Question

A particle moves with velocity ($v = 3t^2 + 2t$). Find acceleration at ($t = 2$) s. ____.

• A. 14 m/s²
• B. 12 m/s²
• C. 16 m/s²
• D. 10 m/s²

Hint:

Since velocity is already given as a function of time, acceleration is its rate of change with respect to time. Differentiate the velocity expression once (a second differentiation is only needed when position is given instead of velocity), then substitute the given time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Acceleration is defined as the instantaneous rate of change of velocity with respect to time. Mathematically, it is the first derivative of the velocity function.

Step 2: Key Formula or Approach:
\[ a = \frac{dv}{dt} \]

Step 3: Detailed Explanation:
Given: $v = 3t^2 + 2t$. 1. Differentiate $v$ with respect to $t$: \[ a = \frac{d}{dt}(3t^2 + 2t) \] \[ a = 6t + 2 \] 2. Substitute $t = 2$ into the acceleration equation: \[ a = 6(2) + 2 \] \[ a = 12 + 2 = 14 \text{ m/s²} \]

Step 4: Final Answer:
The acceleration at $t = 2$ s is 14 m/s².

Answer 2

Concept:

• Acceleration can be found directly from the first-principles definition of a derivative, without applying the power rule, by evaluating the limit of the average rate of change of velocity as the time interval shrinks to zero.

Step 1: Write the first-principles definition of acceleration.
$a = \displaystyle\lim_{\Delta t \to 0} \dfrac{v(t+\Delta t) - v(t)}{\Delta t}$

Step 2: Expand $v(t+\Delta t)$ for $v = 3t^2 + 2t$.
$v(t+\Delta t) = 3(t+\Delta t)^2 + 2(t+\Delta t) = 3t^2 + 6t\Delta t + 3\Delta t^2 + 2t + 2\Delta t$

Step 3: Substitute and simplify.
$v(t+\Delta t) - v(t) = 6t\Delta t + 3\Delta t^2 + 2\Delta t$
$\dfrac{v(t+\Delta t) - v(t)}{\Delta t} = 6t + 3\Delta t + 2$
Taking the limit as $\Delta t \to 0$: $a = 6t + 2$

Step 4: Substitute $t = 2\,s$.
$a = 6(2) + 2 = 14\,m/s^2$

Final Answer: $a = 14\,m/s^2$