Example 8.6

A car accelerates uniformly from [tex]\(18 \, \text{km/h}\)[/tex] to [tex]\(36 \, \text{km/h}\)[/tex] in 5 seconds.

Calculate:
(i) The acceleration
(ii) The distance covered by the car in that time

Solution:



Answer :

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!

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