A particlesstarts from rest and has an acceleration of2m//s^(2)for 10 sec. After that , it travels for 30 sec with constant speed and then undergoes a retardation of4m//s^(2) and comes back to rest.The total distance covered by the particle is

A particlesstarts from rest and has an acceleration of2m//s^(2)for 10

1.	A particlesstarts from rest and has an acceleration of2m//s^(2)for 10 sec. After that , it travels for 30 sec with constant speed and then undergoes a retardation of4m//s^(2) and comes back to rest.The total distance covered by the particle is
Answer» Solution :Here, `U = 0, a = 2 ms^(–2), t = 10 s" USING "v = u +" at "= 0 +2 xx 10 = 20" ms"^(–1)` For next `t = 30 s, v = 20" ms"^(–1)` and finally the particle comes to rest in next 5S. `because" "v=u + at" or "0=20-4t" or "t=5s` Now, distance = AREA of the TRAPEZIUM OABC `=(1)/(2) xx (30+45) xx20 =750 m`

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A particlesstarts from rest and has an acceleration of2m//s^(2)for 10 sec. After that , it travels for 30 sec with constant speed and then undergoes a retardation of4m//s^(2) and comes back to rest.The total distance covered by the particle is

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