A car is driving along a straight road. Its position $s$ in meters from a fixed starting point after $t$ seconds is given by $s = 5t^3 - 3t^2 + 2t$. Find the car's velocity and acceleration after $2 \text{ s}$.

$v = 50 \text{ m/s}, a = 54 \text{ m/s}^2$

The correct answer is Option (2) → $v = 50 \text{ m/s}, a = 54 \text{ m/s}^2$ ##

The velocity $v$ is the rate of change of position with respect to time. So,

$v = \frac{ds}{dt}$

Differentiate $s = 5t^3 - 3t^2 + 2t$:

$v = 15t^2 - 6t + 2$

The acceleration $a$ is the rate of change of velocity with respect to time. So,

$a = \frac{dv}{dt}$

Differentiate $v = 15t^2 - 6t + 2$

$a = 30t - 6$

Evaluate velocity and acceleration at $t = 2$:

Velocity at $t = 2$:

$v = 15(2)^2 - 6(2) + 2$

$v = 15 \cdot 4 - 12 + 2$

$v = 60 - 12 + 2 = 50 \text{ m/s}$

Acceleration at $t = 2$:

$a = 30(2) - 6 = 60 - 6 = 54 \text{ m/s}^2$

The car's velocity after $2 \text{ s}$ is $50 \text{ m/s}$, and its acceleration is $54 \text{ m/s}^2$.