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 = 8s.
B. The distance travelled in 4 s is approximately 59 m.
C. The distance travelled by the particle in 10 s is 94 m.
D. The velocity at `t = 10 s is 9 ms^(-1)`.

Answer

A, B, C

Explanation

a,b,c. This is the example of non-uniform acceleration.
`a = (dv)/(dt) = -0.5 t`
`int_(16)^(v) dv = -int_(0)^(t) 0.5 t dt rArr 16 - (0.5t^2)/2`
Direction of velocity changes at the moment when it becomes zero momentarily.
`0 = 16 - (0.5t^2)/2 rArr t = 8s`
`(dx)/(dt) = v = 16 - (0.5t)/2`
Let us consider that at `t=0`, particle is at x `= 0`
`int _(0)^(x) dx = int_(0)^(t) (16-(0.5t^2)/2), x = 16t - (0.5t)/6`,
Distance travelled = |displacement for `tle 8s`. So, distance
travelled in `4 s`.
`x = 16 xx 4 - (0.5 xx 4^3)/(6) = 59m`
Distance travelled in `10 s`.
`= |Displacement in 8 s| xx 2 - Displacement in 10 s`
`85.33 xx 2 - 76.55 = 94m`
`v(t = 10s) = -9 ms^(-1)`.