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Understanding Physics Topics is essential for solving complex problems in many fields, including engineering and medicine.

## Derive the Three Equations of Uniformly Accelerated Motion

There are three equations for the motion of those bodies which travel with a uniform acceleration. These equations give relationship between initial velocity, final velocity, time taken, acceleration and distance travelled by the bodies. We will study these equations one by one.

### 1. First Equation of Motion

The first equation of motion is : v = u + at. It gives the velocity acquired by a body in time t. We will now derive this first equation of motion.

Consider a body having initial velocity ‘u’. Suppose it is subjected to a uniform acceleration V so that after time T its final velocity7 becomes V. Now, from the definition of acceleration we know that:

Acceleration = \(\frac{\text { Change in velocity }}{\text { Time taken }}\)

or Acceleration = \(\frac{\text { Final velocity }- \text { Initial velocity }}{\text { Time taken }}\)

So, a = \(\frac{v-u}{t}\)

at = v – u

and, v = u + at

where v = final velocity of the body

u = initial velocity of the body

a = acceleration

and t = time taken

The equation v = u + at is known as the first equation of motion and it is used to find out the velocity ‘v’ acquired by a body in time ‘t’, the body having an initial velocity ‘u’ and a uniform acceleration ‘a’.

In fact, this equation has four values in it, if any three values are known, the fourth value can be calculated. By paying due attention to the sign of acceleration, this equation can also be applied to the problems of retardation.

### 2. Second Equation of Motion

The second equation of motion is : s = ut + \(\frac{1}{2}\) at^{2}. It gives the distance travelled by a body in time t. Let us derive this second equation of motion.

Suppose a body has an initial velocity ‘u’ and a uniform acceleration ‘a’ for time ‘t’ so that its final velocity becomes ‘v’. Let the distance travelled by the body in this time be ‘s’. The distance travelled by a moving body in time ‘t’ can be found out by considering its average velocity. Since the initial velocity of the body is ‘u’ and its final velocity is ‘v’, the average velocity is given by :

Average velocity = \(\frac{\text { Initial velocity }+ \text { Final velocity }}{2}\)

That is, Average velocity = \(\frac{u+v}{2}\)

Also, Distance travelled = Average velocity × Time

So, s = \(\frac{(u+v)}{2}\) × t

From the first equation of motion we have, v = u + at. Putting this value of v in equation (1), we get:

s = \(\frac{(u+u+a t) \times t}{2}\)

or s = \(\frac{(2 u+a t) \times t}{2}\)

or s = \(\frac{2 u t+a t^2}{2}\)

or s =ut + \(\frac{1}{2}\) at^{2}

where s = distance travelled by the body in time t

u = initial velocity of the body

and a = acceleration

This is the second equation of motion and it is used to calculate the distance travelled by a body in time t. This equation should also be memorized because it will be used to solve numerical problems.

### 3. Third Equation of Motion

The third equation of motion is : v^{2} = u^{2} + 2as. It gives the velocity acquired by a body in travelling a distance s. We will now derive this third equation of motion.

The third equation of motion can be obtained by eliminating t between the first two equations of motion. This is done as follows.

From the second equation of motion we have :

s = ut + \(\frac{1}{2}\) at^{2} …………. (1)

And from the first equation of motion we have :

v = u + at ………….. (2)

This can be rearranged and written as :

at = v – u

or t = \(\frac{v-u}{a}\)

Putting this value of t in equation (1), we get :

s = \(\frac{u(v-u)}{a}\) + \(\frac{1}{2} a\left(\frac{v-u}{a}\right)^2\)

or s = \(\frac{u v-u^2}{a}\) + \(\frac{a\left(v^2+u^2-2 u v\right)}{2 a^2}\) [because (v – u)^{2} = v^{2} + u^{2} – 2uv]

or s = \(\frac{u v-u^2}{a}\) + \(\frac{v^2+u^2-2 u v}{2 a}\)

or s = \(\frac{2 u v-2 u^2+v^2+u^2-2 u v}{2 a}\)

or 2as = v^{2} – u^{2}

or v^{2} = u^{2} + 2as

where v = final velocity

u = initial velocity

a = acceleration

and s = distance travelled

This equation gives us the velocity acquired by a body in travelling a distance s.

We will now solve some problems based on motion. To solve the problems on motion we should remember that:

- if a body starts from rest, its initial velocity, u = 0
- if a body comes to rest (it stops), its final velocity, v = 0
- if a body moves with uniform velocity, its acceleration, a = 0

**Example Problem 1.**

A scooter acquires a velocity of 36 km per hour in 10 seconds just after the start. Calculate the acceleration of the scooter.**Solution:**

First of all we should convert the given velocity into proper units, that is, we should convert the velocity of 36 kilometres per hour into metres per second (because the time is given in seconds).

Now, 1 km = 1000 m

So, 36 km = 36 × 1000 m

= 36,000 m …………… (1)

Also, 1 hour = 60 minutes

= 60 × 60 seconds

= 3600 s ………….. (2)

So, 36 km per hour = \(\frac{36,000 \mathrm{~m}}{3600 \mathrm{~s}}\)

= 10 m/s ………. (3)

Thus, the given velocity of 36 km per hour is equal to 10 metres per second.

