Question

Force acting on a particle varies with time \( t \) as \( F = kt \), where \( k \) is a constant. The velocity of the particle after a time \( t \) is

Hint:

Remember the hierarchy: Position $\xrightarrow{\text{differentiate}}$ Velocity $\xrightarrow{\text{differentiate}}$ Acceleration (Force/m).
Acceleration (Force/m) $\xrightarrow{\text{integrate}}$ Velocity $\xrightarrow{\text{integrate}}$ Position.

Answer 1

Step 1: Understanding the Concept:
According to Newton's Second Law of Motion, force is related to acceleration. Acceleration is the rate of change of velocity. To find velocity from a time-varying force, we need to integrate.

Step 2: Key Formula or Approach:
Newton's Second Law: $F = m \cdot a = m \frac{dv}{dt}$
Rearrange to solve for $dv$: $dv = \frac{F}{m} dt$
Integrate both sides to find velocity $v(t)$.

Step 3: Detailed Explanation:
Given force $F = kt$.
Substitute this into the equation:
\[ m \frac{dv}{dt} = kt \]
\[ dv = \frac{k}{m} t \ dt \]
Integrate to find the velocity at time $t$. Typically, for such problems, unless stated otherwise, we assume the particle starts from rest, so initial velocity $v(0) = 0$ at $t=0$.
\[ \int_0^v dv' = \int_0^t \frac{k}{m} t' \ dt' \]
\[ [v']_0^v = \frac{k}{m} \left[\frac{t'^2}{2}\right]_0^t \]
\[ v - 0 = \frac{k}{m} \left(\frac{t^2}{2} - 0\right) \]
\[ v = \frac{kt^2}{2m} \]

Step 4: Final Answer:
The velocity after time t is $\frac{kt^2}{2m}$.