Kinematics-1D

Let initial velocity at $t = 0$ be $u$ and constant acceleration be $a$.
For the first $5\text{ sec}$, distance covered $s_5 = 10\text{ m}$.
Using $s = ut + \frac{1}{2}at^2$:
$10 = 5u + \frac{1}{2}a(5)^2 \implies 10 = 5u + \frac{25}{2}a \implies 2u + 5a = 4$ (Equation 1)
For the total time of first $8\text{ sec}$ ($5\text{ sec} + 3\text{ sec}$), total distance covered $s_8 = 10 + 10 = 20\text{ m}$.
$20 = 8u + \frac{1}{2}a(8)^2 \implies 20 = 8u + 32a \implies 2u + 8a = 5$ (Equation 2)
Subtracting Equation 1 from Equation 2:
$(2u + 8a) - (2u + 5a) = 5 - 4$
$3a = 1 \implies a = \frac{1}{3}\text{ m/s}^2$
Substitute $a = \frac{1}{3}$ into Equation 1 to find $u$:
$2u + 5\left(\frac{1}{3}\right) = 4 \implies 2u = 4 - \frac{5}{3} = \frac{7}{3} \implies u = \frac{7}{6}\text{ m/s}$
Now, calculate the total distance covered in $10\text{ sec}$ ($t = 10\text{ s}$):
$s_{10} = \left(\frac{7}{6}\right)(10) + \frac{1}{2}\left(\frac{1}{3}\right)(10)^2$
$s_{10} = \frac{70}{6} + \frac{100}{6} = \frac{170}{6} = \frac{85}{3} \approx 28.33\text{ m}$
Distance travelled in the next $2\text{ sec}$ (between $t = 8\text{ s}$ and $t = 10\text{ s}$):
$\Delta s = s_{10} - s_8 = \frac{85}{3} - 20 = \frac{25}{3} \approx 8.33\text{ m}$
Thus, the distance travelled in the next $2\text{ sec}$ is $8.3\text{ m}$.

Since the body starts from rest, initial velocity $u = 0$.
Distance covered in the first $5\text{ seconds}$:
$S_1 = \frac{1}{2}a(5)^2 = \frac{25a}{2}$
Total distance covered in the first $10\text{ seconds}$:
$S_1 + S_2 = \frac{1}{2}a(10)^2 = \frac{100a}{2}$
Distance covered in the second $5\text{ seconds}$:
$S_2 = \frac{100a}{2} - S_1 = \frac{100a}{2} - \frac{25a}{2} = \frac{75a}{2} = 3S_1$
Total distance covered in the first $15\text{ seconds}$:
$S_1 + S_2 + S_3 = \frac{1}{2}a(15)^2 = \frac{225a}{2}$
Distance covered in the third $5\text{ seconds}$:
$S_3 = \frac{225a}{2} - (S_1 + S_2) = \frac{225a}{2} - \frac{100a}{2} = \frac{125a}{2} = 5S_1$
Therefore, $S_1 = \frac{S_2}{3} = \frac{S_3}{5}$, which gives $S_1 = \frac{1}{3}S_2 = \frac{1}{5}S_3$.

Given initial velocity $u = 0$.
Given uniform acceleration $a = 8\text{ m/sec}^2$.
Time index $n = 5$ for the fifth second.
Using the formula $S_n = u + \frac{1}{2}a(2n - 1)$:
Substitute the given values into the equation:
$S_5 = 0 + \frac{1}{2}(8)(2 \times 5 - 1)$
$S_5 = 4 \times (10 - 1)$
$S_5 = 4 \times 9 = 36\text{ metres}$

Let $m$ be the mass of the single car.
Initial mass $m_1 = m$ and initial acceleration $a_1 = 4\text{ m/s}^2$.
When the car pulls another car of the same mass, the total effective mass becomes $m_2 = m + m = 2m$.
Since the force $F$ produced by the engine remains constant, $F = m_1 a_1 = m_2 a_2$.
$\frac{a_2}{a_1} = \frac{m_1}{m_2} = \frac{m}{2m} = \frac{1}{2}$
$a_2 = \frac{a_1}{2} = \frac{4}{2} = 2\text{ m/s}^2$