A train starts from rest and then accelerates at a constant acceleration of 1.4ms-2 for 10s. The train then moves with a constant velocity for a further 20s before the brakes are applied bringing the train to a stop with a constant deacceleration of 2.8ms-2 Calculate the total distance covered by the train.
Please explain and help
Graph velocity vs time. That area below the graph is the distance covered. You can easily determine the area from basic geometry.
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V1 = a*t = 1.4 * 10 = 14 m/s.
d1 = V1*t = 14 * 20 = 280 m.
V2^2 = V1^2 + 2a*d2 = 0.
14^2 - 5.6d2 = 0
d2 = 35 m.
d = d1+d2 = 280 + 35 = 315 m. = Total distance.
To calculate the total distance covered by the train, we need to break down the motion into different phases and calculate the distance traveled in each phase.
Phase 1: Acceleration
The train starts from rest and accelerates at a constant acceleration of 1.4 m/s^2 for a duration of 10 seconds.
To find the distance covered during this phase, we can use the kinematic equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the train starts from rest, the initial velocity is 0, and the equation simplifies to:
distance = 0.5 * acceleration * time^2
Plugging in the values, we get:
distance1 = 0.5 * 1.4 * (10^2) = 0.5 * 1.4 * 100 = 70 meters
Phase 2: Constant Velocity
After accelerating, the train moves with a constant velocity for 20 seconds. Since the velocity remains constant, there is no change in distance during this phase. Therefore, the distance traveled in this phase is equal to the distance covered at the end of the previous phase, which is 70 meters.
Phase 3: Deceleration
The brakes are applied, and the train comes to a stop with a constant deceleration of 2.8 m/s^2. We need to find the time taken to bring the train to a stop first. Since the initial velocity is the same as the final velocity (0 m/s), we can use the equation:
final velocity = initial velocity + (acceleration * time)
0 = constant velocity + (-2.8 * time)
Solving for time, we get:
time = constant velocity / 2.8
Since the constant velocity was maintained for 20 seconds, we can substitute it in:
time = 20 / 2.8 = 7.14 seconds (approx.)
Now, we can use the kinematic equation again to find the distance covered during deceleration:
distance2 = (constant velocity * time) + (0.5 * deceleration * time^2)
Substituting the values, we get:
distance2 = (0 * 7.14) + (0.5 * (-2.8) * (7.14^2)) = -0.5 * 2.8 * 51.06
Note that the negative sign is used because the displacement is in the opposite direction of the acceleration. This gives us:
distance2 ≈ -71.47 meters
The negative sign indicates that the train traveled in the opposite direction during deceleration. To get the actual distance traveled, we take the absolute value:
distance2 = 71.47 meters
Total distance covered by the train = distance1 + distance2
= 70 + 71.47
≈ 141.47 meters
Therefore, the total distance covered by the train is approximately 141.47 meters.