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A train 200m long enters at 5m/s and exits at 3m/s from a bridge 300m long. The time taken by it is?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The time taken by it is 125s. Consequently, the distance traveled overall by train to exit is (200 m + 300 m) = 500
vti = u = 5 m/s
vtf = v = 3 m/s
Assuming that the train is going at a constant rate of acceleration:
Using v2 = u2 + 2as
Substituting the value we get
32 = 52 + 2a x 500
1000a = 16
a = 16/1000 m/s2 = -4/250 = -2/125 m/s2 (-ve acceleration means retardation)
Now using:
v = u + at
Substituting the value we get:
3 = 5 – 2/125 t
2/125 t = 2
t = (2 x 125)/2
t = 125 s
Since acceleration has both a magnitude and a direction, it is a vector quantity. Additionally, it is the first derivative of velocity with respect to time or the second derivative of position with respect to time.
Table of content
Types of acceleration
Average acceleration: The slope of the line connecting time points t1 and t2 in the velocity-time graph above represents the average value for the rate at which the object’s velocity changed between t1 and t2.
In a velocity-time curve, the instantaneous acceleration is determined by the slope of the tangent at any given time.
Summary:
A train 200m long enters at 5m/s and exits at 3m/s from a bridge 300m long. The time taken by it is?
A train 200m long enters at 5m/s and exits at 3m/s from a bridge 300m long. The time taken by it is 125s. Time can be defined as the dimension based on which the evolution of any system takes place.