AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Consider the diagram shown below.
The velocity at point B after 2s is,
v1=u+at1
v1=0+a(2)
v1=2a
The time t2 is,
v2=v1−at2
0=2a−at2
t2=2s
Total time of travel from A to C is,
t=t1+t2
t=(2+2)s
t=4s
The distance between A and B is,
AB=ut1+‌

1

2

at12
AB=(0)(2)+‌

1

2

a(2)2
AB=2a
Thus, AB=BC=2a, and AC=4a
When body return from C to A;u1=0 and Ac=4a
The time taken to return from C to A is,
AC‌‌=u1t′+‌

1

2

at′2
4a‌‌=(0)t′+‌

1

2

at′2
t′‌‌=2√2s
Thus total time taken by body is,
t0=t+t′
t0=(4+2√2)