Question

A train travelling at a speed of 90km h weight is applied so as to produce a uniform acceleration of -0.5 m/s square .find how far the train will go before it is brought to rest

Answer 1

Answer:

please refer the above pic

Answer 2

Answer:

Using equation of motion for calculating the time in which the train is brought to rest:

v

2

=u

2

+2as

⇒a=−54000km/h

2

0

2

=90

2

+2a(75×10

−3

)

v=u+at

0=90+(−54000)t

⇒t=0.00166h=0.00166×60min=0.1min=0.1×60s=6s

So, the train comes to rest in 6 s.

Distance traveled in the first half, i.e., distance traveled in 3s:

v=90+(−54000)(

3600

3

)=45km/h

(45)

2

=90

2

+2(−54000)s

s=56.25m

Thus, distance travelled in second half = (75-56.25)m = 18.75 m