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

Answer

c

Explanation

From O to A, `a=2m//s^(2),t_(1)=10s`
`v=u+at_(1)`
`=0+2xx10`
=20m/s
From A to B, v=20m/s till `t_(2)=40 s`. At point C, v=0 and from B to C, a`=-4m//s^(2)`
`therefore v=u+at_(3)`
`0=20-4t_(3)`
`t_(3)=5` seconds
Total distance covered by the particle
Area under v-t graph
`=(1)/(2)xx20xx10+30xx20+(1)/(2)xx20xx5`
`100+600+50`