Question

The acceleration time graph of a particle moving along a straight line is shown in figure. Initial velocity is zero.

• A. Maximum velocity is 4 m/s
• B. Maximum velocity is 10 m/s
• C. Average acceleration during t = 0 to t = 5 s is 2 m/s²
• D. Distance covered in first 2 s is $\frac{8}{3}$m

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

The problem provides an acceleration-time (a-t) graph for a particle moving along a straight line, with an initial velocity of zero ($u=0$). We need to evaluate the given options.

1. Analyze the a-t graph:

• From $t=0$ to $t=2$ s, the acceleration increases linearly from $0$ to $4 \text{ m/s}^2$. The equation for acceleration in this interval is $a(t) = \frac{4-0}{2-0}t = 2t \text{ m/s}^2$.

• From $t=2$ to $t=5$ s, the acceleration decreases linearly from $4 \text{ m/s}^2$ to $0 \text{ m/s}^2$. The equation for acceleration in this interval can be found using two points $(2,4)$ and $(5,0)$:
  Slope $m = \frac{0-4}{5-2} = -\frac{4}{3} \text{ m/s}^3$.
  Equation: $a - 0 = -\frac{4}{3}(t - 5) \implies a(t) = -\frac{4}{3}(t - 5) \text{ m/s}^2$.

2. Evaluate Option A and B (Maximum velocity):

The change in velocity ($\Delta v$) is given by the area under the a-t graph. Since the initial velocity is zero, the velocity at any time $t$ is equal to the area under the a-t graph from $0$ to $t$.
Since the acceleration is always non-negative ($a \ge 0$), the velocity will continuously increase or remain constant. Therefore, the maximum velocity will occur at the end of the motion, i.e., at $t=5$ s.
The total area under the a-t graph from $t=0$ to $t=5$ s is the area of the triangle:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (5 \text{ s}) \times (4 \text{ m/s}^2) = 10 \text{ m/s}$.
Since $u=0$, the maximum velocity $v_{max} = 0 + 10 \text{ m/s} = 10 \text{ m/s}$.

• Option A: Maximum velocity is 4 m/s (Incorrect).

• Option B: Maximum velocity is 10 m/s (Correct).

3. Evaluate Option C (Average acceleration):

Average acceleration is defined as the total change in velocity divided by the total time interval.
Average acceleration $= \frac{\Delta v}{\Delta t} = \frac{v(5) - v(0)}{5 \text{ s} - 0 \text{ s}}$.
From step 2, $v(5) = 10 \text{ m/s}$ and $v(0) = 0 \text{ m/s}$.
Average acceleration $= \frac{10 \text{ m/s} - 0 \text{ m/s}}{5 \text{ s}} = \frac{10}{5} \text{ m/s}^2 = 2 \text{ m/s}^2$.

• Option C: Average acceleration during $t = 0$ to $t = 5$ s is 2 m/s² (Correct).

4. Evaluate Option D (Distance covered in first 2 s):

To find the distance covered, we first need to find the velocity function $v(t)$ for $0 \le t \le 2$ s, and then integrate it.
For $0 \le t \le 2$ s, $a(t) = 2t$.
Velocity $v(t) = \int a(t) dt = \int 2t dt = t^2 + C_1$.
Given initial velocity $v(0) = 0$, so $0^2 + C_1 = 0 \implies C_1 = 0$.
Thus, $v(t) = t^2$ for $0 \le t \le 2$ s.
Distance covered in the first 2 s is $x(2) = \int_0^2 v(t) dt = \int_0^2 t^2 dt$.
$x(2) = \left[ \frac{t^3}{3} \right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \text{ m}$.

• Option D: Distance covered in first 2 s is $\frac{8}{3}$m (Correct).

Conclusion:
Options B, C, and D are all correct.