Question

A particle starts from rest. Its acceleration ‘a’ versus time ‘t’ is shown in the figure. The maximum speed of the particle will be:

• A. 80 ms^-1
• B. 40 ms^-1
• C. 18 ms^-1
• D. 2 ms^-1

Answer 1

The Correct Option is B

1. Understand the Graph:
Maximum acceleration of $ 8 \, \text{m/s}^2 $ at $ t = 0 $,
Decreasing linearly to $ 0 \, \text{m/s}^2 $ at $ t = 10 \, \text{s} $.
This suggests that the acceleration decreases uniformly from $ 8 \, \text{m/s}^2 $ to $ 0 \, \text{m/s}^2 $ over the time interval $ t = 0 $ to $ t = 10 \, \text{s} $.

2. Use the Formula for Maximum Velocity: 
The velocity is the integral of acceleration with respect to time: $$ v = \int a \, dt $$ The area under the acceleration-time graph gives the total change in velocity. Since the graph is a right triangle, we can compute the area as: $$ \text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height}. $$ Here:

• The base is the time interval, $ t = 10 \, \text{s} $,
• The height is the maximum acceleration, $ a = 8 \, \text{m/s}^2 $.

Thus, the area is: $$ \text{Area} = \frac{1}{2} \times 10 \, \text{s} \times 8 \, \text{m/s}^2 = 40 \, \text{m/s}. $$

3. Final Answer: 
The maximum speed of the particle will be $40 \, \text{m/s}$

Answer 2

The area under the acceleration-time graph represents the change in velocity. The particle starts from rest, so its initial velocity ($v_i$) is 0 m/s. The graph shows a triangle with a base of 10 seconds and a height of 8 m/s².

The formula for the area of a triangle is:

Area = $\frac{1}{2} \times \text{base} \times \text{height}$

In this case, the area represents the change in velocity ($\Delta v$):

$\Delta v = \frac{1}{2} \times 10\,\text{s} \times 8\,\text{m/s}^2 = 40\,\text{m/s}$

Since the initial velocity was 0 m/s, the final velocity ($v_f$), which is also the maximum speed, is equal to the change in velocity:

$v_f = v_i + \Delta v = 0\,\text{m/s} + 40\,\text{m/s} = 40\,\text{m/s}$

The correct answer is (B) 40 ms^-1.