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 Ritesh|Updated : November 13th, 2022

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.

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.