TS EAMCET 4 May 2018 Shift 2 Solved Paper

Given that,
The deceleration of an object is −3v2∕3m∕s2
since,
a=

dv

dt

=

dv

ds

.

ds

dt

Therefore,
ads=vdv
−3v2∕3ds=vdv
ds=−

1

3

v1∕3dv
Integrate both the sides.
∫ds=∫−

1

3

v1∕3dv
s=−

1

4

v4∕3+C
Substitute the values.
0=−

1

4

(8)4∕3+C
C=4
Therefore, the equation will become as,
s=−

1

4

v

4

3

+4
The object stops when v=0. Therefore,
s=−

1

4

(0)4∕3+4
=4m