Question

Displacement of a particle is given by \( x = t^{3/2} + 2 \). Find the time at which velocity becomes zero.

Hint:

Always double-check the given function. If the source intended a more typical exam problem like $x = t^2 - 2t + 2$, the derivative would be $v = 2t - 2$, yielding $t=1$ s when $v=0$.

Answer 1

Step 1: Understanding the Concept:
Velocity is the rate of change of displacement with respect to time. Mathematically, it is the first derivative of the position function $x(t)$. To find when the velocity is zero, we must find the derivative and set it to zero.

Step 2: Key Formula or Approach:
Velocity $v(t) = \frac{dx}{dt}$.
Set $v(t) = 0$ and solve for $t$.

Step 3: Detailed Explanation:
Given displacement function:
\[ x(t) = t^{3/2} + 2 \]
Find the velocity by differentiating $x$ with respect to $t$:
\[ v(t) = \frac{d}{dt}(t^{3/2} + 2) \]
Using the power rule $\frac{d}{dt}(t^n) = n t^{n-1}$ and knowing the derivative of a constant is 0:
\[ v(t) = \frac{3}{2} t^{(3/2 - 1)} + 0 \]
\[ v(t) = \frac{3}{2} t^{1/2} \]
We need to find the time $t$ when velocity becomes zero:
\[ v(t) = 0 \]
\[ \frac{3}{2} t^{1/2} = 0 \]
Divide both sides by $3/2$:
\[ t^{1/2} = 0 \]
Squaring both sides yields:
\[ t = 0 \]
The velocity is zero at the very beginning of the motion.

Step 4: Final Answer:
The velocity becomes zero at time $t = 0$.