We have seen in a previous post the first equation of motion.

Today we are going to see the second equation of motion.

You would remember that the first equation dealt with acceleration, time taken, initial velocity and final velocity.

So how do we derive the second equation of motion?

Let us take an object that starts from an initial velocity u and then accelerates during a time t until it reaches a final velocity v.

Now during the time the object was accelerating the object has moved a distance s. In the second equation of motion we will try to derive the equation to calculate this distance s.

Fig 2 below shows what is happening to the object.

Fig 1

Now you would remember that the area under a velocity-time graph is the distance travelled by the object. Hence we are going to calculate the area under the graph as indicated by the shaded area below.

Fig 2

Now if s the distance travelled is the area under a velocity-time graph then we can say that

Area under graph = ^{1}/_{2}( v + u)*t

hence s = ^{1}/_{2}( v + u)*t

Now if you can remember the first equation of motion is

v = u +at

Hence if we replace v = u +at into s = ^{1}/_{2}( v + u)*t we get

s = ^{1}/_{2}( v + u)*t

s = ^{1}/_{2}( [u +at] + u)*t

s = ^{1}/_{2}( u + at + u)*t

s = ^{1}/_{2}(2 u + at )*t

s =^{1}/_{2}(2ut +at^{2})

S = ut + ^{1}/_{2}at^{2}

Hence as you can see above the second equation of motion is

**S = ut + ^{1}/_{2}at^{2}**

** **We are now going to look at two examples where the second equation of motion can be used.

Example 1

A bus starts from rest and accelerates at a rate of 2.5 m/s^{2} for a time of 50 s. Determine the distance travelled by the bus during the acceleration phase.

Now let us identify the different variables involved.

u the initial speed = 0 m/s

t time taken = 50 s

a acceleration = 2.5 m/s^{2}

s distance travelled = ???

If we want calculate s the distance travelled we will have to use the second equation of motion

**S = ut + ^{1}/_{2}at^{2}**

Substituting the different values we get

S = ut + ^{1}/_{2}at^{2}

= o*50 + 0.5*2.5 *50^{2}

= 0 + 1.25*2500

=3125 m

Example 2

An aero plane travelling at a speed of 300 m/s lands on a track 2000 m long and decelerates for a time of 25 s until it comes to rest. Calculate the deceleration needed to stop the plane.

We must now identify the different variables involved as in the first example.

initial speed u = 300 m/s

time taken t = 25 s

Distance travelled s = 2000 m

acceleration a = ????

If we want calculate the acceleration a we will have to use the second equation of motion

**S = ut + ^{1}/_{2}at^{2}**

and make a the subject of formula.

s – ut = ^{1}/_{2}at^{2}

a =2(s - ut)/t^{2}^{ }

= 2(2000 – 300*25)/25^{2}

= –8.8 m/s^{2}

As you can see using the second equation of motion is not that difficult. Should you enter into any difficulties just post the question into the comment section.

Example 1 is wrong,

ReplyDeleteit should be = 0*50 + 0.5* 2.5 *50(squared), not = 0*50 + 0.5 *50(squared) and the answer should be 3125m

Have a look at it again, its correct..

Deleteut+1/2at^2

0*50+ 0.5* 2.5 *50^2

The error has been corrected. Thanks

ReplyDeleteYou forget to use half of the acceleration constant. I believe the equations should be:

ReplyDelete1/2 x 2.5 x 2500 = 3125 and NOT:

S = ut + 1/2at2

= o*50 + 0.5 *502

= 0 + 0.5*2500

=1250 m

This error have been corrected several times but it seems blogger just leave the post as it is and create a second post which is corrected. I have now successfully corrected the post. Thanks every one for their support.

ReplyDeletepls can u make t the subject of the formula for me in the second equation of motion

ReplyDeleteIt is possible to make t the subject of formula so that you can calculate it using s, u and a.

ReplyDeleteYou will have to write the equation s = ut + 0.5at^2

as being 0.5at^2 + ut - s = 0

where A = 0.5a

B = u

C= -s

making t the subject of formula will give

t = [-B +(B^2 -4AC)^0.5]/2A

or

t = [-B - (B^2 -4AC)^0.5]/2A

Substitute the values for A, B and C and you will obtain the answer for t.

You will have two answers. Just disregard the time that is negative.

Thankyou very much, this helped me for a test, very useful site

ReplyDeleteI am an 8 grade student I want to know the second equation of motion in the simplest form

ReplyDeletecan u suggest me some practical to prove s=ut+1/2at2

ReplyDeleteIt is possible to do two practicals to prove this equation.

ReplyDeleterelease a steel ball in air. Using a pair of light gates measure the time t to move a certain distance s. this distance s is usually 1 m. the ball is released near the top light gate to make u-0 m/s. you can the calculate g.

I would write a post on the topic in the future.

http://www.nuffieldfoundation.org/practical-physics/investigating-free-fall-light-gate

ReplyDeletePleae see this experiment that can be used to prove the second equation of motion

I keep getting -17.6 m/s^2 when plugging it into the formula. I realize this is the answer multiplied by 2 but don't understand how halving it works into the equation. I understand this must be very elementary but I wanted to ask for a more detailed explanation

ReplyDeleteI keep getting -17.6 m/s^2 when plugging it into the formula. I realize this is the answer multiplied by 2 but don't understand how halving it works into the equation. I understand this must be very elementary but I wanted to ask for a more detailed explanation

ReplyDeleteI have the same question as JustDolt55, I have done everything the way that was shown, but i continue to get -17.6 m/s^2 as well.

ReplyDeleteA bullet covers a distance of 120m in 10seconds.Find its acceleration at the end of 10seconds

ReplyDelete0

Delete0

Deletei am also getting -17.6m/s^2.

ReplyDeleteAn stone is dropped from height of 40m. Find the vf and t?

ReplyDeleteAn stone is dropped from a height of 40m.find the vf and time.

ReplyDelete