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109, For a train engine moving with speed of 20 ms, the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed Vx ms. The value of x is
(Assuming same retardation is produced by brakes) [JEE (Main) -2023]?
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109, For a train engine moving with speed of 20 ms, the driver must ap...
Problem Overview
To determine the speed of the train (Vx) when brakes are applied at half the required distance, we will use the equations of motion.
Initial Conditions
- Speed (u) = 20 m/s
- Distance (s) = 500 m
- Final speed (v) = 0 m/s (at the station)
Using the Equation of Motion
We can use the equation:
v² = u² + 2as
Where:
- v = final velocity (0 m/s)
- u = initial velocity (20 m/s)
- a = acceleration (deceleration in this case)
- s = distance (500 m)
Rearranging gives:
0 = (20)² + 2a(500)
Solving for a:
400 = -1000a
=> a = -0.4 m/s²
Braking at Half Distance
When the brakes are applied at half the distance (250 m):
Using the same equation again:
v² = u² + 2as
Let s = 250 m, u = 20 m/s, and a = -0.4 m/s²
We can substitute:
v² = (20)² + 2(-0.4)(250)
Calculating:
v² = 400 - 200
=> v² = 200
=> v = √200
Final Calculation
Therefore, Vx = √200 = 10√2 m/s.
The value of x is 10√2, which is approximately 14.14 m/s.
Conclusion
The train will cross the station with a speed of approximately 14.14 m/s if the brakes are applied at half the original distance.
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109, For a train engine moving with speed of 20 ms, the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed Vx ms. The value of x is(Assuming same retardation is produced by brakes) [JEE (Main) -2023]?
Question Description
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