Question

A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/s when it passes a railway worker who is standing 195m from where the front of the train started (see the figure below).a.What will be the speed of the last car as it passes the worker?b.Express your answer to two significant figures and include the appropriate units.

Answer 1

The speed of the last car is 21.18 m/s.

Step-by-step explanation:

Given that,

Length of train = 75 m

Speed of train= 18 m/s

Distance 195 m

Suppose the acceleration is constant.

We need to calculate the speed of the last car

Using equation of motion

`v^2=u^2+2as`

`a=(v^2-u^2)/(2s)`

For train and car,

`(v^2-u^2)/(2s)=(v'^2-u'^2)/(2s')`

Where, v = final velocity of train

u = initial velocity

s = distance

u'= speed of train

s'=distance

Put the value into the formula

`(18^2-0)/(2*195)=(v'^2-18^2)/(2*75)`

`v'^2=(18^2*2*75)/(2*195)+18^2`

`v'=√(448.61)`

`v'=21.18\ m/s`

Hence, The speed of the last car is 21.18 m/s.