[" A "C" ,A body starts from rest has an "],[" acceleration "4m/sec" for a time "],[2sec" and comes to rest under "],[" a retardation "2m/sec^(2)" for a time "],[" 4sec.then average velocity of "],[" the whole trip is "]

A body starts from rest with an acceleration 2m//s^(2) till it attains the maximum velocity then retards to rest with 3m//s^(2) . If total time taken is 10 second, then maximum speed attained is

A bus starts from rest and acceleration at a uniform rate 4 m//s^(2) for certain time. It then moves with a constant speed for some time and finally retards at 4 m//s^(2) to come to rest. Average speed of the bus during the total journey is 15 m/s and the total time is 20 sec. Time duration for which the car moves with constant speed is:

A body moves from rest with a constant acceleration of 5m//s^(2) . Its instantaneous speed (in m/s) at the end of 10 sec is

A particla starting from rest undergoes acceleration given by a=|t-2|" m/s"^(2) where t is time in sec. Find velocity of particle after 4 sec. is :-

A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th sec to that covered in 5 sec is

A particle comes round a circle of radius 1 m once. The time taken by it is 10 sec . The average velocity of motion is

A particles starts from rest and has an acceleration of 2m//s^(2) for 10 sec. After that , it travels for 30 sec with constant speed and then undergoes a retardation of 4m//s^(2) and comes back to rest. The total distance covered by the particle is

A particle starts from rest with a time varying acceleration a=(2t-4) . Here t is in second and a in m//s^(2) Particle comes to rest after a time t= ………… second