A metro train starts from a station with an acceleration of 2 m/s2 for 8 seconds. It then runs for 12 seconds with the speed acquired. Finally, it s at a uniform rate of 4 m/s2 and stops. Find the distance travelled.
32
188
224m
520m
288 m
288
९२७२४८५४६८१८५७७
paagal hai kya
To find the total distance traveled by the metro train, we need to calculate the distance traveled during each phase of its motion separately.
1. Acceleration phase:
During this phase, the metro train undergoes constant acceleration. We can use the following kinematic equation to find the distance traveled during this phase:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Given:
initial velocity (u) = 0 m/s (the metro train starts from rest)
acceleration (a) = 2 m/s^2
time (t1) = 8 s
Plugging in the values, we have:
distance1 = (0 * 8) + (0.5 * 2 * 8^2)
distance1 = 0 + 64
distance1 = 64 meters
2. Constant speed phase:
During this phase, the metro train runs for 12 seconds with the speed acquired. Since the speed remains constant, we can use the formula:
distance = speed * time
Given:
speed (v) = acquired speed (which we need to find)
time (t2) = 12 s
We need to find the acquired speed. Since the acceleration in this phase is not specified, we can assume that the speed acquired during the acceleration phase is maintained. Therefore, the acquired speed is the final velocity (v) at the end of the acceleration phase.
Using the formula:
v = u + (acceleration * time)
v = 0 + (2 * 8)
v = 16 m/s
Now, we can calculate the distance traveled during this phase:
distance2 = speed * time
distance2 = 16 * 12
distance2 = 192 meters
3. Retardation phase:
During this phase, the metro train s (decelerates) at a uniform rate. The deceleration is given as 4 m/s^2. We can use the kinematic equation again:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Given:
initial velocity (u) = acquired speed (v) from the previous phase = 16 m/s
deceleration (a) = -4 m/s^2 (negative sign because it is deceleration)
time (t3) = unknown
We need to find the time (t3) it takes for the train to stop. The final velocity (v) at the end of this phase is 0 m/s. Thus, the equation becomes:
0 = 16 + (-4 * t3)
Solving for t3, we have:
4t3 = 16
t3 = 16/4
t3 = 4 seconds
Now, we can calculate the distance traveled during this phase using the same kinematic equation:
distance3 = (16 * 4) + (0.5 * -4 * 4^2)
distance3 = 64 - 32
distance3 = 32 meters
Finally, to find the total distance traveled, we add up the distances from all three phases:
total distance = distance1 + distance2 + distance3
total distance = 64 + 192 + 32
total distance = 288 meters
Therefore, the metro train has traveled a total distance of 288 meters.