A trolley moves down a slope from rest with constant acceleration. In the first second of its motion it travels 1.6 m. Its acceleration (in `m//s^(2)`) is

Options

A. `3.2`
B. `1.6`
C. `0.8`
D. `2.4`

Answer

To find the acceleration of the trolley moving down the slope, we can use the kinematic equation for uniformly accelerated motion:

\[ s = ut + \frac{1}{2} a t^2 \]

where:
- \( s \) is the distance traveled,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.

### Step-by-step Solution:

1. **Identify Given Values**:
- The trolley starts from rest, so the initial velocity \( u = 0 \, \text{m/s} \).
- The distance traveled in the first second \( s = 1.6 \, \text{m} \).
- The time taken \( t = 1 \, \text{s} \).

2. **Substitute Values into the Equation**:
Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ s = 0 \cdot t + \frac{1}{2} a t^2 \]
This reduces to:
\[ s = \frac{1}{2} a t^2 \]

3. **Plug in the Known Values**:
Substitute \( s = 1.6 \, \text{m} \) and \( t = 1 \, \text{s} \):
\[ 1.6 = \frac{1}{2} a (1^2) \]

4. **Simplify the Equation**:
This simplifies to:
\[ 1.6 = \frac{1}{2} a \]

5. **Solve for Acceleration \( a \)**:
Multiply both sides by 2 to isolate \( a \):
\[ a = 1.6 \times 2 = 3.2 \, \text{m/s}^2 \]

6. **Conclusion**:
The acceleration of the trolley is \( 3.2 \, \text{m/s}^2 \).

### Final Answer:
The acceleration of the trolley is \( 3.2 \, \text{m/s}^2 \).

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