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 $2 \text{ s}$.
Answer & explanation
Correct answer: option 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 $54 \text{ m/s}^2$.