TS EAMCET 5 May 2018 Shift 2 Solved Paper

The velocity attained by the particle at the end of the time is calculated as,
v1=u+at1
=4t1
Therefore, the displacement in first t1 seconds is given by,
s1=

1

2

at12
=

1

2

×4×t12
=2t12
The displacement in the next t2 seconds is given by,
s2=v1t2
=4t1t2
For third part of journey,
Initial velocity is v1=4t1, final velocity is 0 and time interval t1
Thus, acceleration is given by,
a=

0−4t1

t1

=−4ms−2
Thus, displacement in the third part is,
s3=

−16t12

−8

=2t12
Therefore, v22−v12=2as3
Now, average velocity is calculated as,
vavg =

Total displacement

Total time

5.1=

s1+s2+s3

t1+t2+t1

5.1=

2t12+4t1t2+2t12

10

51=4t12+4t1(10−2t1)
Solve further.
4t12+40t1+51=0
(2t1−3)(2t1−17)=0
t1=

3

2

or t1=

17

2

t1=1.5s
As, t1=8.5s is not possible. It exceeds total time.