An object is moving with speed of 6.25 m/s, is decelerated at a rate given by dv/dt = −2.5√v, where v is instantaneous speed. The time taken by the object to come to rest would be in ______ seconds.
Physics · Mixed Chapter
JEE Mains

Question

An object is moving with speed of 6.25 m/s, is decelerated at a rate given by dv/dt = −2.5√v, where v is instantaneous speed. The time taken by the object to come to rest would be in ______ seconds.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The differential equation is dv/dt=−2.5√v and the initial speed is v₀=6.25 m s⁻¹. Step 2: Separate variables: dv/√v=−2.5 dt. Integrate: 2√v=−2.5t+2√v₀. Step 3: At rest v=0; therefore 2.5t=2√6.25=5. Step 4: t=2 s. Answer: 2.