TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

The given situation of entire motion of motor bike is shown in the figure.
If time taken by the motorbike to reach from point A to B is tAB, then for A to B,
vA=0,vB=10ms−1,a=0.5ms−2
∴ From equation,v=u+at
vB=vA+atAB⇒10=0+0.5tAB
⇒
tAB=‌

10

0.5

=20s
Since, motorbike moves from point B to C with constant velocity.
Hence, time taken to travel distance BC is given as
tBC=‌

BC

vB

=‌

10km

10ms−1

=‌

10000m

10ms−1

=1000s
For motion of motorbike from point C to D, time =tCD,vB=10ms−1,vf=0,a′=0.2ms−2
From equation, v=u−at,
vf=vB−a′tCD
0=10−0.2×tCD⇒tCD=50s
∴‌ Total time ‌=‌‌tAB+tBC+tCD
=‌‌20s+1000s+50s=1070s