Question

A particle moving with an initial velocity of $1\ \text{m s}^{-1}$ has a uniform acceleration of $2\ \text{m s}^{-2}$. The distances travelled by the particle in the first two intervals of $5\ \text{s}$ are respectively:

• A. 30 m and 110 m
• B. 50 m and 110 m
• C. 40 m and 80 m
• D. 30 m and 80 m
• E. 60 m and 160 m

Hint:

To find the distance in the "next" interval, subtract the total distance of the previous intervals from the new total distance.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves kinematics with constant acceleration. "The first two intervals of 5 s" means we need to find the distance covered in the time interval from \( t=0 \) to \( t=5 \) seconds, and the distance covered in the time interval from \( t=5 \) to \( t=10 \) seconds.
Step 2: Key Formula or Approach:
We will use the equation of motion for displacement under constant acceleration:
\[ s = ut + \frac{1}{2}at^2 \] where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Step 3: Detailed Explanation:
Given values:
- Initial velocity, \( u = 1 \text{ m/s} \)
- Uniform acceleration, \( a = 2 \text{ m/s}^2 \)
First 5-second interval (from t=0 to t=5 s):
Let \( s_1 \) be the distance travelled in the first 5 seconds.
\[ s_1 = u(5) + \frac{1}{2}a(5)^2 \] \[ s_1 = (1)(5) + \frac{1}{2}(2)(25) = 5 + 25 = 30 \text{ m} \] Second 5-second interval (from t=5 s to t=10 s):
To find the distance travelled in this specific interval, we can calculate the total distance travelled in 10 seconds and subtract the distance travelled in the first 5 seconds.
First, calculate the total distance travelled in the first 10 seconds (\( s_{10} \)).
\[ s_{10} = u(10) + \frac{1}{2}a(10)^2 \] \[ s_{10} = (1)(10) + \frac{1}{2}(2)(100) = 10 + 100 = 110 \text{ m} \] Now, let \( s_2 \) be the distance travelled in the second 5-second interval.
\[ s_2 = s_{10} - s_1 = 110 \text{ m} - 30 \text{ m} = 80 \text{ m} \] The distances for the first and second 5-second intervals are 30 m and 80 m, respectively.
Step 4: Final Answer:
The distances are 30 m and 80 m.