# Physics 9th: Motion | Equations of Motion and Some Solved Examples

## Physics 9th: Motion | Equations of Motion and Some Solved Examples

### Equations of Motion

When the body is moving along a straight line with uniform acceleration, a relation can be established between velocity of the body, acceleration of the body and the distance travelled by the body in a specific time by a set of equation. These equations are called equations of motion.

The three equations are:

First Equation of motion     :    v = u + at

Second Equation of motion    :    s = ut + $\displaystyle \frac{1}{2}a{{t}^{2}}$

Third Equation of motion     :    $\displaystyle {{v}^{2}}-{{u}^{2}}=2as$

Where u = initial velocity of the body

v = final velocity of the body

a = uniform acceleration of the body

t = time taken

s = distance travelled

### Question:     A particle is moving with a uniform acceleration. Will it surely moving in a straight line?

Solution:

No, we are not sure about straight line motion.

The motion of the particle will be along a straight line only if the uniform acceleration is in the same direction as that of its velocity. If velocity and acceleration are in different directions then motion will not be along a straight line.

### Question:     A car covers distance s1 with a constant speed v1 and then covers a further distance s2 with a constant speed v2. Find an expression for his average speed.

Solution:

When a car covers a distance s1 with a constant speed v1, it takes time $\displaystyle {{t}_{1}}=\frac{{{s}_{1}}}{{{v}_{1}}}$. Similarly for covering a distance s2 with a constant speed v2 the time taken $\displaystyle {{t}_{2}}=\frac{{{s}_{2}}}{v}.$

\ Average speed $\displaystyle {{v}_{av}}=\frac{s}{t}=\frac{{{s}_{1}}+{{s}_{2}}}{{{t}_{1}}+{{t}_{2}}}=\frac{{{s}_{1}}+{{s}_{2}}}{\left( \frac{{{s}_{1}}}{{{v}_{1}}}+\frac{{{s}_{2}}}{{{v}_{2}}} \right)}=\frac{\left( {{s}_{1}}+{{s}_{2}} \right){{v}_{1}}{{v}_{2}}}{\left( {{s}_{1}}{{v}_{2}}+{{s}_{2}}{{v}_{1}} \right)}$

### Question:     What is the direction of acceleration acting on a particle having uniform circular motion?

Solution:

Acceleration is directed perpendicular to the direction of motion.

### Question:     Can a particle be accelerated (i) if its speed is constant, (ii) if its velocity is constant? Give reason.

Solution:

(i)    A particle can be accelerated even if its speed is constant because it means that only the magnitude of velocity is constant and its direction of motion may change.

(ii) No acceleration is possible if velocity of a particle remains constant.

### (ii) What is the resultant displacement?

Solution:

Distance travelled     = AB + BC + CD

= 3 + 4 + 9 = 16 km

=$\displaystyle \sqrt{A{{E}^{2}}+E{{D}^{2}}}$

= $\displaystyle \sqrt{{{\left( 4 \right)}^{2}}+{{\left( 6 \right)}^{2}}}$

= $\displaystyle \sqrt{16+36}$

= $\displaystyle \sqrt{52}$

= $\displaystyle 2\sqrt{13}$ km

= $\displaystyle 2\times 3.6$ km

= 7.2 km

### Question:     A moving train is brought to rest within 20 seconds by applying brakes. Find the initial velocity, if the retardation due to breaks is 5 ms-2.

Solution:

Here, Final velocity,    v = 0

Acceleration,         a = $\displaystyle -5\,m{{s}^{-2}}$

Time,               t = 20 s

Initial velocity,         u =?

Now, putting these values in the first equation of motion

v = u + at

We get:            0 = u + (5) ´ 20

0 = u 100

u = 100 $\displaystyle m{{s}^{-1}}$

Thus, the initial velocity of the train is 100 meters per second.

### Question:     A train has a uniform acceleration of 14 m/s2. What distance will it cover in 10 seconds after the start?

Solution:

Here, Initial velocity,    u = 0

Time,      t = 10 s

Acceleration,        a =$\displaystyle 14m/{{s}^{2}}$

And,     Distance,     s = ?

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

$\displaystyle s=ut+\frac{1}{2}a{{t}^{2}}$

We get:        $\displaystyle s=0\times 10+\frac{1}{2}\times 14\times {{(10)}^{2}}$

$\displaystyle s=700$m

Thus, the distance covered by the train in 10 seconds is 700 metres.

### Question:     A vehicle moving at a speed of 20 m/s is stopped by applying brakes which produces a uniform acceleration of, –0.5 m/s2. How much distance will be covered by the vehicle before it stops?

Solution:

Here,     Initial speed,      $\displaystyle u=20$ m/s

Final speed,          $\displaystyle v=0$

Acceleration,         $\displaystyle a=-0.5$ m/s-2

And, Distance covered,$\displaystyle s=?$

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

$\displaystyle {{v}^{2}}={{u}^{2}}+2as$

We get :     $\displaystyle {{\left( 0 \right)}^{2}}={{\left( 20 \right)}^{2}}+2\times \left( -0.5 \right)\times s$

$\displaystyle 0=400\,\,m-s$

s = 400 m

Thus, the distance covered is 400 metres.

