Solution:

We can use the work-energy principle to find the force.

Given:
- Initial velocity, \( u = 20 \, \text{m/s} \)
- Final velocity, \( v = 5 \, \text{m/s} \)
- Distance, \( s = 100 \, \text{m} \)
- Mass, \( m = 20 \, \text{kg} \)

Using the equation of motion:
\[
v^2 = u^2 + 2as
\]
\[
(5)^2 = (20)^2 + 2a(100)
\]
\[
25 = 400 + 200a
\]
\[
200a = -375 ;a = -\frac{375}{200} = -1.875 \, \text{m/s}^2
\]

Now, force \( F = ma \):
\[
F = 20 \times (-1.875) = -37.5 \, \text{N}
\]

Thus, the force on the body is -37.5 N (opposite to the direction of motion).

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Solution:

We will use the formula for the distance covered in the \(n\)-th second:

\[
s_n = u + \frac{a}{2}(2n - 1)
\]

Given:
- Distance in the 5th second: \( s_5 = 65 \, \text{m} \)
- Distance in the 9th second: \( s_9 = 105 \, \text{m} \)

Using the formula for both:
1. \( s_5 = u + \frac{a}{2}(9) = 65 \)
\[
u + \frac{9a}{2} = 65 \tag{1}
\]

2. \( s_9 = u + \frac{a}{2}(17) = 105 \)
\[
u + \frac{17a}{2} = 105 \tag{2}
\]

Now, subtract equation (1) from equation (2):
\[
\left( u + \frac{17a}{2} \right) - \left( u + \frac{9a}{2} \right) = 105 - 65
\]
\[
\frac{8a}{2} = 40 \implies 4a = 40 \implies a = 10 \, \text{m/s}^2
\]

Substitute \( a = 10 \) into equation (1):
\[
u + \frac{9 \times 10}{2} = 65 \implies u + 45 = 65 \implies u = 20 \, \text{m/s}
\]

Now, to find the total distance covered in 20 seconds:
\[
S = ut + \frac{1}{2} a t^2 = 20 \times 20 + \frac{1}{2} \times 10 \times 20^2
\]
\[
S = 400 + 0.5 \times 10 \times 400 = 400 + 2000 = 2400 \, \text{m}
\]

Thus, the body will cover 2400 meters in 20 seconds.

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Solution:

For a body starting from rest with constant acceleration:

- Distance covered in the 5th second: \( s_5 = u + \frac{a}{2}(2n - 1) \), where \( u = 0 \) and \( n = 5 \):
\[
s_5 = \frac{a}{2} (9) = \frac{9a}{2}
\]

- Distance covered in 5 seconds: \( S = \frac{1}{2} a t^2 \), where \( t = 5 \):
\[
S = \frac{1}{2} a (5)^2 = \frac{25a}{2}
\]

Ratio of distances:
\[
\frac{s_5}{S} = \frac{\frac{9a}{2}}{\frac{25a}{2}} = \frac{9}{25}
\]

Thus, the ratio is 9:25.

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Solution:

Given:
- Initial velocity: \( u = 20 \, \text{m/s} \)
- Time: \( t = 2 \, \text{seconds} \)
- Distance to the barrier: \( s = 20 \, \text{m} \)

Using the equation of motion:
\[
s = ut + \frac{1}{2} a t^2
\]
\[
20 = 20 \times 2 + \frac{1}{2} a \times 2^2
\]
\[
20 = 40 + 2a
\]
\[
2a = -20
\]
\[
a = -10 \, \text{m/s}^2
\]

Thus, the magnitude of the deceleration is 10 m/s².

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Solution:

To find the resultant velocity using unit vectors:

- Initial velocity: \( \vec{u} = 0.4 \, \hat{i} \, \text{m/s} \)
- Acceleration: \( \vec{a} = 0.15 \, \hat{j} \, \text{m/s}^2 \)
- Time: \( t = 2 \, \text{seconds} \)

Using the equation of motion \( \vec{v} = \vec{u} + \vec{a}t \):

- In the \( x \)-direction: \( v_x = 0.4 \, \text{m/s} \)
- In the \( y \)-direction: \( v_y = 0 + 0.15 \times 2 = 0.3 \, \text{m/s} \)

The resultant velocity is:

\[
|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(0.4)^2 + (0.3)^2} = \sqrt{0.25} = 0.5 \, \text{m/s}
\]

Thus, the resultant velocity is 0.5 m/s

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Solution:

To solve this, we'll use the equation of motion:

\[
v^2 = u^2 + 2as
\]

Where:
- \( v = 0 \) (final velocity, since the car comes to rest)
- \( u = 144 \, \text{km/h} = 40 \, \text{m/s} \) (initial velocity)
- \( s = 200 \, \text{m} \) (distance)
- \( a \) is the acceleration (which we'll solve for)

Rearranging the equation:

\[
0 = (40)^2 + 2 \cdot a \cdot 200
\]
\[
0 = 1600 + 400a
\]
\[
a = - \frac{1600}{400} = -4 \, \text{m/s}^2
\]

Now, to find the time (\( t \)):

\[
v = u + at
\]
\[
0 = 40 + (-4)t
\]
\[
t = \frac{40}{4} = 10 \, \text{seconds}
\]

Thus, it takes 10 seconds for the car to stop.

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Solution:

Using the formula $a = \frac{v_f - v_i}{t}$:

Converting to m/s:
$90$ km/h = $25$ m/s, $54$ km/h = $15$ m/s.

$$a = \frac{15 - 25}{15} = \frac{-10}{15} = -0.67 \text{ m/s}^2$$

Retardation = 0.67 m/s².

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Solution:

Using $t = \frac{s}{v}$, with $v = 54$ km/h = 15 m/s and $s = 5$ m:

$$t = \frac{5}{15} = 0.33 \text{ s}$$

Reaction time = 0.33 s.

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Solution:

Using the stopping distance formula:

$$s \propto u^2$$

If speed is halved, new stopping distance:

$$s_2 = \frac{100}{4} = 25 \text{ m}$$

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Solution:

Using the stopping distance formula:

$$s = \frac{u^2}{2a}$$

For Car A ($u_A = 20$ m/s, $a_A = 4$ m/s²):

$$s_A = \frac{20^2}{2 \times 4} = \frac{400}{8} = 50 \text{ m}$$

For Car B ($u_B = 10$ m/s, $a_B = 2$ m/s²):

$$s_B = \frac{10^2}{2 \times 2} = \frac{100}{4} = 25 \text{ m}$$

Total minimum separation to avoid collision:

$$X = s_A + s_B = 50 + 25 = 75 \text{ m}$$

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