Answer :
Let's solve the problem step-by-step using equations of motion.
### Given:
1. Acceleration, [tex]\( a = 0.8 \, \text{m/s}^2 \)[/tex]
2. Distance, [tex]\( s = 1 \, \text{km} = 1000 \, \text{m} \)[/tex]
3. Initial velocity, [tex]\( u = 0 \, \text{m/s} \)[/tex] (since the bus starts from rest)
### To find:
1. The final velocity [tex]\( v \)[/tex] of the bus.
2. The time [tex]\( t \)[/tex] it takes to cover the distance of [tex]\( 1.0 \, \text{km} \)[/tex].
### Step 1: Finding the final velocity [tex]\( v \)[/tex]
We can use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Since [tex]\( u = 0 \)[/tex]:
[tex]\[ v^2 = 0 + 2 \cdot 0.8 \cdot 1000 \][/tex]
[tex]\[ v^2 = 1600 \][/tex]
[tex]\[ v = \sqrt{1600} \][/tex]
[tex]\[ v = 40 \, \text{m/s} \][/tex]
Therefore, the final velocity [tex]\( v \)[/tex] of the bus is [tex]\( 40 \, \text{m/s} \)[/tex].
### Step 2: Finding the time [tex]\( t \)[/tex]
We can use another kinematic equation:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Since [tex]\( u = 0 \)[/tex]:
[tex]\[ s = \frac{1}{2}at^2 \][/tex]
Rearranging for [tex]\( t \)[/tex]:
[tex]\[ 1000 = \frac{1}{2} \cdot 0.8 \cdot t^2 \][/tex]
[tex]\[ 1000 = 0.4 \cdot t^2 \][/tex]
[tex]\[ t^2 = \frac{1000}{0.4} \][/tex]
[tex]\[ t^2 = 2500 \][/tex]
[tex]\[ t = \sqrt{2500} \][/tex]
[tex]\[ t = 50 \, \text{s} \][/tex]
Therefore, the time [tex]\( t \)[/tex] it takes to cover [tex]\( 1.0 \, \text{km} \)[/tex] is [tex]\( 50 \, \text{s} \)[/tex].
### Summary
- The final velocity of the bus is [tex]\( 40 \, \text{m/s} \)[/tex].
- The time it takes to cover [tex]\( 1.0 \, \text{km} \)[/tex] is [tex]\( 50 \, \text{s} \)[/tex].
### Given:
1. Acceleration, [tex]\( a = 0.8 \, \text{m/s}^2 \)[/tex]
2. Distance, [tex]\( s = 1 \, \text{km} = 1000 \, \text{m} \)[/tex]
3. Initial velocity, [tex]\( u = 0 \, \text{m/s} \)[/tex] (since the bus starts from rest)
### To find:
1. The final velocity [tex]\( v \)[/tex] of the bus.
2. The time [tex]\( t \)[/tex] it takes to cover the distance of [tex]\( 1.0 \, \text{km} \)[/tex].
### Step 1: Finding the final velocity [tex]\( v \)[/tex]
We can use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Since [tex]\( u = 0 \)[/tex]:
[tex]\[ v^2 = 0 + 2 \cdot 0.8 \cdot 1000 \][/tex]
[tex]\[ v^2 = 1600 \][/tex]
[tex]\[ v = \sqrt{1600} \][/tex]
[tex]\[ v = 40 \, \text{m/s} \][/tex]
Therefore, the final velocity [tex]\( v \)[/tex] of the bus is [tex]\( 40 \, \text{m/s} \)[/tex].
### Step 2: Finding the time [tex]\( t \)[/tex]
We can use another kinematic equation:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Since [tex]\( u = 0 \)[/tex]:
[tex]\[ s = \frac{1}{2}at^2 \][/tex]
Rearranging for [tex]\( t \)[/tex]:
[tex]\[ 1000 = \frac{1}{2} \cdot 0.8 \cdot t^2 \][/tex]
[tex]\[ 1000 = 0.4 \cdot t^2 \][/tex]
[tex]\[ t^2 = \frac{1000}{0.4} \][/tex]
[tex]\[ t^2 = 2500 \][/tex]
[tex]\[ t = \sqrt{2500} \][/tex]
[tex]\[ t = 50 \, \text{s} \][/tex]
Therefore, the time [tex]\( t \)[/tex] it takes to cover [tex]\( 1.0 \, \text{km} \)[/tex] is [tex]\( 50 \, \text{s} \)[/tex].
### Summary
- The final velocity of the bus is [tex]\( 40 \, \text{m/s} \)[/tex].
- The time it takes to cover [tex]\( 1.0 \, \text{km} \)[/tex] is [tex]\( 50 \, \text{s} \)[/tex].
