At time `t = 0,` a car moving along a straight line has a velocity of `16 ms^-1.` It slows down with an acceleration of `-0.5t ms^-2,` where t is in second. Mark the correct statement (s).

Options

A. The direction of the velocity changes at
B. The distance travelled by the particle in 4.0s is ` 58 (2)/(3) m `
C. The distance travelled by the particle in 10.0s is approximately 94 m
D. The speed at t = 10 sec is 9.0 m/s

Answer

A, B, C, D

Explanation

` a(t) =-(t)/(2) =(dv)/(dt) , v(0) = 16 ms ^(-1) `
`rArr int _( 16) ^(v(t) ) dv =int _0^(t) a(t) dt `
` rArr v(t) -16 -(t^(2))/(4) `
Put v=0 to get t=-8 ` rArr " "` Direction of motion reverses at t=8
`rArr v(t=10 ) =-9ms ^(-1)`
` int _0^(x) dx =int _0^(t) (16-t^(2) //4) dt rArr " "x( t) = 16 t-(t^(3))/(12) ` (Assuming x (0) =0m )
During ` 0 le t le 4 ` : The Particle does't change direction .
`rArr ` |Displacement |=distance =x(4) -x(0) =`58 (2)/(3) m.`
During ` 0 le t le 10 : ` The particle change direction at t= 8
`rArr `Distance ` =OA +AB =|Delta x_1|+|Delta x_2|`
` =|x(8) -x(0) |x(10) -x(8) | " " =85 (1)/(2) +8(2)/(3) =94 m`