A motor cycle moving with a speed of `5 m//s` is subjected to an acceleration of `0.2 m//s^(2)`. Calculate the speed of the motor cycle after 10 second, and the distance travelled in this time.

Explanation

To solve the problem step by step, we will use the equations of motion.

### Given:
- Initial speed (u) = 5 m/s
- Acceleration (a) = 0.2 m/s²
- Time (t) = 10 seconds

### Step 1: Calculate the final speed (v) after 10 seconds
We will use the first equation of motion:
\[ v = u + at \]

Substituting the values:
\[ v = 5 \, \text{m/s} + (0.2 \, \text{m/s}^2 \times 10 \, \text{s}) \]

Calculating the acceleration term:
\[ 0.2 \, \text{m/s}^2 \times 10 \, \text{s} = 2 \, \text{m/s} \]

Now substituting this back into the equation:
\[ v = 5 \, \text{m/s} + 2 \, \text{m/s} = 7 \, \text{m/s} \]

### Step 2: Calculate the distance travelled (s) in 10 seconds
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]

Substituting the values:
\[ s = (5 \, \text{m/s} \times 10 \, \text{s}) + \frac{1}{2} (0.2 \, \text{m/s}^2) (10 \, \text{s})^2 \]

Calculating the first term:
\[ 5 \, \text{m/s} \times 10 \, \text{s} = 50 \, \text{m} \]

Now calculating the second term:
\[ \frac{1}{2} \times 0.2 \, \text{m/s}^2 \times 100 \, \text{s}^2 = 0.1 \times 100 = 10 \, \text{m} \]

Now substituting these back into the equation:
\[ s = 50 \, \text{m} + 10 \, \text{m} = 60 \, \text{m} \]

### Final Answers:
- Speed of the motorcycle after 10 seconds: **7 m/s**
- Distance travelled in this time: **60 m**

---