Question

The position vector of a particle P of mass 6kg at time t seconds is r where r=[(18t-4t^3)I+ t^2j]m determine to 2 decimal places the value of t for which the velocity and the acceleration vectors of P are at right angles

Answer 1

$\overrightarrow{r}$ = [(18t-4t³)i+ t²j] m

Velocity will be

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}= [ (18 - 12t²)i + 2tj]$ m/s

Acceleration will be

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}= [ (- 24t)i + 2j]$ m/s²

Since velocity vector and acceleration vector are perpendicular $\overrightarrow{v}.\overrightarrow{a}$ = 0

=> $[ (18 - 12t²)i + 2tj].[ (- 24t)i + 2j]=0$

=> [ -24t(18-12t²)+4t ]= 0

=> 4[-6t(18-12t²)+t] = 0

=> 72t³-108t+t = 0

=> 72t³ = 107t

=> 72t² = 107 as t ≠ 0

=> t = $\sqrt{\frac{107}{72}}$

=> t = 1.22 second [ correct up to 2 decimal places]