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## Wednesday, February 3, 2010

### First equation of motion

The first equation of motion is mainly derived from the definition of acceleration.

You would remember that acceleration is the rate of change of velocity with time  and that the equation to calculate it is

acceleration = (change in velocity)/time taken

= (final velocity – initial velocity) /time taken

a  = (vf – vi)/t

Rearranging the equation will give you

vf = vi + a*t

In some books the equation is given as such

v = u + at

If we used the equation above then the symbol v will be the final velocity and the symbol u will be the initial velocity of the object.

Now as we have seen in this post on acceleration if the final velocity is greater than the initial velocity then the acceleration is positive but if the initial velocity is greater than the initial velocity then the acceleration is negative.

Now let us have a look at two examples on the first equation of motion.

Example 1

A boy starts from rest accelerate to a velocity of 10 m/s in a time of 10 s. Determine the acceleration of the boy.

His final velocity v = 10 m/s

His initial velocity u = 0 m/s since he started from rest.

Time taken t = 10 s

Using the equation v = u +at and making  a the subject of formula will give you a = (v-u)/t

Substituting the different values of v, u and t in the equation will give you

a = (v-u )/t

=(10 – 0 ) /10

= 1 m/s2

Example 2

A train travelling at a speed of 100 m/s is decelerated to a speed of 50 m/s in a time of 15 s. Calculate the deceleration that the train is subjected.

final velocity v = 50 m/s

initial velocity u = 100 m/s

time taken t = 15 s

using the equation v = u +at and making a the subject of formula will give you the equation

a = (v-u) /t

Substituting in the equation above the different values of v, u and t

a = (v-u)/t

= (50 –100 )/15

= –50 /15

= - 3.33 m/s2

Since the acceleration is – 3.33 m/s2 then

deceleration  = 3.33 m/s2

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