QuestionA particle $P$ has a velocity of $v=24-4 t^{2}$ m/s. Find: (a) distance in the first second, (b) when it changes direction, (c) when it returns to start.

Studdy Solution
olving the equation gives two solutions, t =0 and t = √18 s. Since t =0 is the starting point, the particle returns to its starting point at t = √18 s.
So, the solutions are(a) The distance travelled by in the first second is $\frac{68}{}$ m. (b) The value of t when changes direction of motion is $\sqrt{6}$ s. (c) The value of t at the instant returns to its starting point is $\sqrt{18}$ s.