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 velocity changes at `t = 8 s`
B. The distance travelled in 4 s is approximately 58.67 m
C. The distance travelled by the particle in 10 s is 94 m
D. The speed of particle at `t = 10` s is `9 ms^-1`

Answer

A, B, C, D

Explanation

`a=-0.5 t=(dv)/(dt)`
`:. int_16^vdv=int_0^t-0.5tdt`
`:. v=16-0.25t^2`
`s=int vdt=16t-(0.25 t^3)/3`
`v=0` when `16-0.25t^2=0`
or `t=8s`
So direction of velocity changes at 8s. Up to 8s
distance =displacement
`:.` At `4s`
`d=16xx4-(0.25xx(4)^3)/3=58.67m`
`S_(8s)=16xx8-((0.25)(8)^3)/3`
`=85.33 m`
`S_(10s)=16xx10-((0.25)(10)^3)/3`
`=76.67 cm`

Distance travelled in 10s,
`d=(85.33)+(85.33-76.67)`
`=94 m`
At 10s
`v=16-0.25(10)^2=-9 m//s`
`:. Speed =9 m//s`