A particle starts from rest and has acceleration of 2 ms-2 for 10 s. After that it travels for 30 s. with constant speed and then undergoes a constant retardation of 4 ms and comes back to rest. The total distance covered by the particle is:

Options

A. 650 m
B. 700 m
C. 750 m
D. 800 m

Answer

c

Explanation

Let`v_1` be the final vel. In first case, then applying
`v=u+at`
`v_1=0+2xx10 =20 ms^(-1)`
Also `s_1=(V_(1)^(2)-u_(1)^(2))/(2a)=(20xx20-0)/(2xx2)=100m`
In second case ,`v_(2)`be the final vel.Which remains
constant `s_(2)=v_(2)xxt(2)`
`=20xx30=600 m`
In third case `v_(3)=0,u_(3)=20 ms^(-1)`
`a=-4 ms^(-2)`
Now `t=(v_(3)-u_(3))/(a)=(-20)/(-4)`
Total distance =`S_1+S_2+S_3`=100+600+50=750 m