To solve this problem of a bus accelerating from rest with uniform acceleration, follow the steps outlined below:
### Given Data: 1. Initial speed of the bus ([tex]\(u\)[/tex]): 0 m/s (since the bus starts from rest) 2. Acceleration ([tex]\(a\)[/tex]): 0.1 m/s² 3. Time ([tex]\(t\)[/tex]): 2 minutes
First, we need to convert the time from minutes to seconds because the acceleration is given in meters per second squared.
To find the speed acquired by the bus ([tex]\(v\)[/tex]) after 2 minutes of acceleration, we use the first equation of motion: [tex]\[ v = u + at \][/tex]
Substitute the given values: [tex]\[ u = 0 \text{ m/s} \][/tex] [tex]\[ a = 0.1 \text{ m/s}^2 \][/tex] [tex]\[ t = 120 \text{ seconds} \][/tex]
Calculate the speed: [tex]\[ v = 0 + (0.1 \times 120) \][/tex] [tex]\[ v = 12 \text{ m/s} \][/tex]
So, the speed acquired by the bus is [tex]\(12 \text{ m/s}\)[/tex].
### (b) Calculate the Distance Travelled:
To find the distance travelled by the bus ([tex]\(s\)[/tex]), we use the second equation of motion: [tex]\[ s = ut + \frac{1}{2} a t^2 \][/tex]
Again, substitute the given values: [tex]\[ u = 0 \text{ m/s} \][/tex] [tex]\[ a = 0.1 \text{ m/s}^2 \][/tex] [tex]\[ t = 120 \text{ seconds} \][/tex]
Calculate the distance: [tex]\[ s = (0 \times 120) + \frac{1}{2} \times 0.1 \times (120)^2 \][/tex] [tex]\[ s = 0 + 0.05 \times 14400 \][/tex] [tex]\[ s = 720 \text{ meters} \][/tex]
So, the distance travelled by the bus is [tex]\(720 \text{ meters}\)[/tex].
### Summary: (a) The speed acquired by the bus after 2 minutes is [tex]\(12 \text{ m/s}\)[/tex]. (b) The distance travelled by the bus after 2 minutes is [tex]\(720 \text{ meters}\)[/tex].
