Question

A train is accelerated uniformly from rest and attain a maximum speed of 40m/s in 20sec. If it travels at this speed for 20sec and is brought to rest with uniform retardatipn in 40 sec. The average velocity during this period is?

Answer 1

refer the attachment

Answer 2

Solution:-
Initial velocity=0
Final velocity=40m/s
Time=20seconds

So,
Acceleration=(v-u)/t
$= > \frac{(40 - 0)}{20}$
=>2m/s^2
Now,
Distance travelled=
$s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times 20 + \frac{1}{2} \times 2 \times {(20)}^{2} \\ s = 0 + 400 \\ s = 400m$
It then travelled with 40m/s for 20 seconds.
So,
Distance travelled=40×20=800m
Then,
It comes to rest with Retardation in40 seconds.
Retardation=
$a \: = \frac{(v - u)}{t } \\ a = \frac{40 - 0}{40}$
a=1m/s
By using equation =v^2=u^2+2as
Distance again travelled during this time=
$0 = 1600 + 2 \times 2 \times s \\ s = 800m$
Now,
The total distance travelled=400+800+800
=2000m

Total time taken=20+20+40=80second
Since we know that,
Average velocity=total displacement/total time
$= > \frac{2000}{80} \\ = > 25$

Hence,the average speed =25m/s