UPSEE Solved Model Paper 7

Given :

dv

dt

= −2.5√v
Re-arranging, we get

dv

√v

= - 2. dt
Integrating the above equation, we have
2√v = −2.5t+x ... (i)
where. C constant of integration
Al t = 0. v = 6.25 m/s [Given]
Thus,
2√6.25 = ( - 2.5 × 0) + C
⇒ C = 2√6.25 = 2 × 2.5
⇒ C = 5
Thus, eq (i) reduces to
2√v = - 2.5 t + 5 ... (ii)
When the object comes to rest, its speed becomes zero, i.e. v = 0 Substituting the value, v = 0 in eq (ii),
we obtain
0 = - 2.5 t + 5
⇒ t =

5

2.5

s
⇒ t = 2 s
This represents the time at which the object comes to rest.
Hence, option 'A' is correct