A scooter can produce a maximum acceleration of 5ms^(-2). Its brakes can produce a maximum retardation of 10ms^(-2). The minimum time in which it can cover a distance of 1.5 km is?

A scooter can produce a maximum acceleration of 5ms^(-2). Its brakes c

1.	A scooter can produce a maximum acceleration of 5ms^(-2). Its brakes can produce a maximum retardation of 10ms^(-2). The minimum time in which it can cover a distance of 1.5 km is?
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP1_C03_SLV_008_S01.png" width="80%"/> <br/> If v is the maximum velocity attained, <br/> then during acceleration is between A, B <br/> `v^(2)-o^(2)=2xx5xxS_(1)impliesS_(1)=(v^(2))/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)`, <br/> Also, during <a href="https://interviewquestions.tuteehub.com/tag/retardation-1187717" style="font-weight:bold;" target="_blank" title="Click to know more about RETARDATION">RETARDATION</a> <br/> `o^(2)-v^(2)=2xx10xxS_(2)impliesS_(2)=(v^(2))/(20)` <br/> `S=S_(1)+S_(2)implies1500=(v^(2))/(10)+(v^(2))/(20)=(3v^(2))/(20)or` <br/> `v^(2)=(1500xx20)/(3)=10000 or v=100ms^(-1)` <br/> `v=alphat_(1)impliest_(1)=(100)/(5)=<a href="https://interviewquestions.tuteehub.com/tag/20sec-293103" style="font-weight:bold;" target="_blank" title="Click to know more about 20SEC">20SEC</a>` <br/> `v=betat_(2)impliest_(2)=(100)/(10)=10sec` <br/> Total time `=20+10=30sec`.</body></html>