1. It can change the shape of the body.
2. It can change the direction of motion of the body.
3. It can change the speed of the body.

1N = 10⁵ dyne

Though action and reaction are equal in magnitude and opposite in direction, they act on different bodies. Thus the forces acting on different bodies don't cancel each other. Hence, as they act on different bodies, action and reaction don't cancel each other.

A) Let d the distance between his home and school in a straight line.The time taken by the boy to run from home to school (t1) = \(\frac{d}{3}\)

(speed= \(\frac{distance}{time}\))

The time taken by him to walk from school to home (t2) = \(\frac{d}{5}\)

Total time taken (t) = t1+t2
= \(\frac{d}{3}\) + \(\frac{d}{5}\) = \(\frac{8}{15}\) d

Total Distance Travelled = d+d =2d

Average speed = \(\frac{Total Distance}{Total Time}\)

= \(\frac{2d}{8d}\) x 15

= 3.75 m/s

So,Average speed = 3.75 m/s

Again,
Since the boy comes back to the starting point (i.e.,home) his total displacement is zero.

Average velocity = \(\frac{Total Displacement}{Total Time}\)
= \(\frac{s}{t}\) = \(\frac{0}{8}\) = 0

Average velocity = 0m/s

Total distance covered (s) = 400 km = 3x = \(10^{5}\) m [1km=1000m]

Total time taken (t) = 8 hours = 8 x 3600 s [ 1hr=3600sec]

We know that,

Average speed (V) = = \(\frac{Total\;distance\;covered (S) }{Total\;Time\;Taken(T) }\)

= \(\frac{4 X 10{^5}} {8 X 3600}\)

= 13.89 m/s

Therefore,the average speed of the bus is 13.89 m/s.

Here, initial velocity (u) = 6m/s
Final Velocity = 8m/s
Time Taken(t) = 2s
Acceleration(a) = ?
We know that,
a= \(\frac{v-u}{t}\)
= \(\frac{8-6}{2}\)
= \(\frac{2}{2}\) = 1 m/\(s^{2}\)

Therefore, the acceleration of a car is 1 m/\(s^{2}\)

Here, Initial velocity (u) = 0
Acceleration (a) = \(0.5 m/s^{2}\)
Time Taken (t) = 2 minutes = 120 s [1 min=60 sec]
Final Velocity (V) = ?
Distance Covered (S)= ?

We have,
V=u+at
or, v = 0+0.5 x 120
v= 60m/s

Again,
s= \(\frac{u+v}{2}\)

= \(\frac{0+60}{2}\) x 120

= 3600m

Therefore, the final veocity of the bus is 60m/s and it will cover 3600m in 2 minutes

Here, Initial Velocity (u) = 40m/s
For upwoard direction, acceleration (a) = -10m/\(s^{2}\)

The velocity of the ball goes on decreasing as the ball attains height. At the highest point, final velocity (v)=0
We have,
v=u+at
or,0=40+(-10)t
or 10t= 40
or t = \(\frac{40}{10}\)
t= 4s

Again, we have
s= ut+\(\frac{1}{2}\) \(at^{2}\)
or,s= 40 x 2 + \(\frac{1}{2}\) (-10) (\(2^{2}\))
s= 80 + \(\frac{1}{2}\) x (-40)
or, s = 80-20 = 60 m

Since 2s is the time taken by the ball to reach the maximum height, so time taken to return to the initial position is 2t= 2x2 = 4s

Therefore, the maximum height attained is 60 m and the time taken for it to return to the initial position is 4s.

Although the driver sees a baby 60 m ahead on the road , he stops the car at a distance of 15 m, hence,

Distance covered (s) = 15m
Initial Velocity (v) = 0
90km/hr = \(\frac {60 \times 1000m }{60 \times 60s }\)
= 16.66 m/s

Retardation = ?
Time taken (t) = ?

We know ,
\(v^{2}\) =\(u^{2}\) + 2as
or, \(0^{2}\) = ( \(16.66^{2}\) ) + 2a\(\times\) 15
or, -625 = 30a
or, a = \(\frac{277.5}{30}\)
\(\therefore\) a= -9.25 m/\(s^{2}\)

Again , we have

v= u + at
or, 0 = 16.66 + (-20.83) t
or, -16.66 = -9.25t
or, t = \(\frac{16.66}{9.25}\)
\(\therefore\) t= 1.80 s

Therefore, the acceleration of the car is -9.25 m/\(s^{2}\). i.e. its retardation 9.25 m/\(s^{2}\) and it takes 1.8 to come at rest.

Here Mass (m) = 100g = \(\frac{100}{1000}\) kg = 0.1

Initial Velocity (u) = 15m/s
Final Velocity (v) = 0
Time Taken (t) = 0.05s

We have,

v= u + at
or, 0 = 15 + a \(\times\) 0.05
or, a = \(\frac{15}{0.05}\)
\(\therefore\) a = -300 m/\(s^{2}\)

Again, we have
F= ma
or,F= 0.1 \(\times\) (-300)
\(\therefore\) = -30N

Here,
Mass (m) = 100g = \(\frac{100}{1000}\) kg
= 0.1kg
Distance Covered (s)= 200cm = 2m
Time Taken (t) = 5s
Initial Velocity (u) = 0

We know,
\(\therefore\) s= ut + \(\frac{1}{2}\) \(at^{2}\)
or, 2= 0 \(\times\) t + \(\frac{1}{2}\) a \( (5)^{2}\)
a= 2 \(\frac{4}{25}\) m/ \(s^{2}\)

Again, we have
\(\therefore\) F = ma
or, F= (0.1) \(\times\) \(\frac{4}{25}\) = 0.016 N
Thus, the magnitude of the force is 0.016 N

Solution:
Here, Initial Velocity (U) = 0
Acceleration(a) = 0.5m/s \(^2\)
Time taken (t) = 2 minutes = 120 s
Final velocity (V) = ?
Distance Covered(s) = ?
We have,
v = u + at
or, v = 0 + 0.5 \(\times\) 120
v = 60m/s

Again,
s = \(\frac{u\;+v}{2}\) \(\times\) t
= \(\frac{o\;+60}{2}\) \(\times\) 120 = 3600 m

Therefore, the final velocity of the bus is 60m/s and it will cover 3600 m in 2 minutes.

Here, Initial velocity (u) = 20m/s
Acceleration (a) = -10m/s \(^2\)
Final Velocity (v) = 0

We have,
v = u + at
or, 0 = 20 + (-10) t
or, 10t = \(\frac{20}{10}\)
\(\therefore\) t = 2s
Again, we have
s = ut + \(\frac{1}{2}\) \(at^{2}\)
or, s = 20 \(\times\) 2 + \(\frac{1}{2}\) (-10) (2²⁾
or, s = 40 + \(\frac{1}{2}\) \(\times\) (-40)
or, s = 40 - 20 = 20m

Since, 2s is the time taken by the ball to reach the maximum height, so time taken to return to the initial position is 2t = 2 \(\times\) 2s = 4s.

Therefore, the maximum height attained is 20 m and the time taken is 20m and the time taken for it to return to the initial position is 4s.