An electric train moving at 20kmh accelerates to a speed of 30kmh in 20 seconds.find the distance travelled in metres during the period of acceleration
d = Vi t + (1/2) a t^2
Vi = 20 /3.6 = 5.56 m/s
V final = 30/3.6 = 8.33 m/s
oh well, I do not need the physics equation because
Vaverage = (1/2)(5.56+8.33)
= 6.94 m/s
6.94 m/s * 20 s = 139 meters
Can't solve
I need the solving of the question
Well, you could say the train was really amped up during that acceleration!
To find the distance, we can use the equation: distance = average speed × time
During the acceleration, the train starts at 20 km/h and ends at 30 km/h, so the average speed is halfway between them, which is 25 km/h.
Now, let's convert everything to meters and seconds to make our calculations easier.
20 km/h is equivalent to 20,000 meters per hour, or 20,000/3600 meters per second, which is approximately 5.56 m/s.
30 km/h is equivalent to 30,000 meters per hour, or 30,000/3600 meters per second, which is approximately 8.33 m/s.
The average speed during acceleration is 25 km/h, which is equivalent to 25,000 meters per hour, or 25,000/3600 meters per second, approximately 6.94 m/s.
Now, let's plug in the values into our equation:
distance = average speed × time
distance = 6.94 m/s × 20 s
Multiplying 6.94 m/s by 20 s, we get a distance traveled of approximately 138.8 meters.
So, during the period of acceleration, the electric train traveled around 138.8 meters. It was really buzzing along!
To find the distance traveled during the period of acceleration, you can use the equation:
Distance = Initial Velocity * Time + 0.5 * Acceleration * Time^2
In this case, the initial velocity is 20 km/h, the final velocity is 30 km/h, and the time is 20 seconds. However, we need to convert the velocities from km/h to m/s because the acceleration is usually given in m/s^2.
To convert km/h to m/s, divide by 3.6 (since 1 km/h = 1000 m/3600 s = 1/3.6 m/s).
So, the initial velocity is 20 km/h * (1/3.6) = 5.56 m/s,
and the final velocity is 30 km/h * (1/3.6) = 8.33 m/s.
Now let's substitute these values into the equation:
Distance = (5.56 m/s) * (20 s) + 0.5 * Acceleration * (20 s)^2
Since we are given the initial and final velocities, we can use them to find the acceleration.
Acceleration = (Final Velocity - Initial Velocity) / Time
Acceleration = (8.33 m/s - 5.56 m/s) / 20 s
Acceleration = 0.1385 m/s^2
Now let's plug the acceleration back into the distance equation:
Distance = (5.56 m/s) * (20 s) + 0.5 * (0.1385 m/s^2) * (20 s)^2
Distance = 111.2 m + 0.5 * 0.1385 m/s^2 * 400 s^2
Distance = 111.2 m + 27.7 m
Distance = 138.9 meters
So, the distance traveled during the period of acceleration is 138.9 meters.