### Question:     Arrange the following speeds in increasing order- 10 m/s, 200 m/min, 30 km/hr

Solution:

(i)    10 m/s

(ii)    200 m/min

$\displaystyle \frac{200\,\,m}{1\,\,\min }=\frac{200\,\,m}{60\,\,\,\sec }=\frac{10}{3}m/s=3.33\,\,m/s$

(iii)    30 km/hr

$\displaystyle \frac{30\,\,km}{1\,\,hr}=\frac{30\times 1000}{1\times 60\times 60}=\frac{30000}{3600}=\frac{25}{3}=8.33\,$m/s

200 m/min < 30 km/hr < 10 m/s

### Question:     Joseph jogs from one end A to the other B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Solution:

Here, as shown

Time taken in jogging from A to B = t    1

= 2 min 30 sec

= (2 ´ 60 + 30) sec = 150 s

BC = 100 m

Time taken in jogging from B to C, t2 = 1 min = 60 s

(a)     From A to B

$\displaystyle average\,\,\,speed=\frac{total\,\,distance}{total\,\,time}=\frac{300m}{150s}=2.00$ m/s

$\displaystyle average\,\,velocity=\frac{displacement\,\,(AB)}{time}=\frac{300m}{150s}=2.00$m/s

(b)     From A to C

$\displaystyle average\,\,\,speed=\frac{total\,\,distance}{total\,\,time}=\frac{AB+BC}{{{t}_{1}}+{{t}_{2}}}$

$\displaystyle =\frac{\left( 300+100 \right)m}{\left( 150+60 \right)s}=\frac{400}{210}$ m/s = 1.90 m/s

$\displaystyle average\,\,velocity=\frac{displacement\,\,(AC)}{time}$

$\displaystyle =\frac{\left( 300-100 \right)m}{\left( 150+60 \right)s}$        $\displaystyle \left( \because \,\,\,AC=AB-BC \right)$

$\displaystyle =\frac{200}{210}$ m/s = 0.952 m/s

### Question:     A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms–2 for 8.0 s. How far does the boat travel this time?

Solution:

Here,     initial velocity,         u = 0

acceleration,         a = 30 ms2

time,             t = 8.0 s

distance travelled     s = ?

Now,     putting these values in the second equation of motion

$\displaystyle s=ut+\frac{1}{2}a{{t}^{2}}$

$\displaystyle s=0\times 8+\frac{1}{2}\times 3.0{{\left( 8.0 \right)}^{2}}$= 96 m

### Question:     A ball gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms–2, with what velocity will it strike the ground? After what time will it strike the ground?

Solution:

Here,     Height = distance = s = 20 m

acceleration, a = 10 m/s2

final velocity, v = ?

time taken, t = ?

As the ball is gently dropped, its initial velocity, u = 0

From         $\displaystyle {{v}^{2}}-{{u}^{2}}=2as$

$\displaystyle {{v}^{2}}={{u}^{2}}+2as=0+2\left( 10 \right)\times 20=400$

or          $\displaystyle v=\sqrt{400}=$ 20 m/s

### Question:     Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul’s trip?

Solution:

Let the school be at a distance of x km. If t1 is time taken to reach the school, then

$\displaystyle {{t}_{1}}=\frac{distance}{average\,\,speed}=\frac{x}{20}$

If$\displaystyle {{t}_{2}}$ is time taken to reach back, then

$\displaystyle {{t}_{2}}=\frac{distance}{average\,\,speed}=\frac{x}{30}$

Total time, $\displaystyle t={{t}_{1}}+{{t}_{2}}=\frac{x}{20}+\frac{x}{40}=x\left[ \frac{1}{20}+\frac{1}{30} \right]=\frac{5x}{60}=\frac{x}{12}$

Total distance = $\displaystyle x+x=2x$

\ $\displaystyle average\,\,speed=\frac{total\,\,distance}{total\,\,time}=\frac{2x}{x/12}=24$ km/h

### Question:     The velocity acquired by a body moving with uniform acceleration is 12 ms–1 in 2 s and 18 ms–1 in 4 s. Find the initial velocity of the body.

Solution:

Let initial velocity of the body be u ms1 and its constant acceleration be a

Using first equation of motion; $\displaystyle v=u+at$

velocity after 2 s will be $\displaystyle v=12=u+a\times 2$

\                 $\displaystyle 12=u+2a$            …(i)

and velocity after 4 s will be $\displaystyle 18=u+a\times 4$

$\displaystyle 18=u+4a$            …(ii)

Subtracting (i) from (ii), we have

6 = 2a

Þ                 $\displaystyle a=\frac{6}{2}=3$ ms2

Subtracting this value of ‘a’ in equation (i) we get

$\displaystyle 12=u+2a=u+2\times 3=u+6$

$\displaystyle u=12-6=6$ ms1

### Question:     An object moving with uniform acceleration has the displacement of 9 m in 3 s but a displacement of 16 m in 4 s. Find the acceleration.

Solution:

Let initial velocity and uniform acceleration of the object be u and a, respectively.

It is given that when $\displaystyle {{t}_{1}}=3$s, $\displaystyle {{s}_{1}}=9$ m and for $\displaystyle {{t}_{2}}=4$ s, $\displaystyle {{s}_{2}}=16$m

Putting these values in the second equation of motion

$\displaystyle s=ut+\frac{1}{2}\times a{{t}^{2}}$, we have

$\displaystyle 9=u\times 3+\frac{1}{2}\times a\times {{\left( 3 \right)}^{2}}=3u+\frac{9}{2}a$

or        $\displaystyle 3=u+\frac{3}{2}a$                …(i)

and         $\displaystyle 16=u\times 4+\frac{1}{2}\times a\times {{\left( 4 \right)}^{2}}=4u+8a$

or        $\displaystyle 4=u+2a$                …(ii)

Subtracting (i) from (ii), we have

$\displaystyle 1=\frac{a}{2}$

Þ        $\displaystyle a=2$ ms2

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