Now, Initial velocity, u = 0 (Scooter starts from rest)

Final velocity, v = 10 m/s (Calculated above)

Acceleration, a = ? (To be calculated)

And, Time, t = 10 s

By putting these values in the first equation of motion :

v = u + at

We get: 10 = 0 + a × 10

10 a = 10

a = \(\frac{10}{10}\)

a = 1 m/s^{2}

Thus, the acceleration is 1 m/s^{2}.

**Example Problem 2.**

A moving train is brought to rest within 20 seconds by applying brakes. Find the initial velocity, if the retardation due to brakes is 2 m s^{-2}.**Solution:**

In this problem we have been given the value of retardation but we require the value of acceleration. We know that retardation is negative acceleration. So, if the retardation is, + 2 m s^{-2} (as given here), then the acceleration will be, – 2 m s^{-2} (the minus sign here indicates the negative acceleration). Let us solve the problem now.

Here, Initial velocity, u = ? (To be calculated)

Final velocity, v = 0 (The train stops)

Acceleration, a = -2ms^{-2}

And, Time, t = 20 s

Now, putting these values in the formula :

v = u + at

We get: 0 = u + (- 2) × 20

0 = u – 40

u = 40 m s^{-1}

Thus, the initial velocity of the train is 40 metres per second.

**Example Problem 3.**

A body starts to slide over a horizontal surface with an initial velocity of 0.5 m/s. Due to friction, its velocity decreases at the rate of 0.05 m/s^{2} (acceleration, – 0.05 m/s^{2}). How much time will it take for the body to stop ?**Solution:**

Here, Initial velocity, u = 0.5 m/s

Final velocity, v = 0 (The body stops)

Acceleration, a = – 0.05 m/s^{2}

And, Time, t – ? (To be calculated)

Now, from the first equation of motion, we have :

v = u + at

0 = 0.5 + (- 0.05) × t

or 0.05 t = 0.5

t = \(\frac{0.5}{0.05}\)

t = 10 s

Thus, the body will take 10 seconds to stop.

**Example Problem 4.**

A racing car has a uniform acceleration of 4 m/s^{2}. What distance will it cover in 10 seconds after the start ?**Solution:**

Here, Initial velocity, u – 0

Time, t = 10 s

Acceleration, a = 4 m/s^{2}

And, Distance, s = ? (To be calculated)

Now, putting these values in the second equation of motion :

s = ut +~ at^{2}

We get: s = 0 × 10 + \(\frac{1}{2}\) × 4 × (10)^{2}

s = 200 m

Thus, the distance covered by the car in 10 seconds is 200 metres.

**Example Problem 5.**

A scooter moving at a speed of 10 m/s is stopped by applying brakes which produce a uniform acceleration of, – 0.5 m/s^{2}. How much distance will be covered by the scooter before it stops ? ,**Solution:**

Here, Initial speed, u = 10 m/s

Final speed, v = 0 (Scooter stops)

Acceleration, a = – 0.5 m/s^{2}

And, Distance covered, s = ? (To be calculated)

Now, putting these values in the third equation of motion :

v^{2} = u^{2} + 2as

We get: (0)^{2} = (10)^{2} + 2 × (- 0.5) × s

0 = 100 – s

s = 100 m

Thus, the distance covered is 100 metres.

**Example Problem 6.**

A car travelling at 20 km/h speeds up to 60 km/h in 6 seconds. What is its acceleration ?**Solution:**

Here, Initial speed, u = 20 km/h

= \(\frac{20 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\)

= 5.55 m/s ……………. (1)

Final speed, v = 60 km/h

= \(\frac{60 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\)

= 16.66 m/s

Acceleration, a = ? (To be calculated)

And, Time, t = 6 s

We know that: . v = u + at

So, 16.66 = 5.55 + a × 6

6 a = 16.66 – 5.55

6a = 11.11

a = \(\frac{11.11}{6}\)

Acceleration, a = 1.85 m/s^{2}

**Example Problem 7.**

A bus increases its speed from 20 km/h to 50 km/h in 10 seconds. Its acceleration is :

(a) 30 m/s^{2}

(b) 3 m/s^{2}

(c) 18 m/s^{2}

(d) 0.83 m/s^{2}

Choose the correct answer.**Solution:**

Here, Initial speed, u = 20 km/h

= \(\frac{20 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\)

= 5.5 m/s …………….. (1)

Final speed, v = 50 km/h

= \(\frac{50 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\)

= 13.8 m/s

Acceleration, a =? (To be calculated)

And, Time, t = 10 s

Now, v = u + at

So, 13.8 = 5.5 + a × 10

10 a = 13.8 – 5.5

10 a = 8.3

a = \(\frac{8.3}{10}\)

a = 0.83 m/s^{2}

Thus, the acceleration is 0.83 m/s^{2}. The correct answer is (d).