Certainly! To find the distance travelled by the electric train during its acceleration period, follow these steps:
### Step 1: Convert speeds from km/h to m/s
First, we need to convert the initial and final speeds from kilometers per hour (km/h) to meters per second (m/s).
1. Initial speed conversion:
[tex]\[
\text{Initial speed (in m/s)} = \frac{\text{Initial speed (in km/h)} \times 1000}{3600}
\][/tex]
[tex]\[
= \frac{20 \times 1000}{3600} = 5.56 \text{ m/s}
\][/tex]
2. Final speed conversion:
[tex]\[
\text{Final speed (in m/s)} = \frac{\text{Final speed (in km/h)} \times 1000}{3600}
\][/tex]
[tex]\[
= \frac{30 \times 1000}{3600} = 8.33 \text{ m/s}
\][/tex]
### Step 2: Calculate acceleration
Next, we calculate the acceleration using the formula:
[tex]\[
a = \frac{v - u}{t}
\][/tex]
where:
- [tex]\( v \)[/tex] is the final speed (8.33 m/s),
- [tex]\( u \)[/tex] is the initial speed (5.56 m/s),
- [tex]\( t \)[/tex] is the time in seconds (20 s).
[tex]\[
a = \frac{8.33 - 5.56}{20} = 0.14 \text{ m/s}^2
\][/tex]
### Step 3: Calculate the distance travelled
Now, we use the formula for distance travelled under uniform acceleration:
[tex]\[
s = ut + \frac{1}{2}at^2
\][/tex]
where:
- [tex]\( u \)[/tex] is the initial speed (5.56 m/s),
- [tex]\( a \)[/tex] is the acceleration (0.14 m/s²),
- [tex]\( t \)[/tex] is the time in seconds (20 s).
1. Calculate [tex]\( ut \)[/tex]:
[tex]\[
ut = 5.56 \times 20 = 111.11 \text{ m}
\][/tex]
2. Calculate [tex]\( \frac{1}{2}at^2 \)[/tex]:
[tex]\[
\frac{1}{2}at^2 = \frac{1}{2} \times 0.14 \times 20^2 = 0.07 \times 400 = 27.78 \text{ m}
\][/tex]
3. Sum them up:
[tex]\[
s = 111.11 + 27.78 = 138.89 \text{ m}
\][/tex]
Thus, the distance travelled by the electric train during the period of acceleration is 138.89 meters.
