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

Options

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

Answer

B

Explanation

Distance covered in first 10 seconds
`=1/2at^(2) =1/2 xx 2 xx 10^(2) = 100 m ( :. u = 0) `
velocity after 10 seconds ,
v = u + at `= 0+2xx2 10 = 20 ms^(-1)`
Distance covered in 30 seconds `= 20 xx 30 = 600 m`
During retardation , distance travelled
`= (v^2-u^2)/(2a)=(0-(20)^2)/(2(-4))=50m`
Total distance ` = 100 + 600 + 50 = 750 m`