as which gives equation 3.

DERIVATION OF EQUATION 4

Equation 4 is obtained by equating the two formulae of average velocity.

The two formulae for average velocity are:

                  Vaverage = s/t 

                  Vaverage = (v+u)/2

Therefore,

                  Vaverage = Vaverage

s/t = (v+u)/2

you make s the subject of the formula:

                  s = (v+u)t / 2   which gives equation 4.

APPLICATION OF EQUATIONS OF MOTIONS

Applications of equations of motion means using the equations of motions to solve motion related problems. This I will show and explain using work examples as shown below.

Worked examples:

1.  An object thrown vertically upward with an initial velocity of 30m/s has a deceleration of 10m/s sqr. Calculate its velocity after 2s and time to reach maximum height. (g = 10m/s sqr.)

Solution:

                        Data given in the question:

Initial velocity u = 30m/s; time t = 2s; velocity at time 2s = ?; deceleration = 10m/s sqr (deceleration = negative acceleration).

Formula for calculation:

You must know that the formula that you use for calculation is determined by the data that you are given in the question. In other words, the formula that contains the data that you are given in the question is the correct formula for the calculation.

i.             Velocity after 2s.

Formula:

                        V = u + at.

Substitution:

                        V = 30+(-10)*2

                        V = 30 – 20

                        V = 10m/s

ii.            Time taken to reach maximum height.

To calculate time taken to reach maximum height, you must remember that at maximum height, final velocity v = 0m/s.

Formula:

                        V = u + at.

Substitution:

                        0 = 30 + (-10)*t

                        0 = 30 – 10t

Collect like terms and make t the subject of the formula.

                        10t = 30

                        T = 30/10

Time to reach maximum height = 3s       

2.  A car accelerates uniformly at 1m/s sqr for 12s from an velocity of 5m/s. Calculate the distance travelled. ( g = 10m/s sqr.)

Solution:        

i.             Distance travelled:

                        Data given in the question:

                        A = 1m/s sqr.; u = 5m/s; t = 12s; g = 10m/s sqr.   ; s = ?

Note:

You use the formula that contains the data that you are given in the question and the one that you are to calculate.

Formula:

                  s = ut+½at²

substitution:

                  s = 5*12+½*1*(12)² 

                  s = 60+1*144

                  s = 60+145.; s = 204m

3.  A stone is thrown vertically upward from the ground with a velocity of 20m/s.

(g = 10 m/s sqr.). calculate:

i.      The maximum height reached.

j.      The time to reach the maximum height;

k.    The time to reach the ground again;

l.      The velocity reached half way up.

Solutions:

Recall that at maximum height, v = 0m/s

                        Data given in the question:

                        U = 20m/s; v = 0m/s; g = 10m/s sqr.

i.                 Maximum height reached:

Formula:

                        v² = u²+2as

substitution:

                        0² = 20²+2*(-10)*s

                        0 = 400+(-20s)

                        0 = 400 - 20s

Collect like terms and make s the subject of the formula.

                        20s = 400

                        S = 400/20;

            Maximum height reached = 20m.

ii. the time taken to reach maximum height:

           

formula:

                  s = ut+½at²

substitution:

                        20 = 20*t+½*(-10)*t²

                        20 = 20t - 5t²

You rearrange the quadratic equation and find t using any formula that is convenient with you.

Therefore,

                        0 = 20t – 5t² - 20

                        0 = -5t² + 20t – 20

You now factorise or use general formula to find t.

Using general formula:

                        T = (- b ± √(b² - 4ac)) / 2a

Where,

                        A = - 5; b = 20; c -20

Substitution:

                        T = (- 20 ± √(20² - 4*(-5)*(-20)) / 2*(- 5)

                        T = (-20 ± √(20² - 4*(-5)*(-20)) / -10

                        T = (-20 ± √(400 – 400)) / -10

                        T = (- 20 ± √0) / -10

                        T = -20/(-10)

            Time to reach maximum height = 2s

iii. the time to reach the ground again:

The time to reach the ground is the same as the time taken to reach maximum height.

Therefore,

                        Time to reach ground = 2* time taken to reach maximum height.

                        Time to reach ground = 2 * 2s

                        Time to reach ground = 4s.

iv. the velocity reached half way up:

to calculate the velocity of the object half way up, the time taken to travelled half distance is half of the time taken to reached maximum height. Or you use half the distance travelled to calculate its velocity half way up.

Therefore,

            The distance half way up = 20m/2 = 10m

Formula:

                        v² = u²+2as

substitution:

                        v² = 20²+2*(-10)*10

                        v² = 400 - 200

                        v² = 200;         v = √200

            velocity half way = 14.14m/s

4. a train travelling at 54km/h accelerates uniformly at 3m/s sqr. for ¼ minutes. calculate:

            i. the distance travelled in this time.

Solution:

                        Data given in the question:

u = 54km/h = 54*(1000/3600) = 15m/s (v = 15m/s); acc = 3m/s sqr; time = ¼ min = ¼ * 60 = 12sec;

Formula:

                  s = ut+½at²

substitution:

                  s = 15*12+½*3*15²

                  s = 180+½*3*225

                  s = 180+337.5

                  s = 517.5m

***more worked example***