Sure, let's solve the problem step-by-step:
1. Bus starting from rest with uniform acceleration:
Given:
- Initial velocity, [tex]\( u = 0 \)[/tex] m/s (since the bus starts from rest)
- Acceleration, [tex]\( a = 0.1 \)[/tex] m/s[tex]\(^2\)[/tex]
- Time, [tex]\( t = 2 \)[/tex] minutes
First, we need to convert the time into seconds:
[tex]\[ t = 2 \times 60 = 120 \text{ seconds} \][/tex]
(a) Speed acquired (Final velocity):
We use the first equation of motion:
[tex]\[ v = u + at \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( t \)[/tex] is the time
Plugging in the values:
[tex]\[ v = 0 + (0.1 \times 120) \][/tex]
[tex]\[ v = 0 + 12 \][/tex]
[tex]\[ v = 12 \text{ m/s} \][/tex]
So, the speed acquired by the bus is 12 m/s.
(b) Distance travelled:
We use the second equation of motion:
[tex]\[ s = ut + \frac{1}{2} a t^2 \][/tex]
Where:
- [tex]\( s \)[/tex] is the distance travelled
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( t \)[/tex] is the time
Plugging in the values:
[tex]\[ s = (0 \times 120) + \frac{1}{2} \times 0.1 \times 120^2 \][/tex]
[tex]\[ s = 0 + \frac{1}{2} \times 0.1 \times 14400 \][/tex]
[tex]\[ s = 0 + 720 \][/tex]
[tex]\[ s = 720 \text{ meters} \][/tex]
So, the distance travelled by the bus is 720 meters.
