= - 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