Sure, let's solve this problem step by step.
### Given:
1. Initial velocity, [tex]\( u \)[/tex] = 18 km/h
2. Final velocity, [tex]\( v \)[/tex] = 36 km/h
3. Time, [tex]\( t \)[/tex] = 5 seconds
### To find:
1. Acceleration, [tex]\( a \)[/tex]
2. Distance covered, [tex]\( s \)[/tex]
### Step-by-Step Solution:
#### Step 1: Convert velocities from km/h to m/s
- Initial velocity, [tex]\( u \)[/tex]:
[tex]\[
u = 18 \, \text{km/h} = 18 \times \frac{1000}{3600} \, \text{m/s} = 5 \, \text{m/s}
\][/tex]
- Final velocity, [tex]\( v \)[/tex]:
[tex]\[
v = 36 \, \text{km/h} = 36 \times \frac{1000}{3600} \, \text{m/s} = 10 \, \text{m/s}
\][/tex]
#### Step 2: Calculate acceleration, [tex]\( a \)[/tex]
Acceleration is given by the formula:
[tex]\[
a = \frac{v - u}{t}
\][/tex]
Substitute the values:
[tex]\[
a = \frac{10 \, \text{m/s} - 5 \, \text{m/s}}{5 \, \text{seconds}} = \frac{5 \, \text{m/s}}{5 \, \text{seconds}} = 1 \, \text{m/s}^2
\][/tex]
#### Step 3: Calculate the distance covered, [tex]\( s \)[/tex]
The distance covered under uniform acceleration is given by the formula:
[tex]\[
s = ut + \frac{1}{2} a t^2
\][/tex]
Substitute the values:
[tex]\[
s = 5 \, \text{m/s} \times 5 \, \text{seconds} + \frac{1}{2} \times 1 \, \text{m/s}^2 \times (5 \, \text{seconds})^2
\][/tex]
Calculate each part:
- [tex]\( 5 \, \text{m/s} \times 5 \, \text{seconds} = 25 \, \text{meters} \)[/tex]
- [tex]\( \frac{1}{2} \times 1 \, \text{m/s}^2 \times 25 \, \text{seconds}^2 = 0.5 \times 25 = 12.5 \, \text{meters} \)[/tex]
Now, sum these values:
[tex]\[
s = 25 \, \text{meters} + 12.5 \, \text{meters} = 37.5 \, \text{meters}
\][/tex]
### Final Answers:
1. Acceleration, [tex]\( a \)[/tex] = 1 m/s²
2. Distance covered, [tex]\( s \)[/tex] = 37.5 meters
And that's our detailed solution!
