A car initially at rest moves with a constant acceleration of 2m.s⁻² east. Calculate the Magnitude of the velocity after 10 s and Distance covered during the first 10 s.

Initial velocity $u = 0 \text{ m/s}$ (at rest).
Acceleration $a = 2 \text{ m/s}^2$.
Time $t = 10 \text{ s}$.

The magnitude of the velocity after $10 \text{ s}$ is $\boxed{\text{20 m/s}}$.

Initial velocity $u = 0 \text{ m/s}$.
Acceleration $a = 2 \text{ m/s}^2$.
Time $t = 10 \text{ s}$.

The distance covered during the first $10 \text{ s}$ is $\boxed{\text{100 m}}$.

The velocity of car A is $\boxed{\text{21.39 m/s}}$.

Initial velocity $u = 0 \text{ m/s}$ (from rest).
Acceleration $a = 1.5 \text{ m/s}^2$.
Time $t = 6.5 \text{ s}$.

The velocity of car B after $6.5 \text{ s}$ is $\boxed{\text{9.75 m/s}}$.

Initial velocity $u = 0 \text{ m/s}$.
Acceleration $a = 1.5 \text{ m/s}^2$.
Time $t = 6.5 \text{ s}$.

Car B has driven $\boxed{\text{31.69 m}}$ (rounded to two decimal places).

Constant velocity $v_A = 77 \text{ km/h} = 21.39 \text{ m/s}$ (from 6.2).
Time $t = 6.5 \text{ s}$.

Car A traveled $\boxed{\text{139.04 m}}$ (rounded to two decimal places).

Car A's distance $s_A = 139.04 \text{ m}$.
Car B's distance $s_B = 31.69 \text{ m}$.
At $6.5 \text{ s}$, Car B is behind Car A ($s_B < s_A$).

Car A's velocity $v_A = 21.39 \text{ m/s}$ (constant).
Car B's velocity $v_B = 9.75 \text{ m/s}$ (after $6.5 \text{ s}$, it continues at this constant velocity).
Since Car B's constant velocity ($9.75 \text{ m/s}$) is less than Car A's constant velocity ($21.39 \text{ m/s}$), Car B will not be able to catch up with Car A.

Explanation: ❌ No, Car B will not be able to catch up with Car A. After $6.5 \text{ s}$, Car B is still behind Car A ($31.69 \text{ m}$ vs $139.04 \text{ m}$). Furthermore, Car B's velocity ($9.75 \text{ m/s}$) is less than Car A's velocity ($21.39 \text{ m/s}$), and Car B continues at this constant, slower velocity. Therefore, the distance between them will continue to increase.

Final velocity $v = 0 \text{ m/s}$ (comes to a standstill).
Distance $s = 25 \text{ m}$.
Time $t = 2.5 \text{ s}$.

We have $s = ut + \frac{1}{2}at^2$ and $v = u + at$.
From $v = u + at$, we can express $u$ as $u = v - at = 0 - a(2.5) = -2.5a$.
Substitute this into the distance equation:

The acceleration car A experiences is $\boxed{\text{-8 m/s}^2}$. The negative sign indicates deceleration.