Question

Find magnitude of acceleration of particle at \(t = 5\,\text{s}\). If velocity–time graph is given as shown in figure. 

• A. \( \dfrac{u_0}{20} \)
• B. \( \dfrac{u_0}{40} \)
• C. \( \dfrac{u_0}{10} \)
• D. \( u_0 \)

Hint:

Slope of \(v-t\) graph gives acceleration: \[ a = \frac{\Delta v}{\Delta t} \] Straight line in \(v-t\) graph means constant acceleration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Acceleration is defined as the rate of change of velocity with respect to time.
On a velocity-time ($v-t$) graph, the acceleration of a particle is given by the slope of the graph.
Step 2: Key Formula or Approach:
Acceleration is calculated as $a = \frac{dv}{dt} = \text{Slope of } v\text{-}t \text{ graph}$.
Step 3: Detailed Explanation:
The $v-t$ graph is a straight line, which implies that the acceleration is constant throughout the motion from $t = 0$ to $t = 40$ s.
To find the slope, take two points on the line:
Point 1: $(t_1, v_1) = (0, u_0)$
Point 2: $(t_2, v_2) = (40, -u_0)$
\[ \text{Slope } a = \frac{v_2 - v_1}{t_2 - t_1} = \frac{-u_0 - u_0}{40 - 0} = \frac{-2u_0}{40} = -\frac{u_0}{20} \]
The magnitude of acceleration is:
\[ |a| = \left| -\frac{u_0}{20} \right| = \frac{u_0}{20} \]
Since acceleration is constant, its magnitude at $t = 5 \text{ sec}$ is also $\frac{u_0}{20}$.
Step 4: Final Answer:
The magnitude of acceleration is u$_0$/20.