The position vector of the point P at time t is given by (αtant)i + (αsect)k where α is positive constant and 0≤t≤(Π/2). Show that the velocity and acceleration of P when t = 0 are at right angle to each other. If A is the point with position vector αj, obtain the vector equation for the straight line AP at time t. If the point Q divides AP internally in the ratio (cost) : (1 − cost). Show that the acceleration of the point Q is constant in magnitude and is always directed towards a fixed point.


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Question Number 6820 by Tawakalitu. last updated on 29/Jul/16

$T h e p o s i t i o n v e c t o r o f t h e p o i n t P a t t i m e t i s g i v e n b y$ $(α t a n t) i + (α s e c t) k w h e r e α i s p o s i t i v e c o n s t a n t a n d$ $0 ⩽ t ⩽ \frac{Π}{2} . S h o w t h a t t h e v e l o c i t y a n d a c c e l e r a t i o n o f P w h e n$ $t = 0 a r e a t r i g h t a n g l e t o e a c h o t h e r . I f A i s t h e p o i n t w i t h$ $p o s i t i o n v e c t o r α j, o b t a i n t h e v e c t o r e q u a t i o n f o r t h e s t r a i g h t$ $l i n e A P a t t i m e t . I f t h e p o i n t Q d i v i d e s A P i n t e r n a l l y i n t h e$ $r a t i o (c o s t) : (1 - c o s t) . S h o w t h a t t h e a c c e l e r a t i o n o f t h e p o i n t$ $Q i s c o n s t a n t i n m a g n i t u d e a n d i s a l w a y s d i r e c t e d t o w a r d s a$ $f i x e d p o i n t .$
Commented by Yozzii last updated on 29/Jul/16

$p (t) = (\begin{matrix} α t a n t \\ 0 \\ α s e c t \end{matrix}) α = c o n s t a n t > 0, t \in [0, \frac{π}{2}]$ $\Rightarrow \frac{d p}{d t} (t) = v_{p} (t) = (\begin{matrix} α {s e c}^{2} t \\ 0 \\ α s e c t \times t a n t \end{matrix})$ $\Rightarrow \frac{d^{2} p}{{d t}^{2}} (t) = a_{p} (t) = (\begin{matrix} 2 α {s e c}^{2} t \times t a n t \\ 0 \\ α (s e c t \times t a n t \times t a n t + {s e c}^{3} t) \end{matrix})$ $a_{p} (t) = (\begin{matrix} 2 α {t a n t s e c}^{2} t \\ 0 \\ α ({t a n}^{2} t + {s e c}^{2} t) s e c t \end{matrix})$ $v_{p} (0) = (\begin{matrix} α \\ 0 \\ 0 \end{matrix}), a_{p} (0) = (\begin{matrix} 0 \\ 0 \\ α \end{matrix})$ $∴ v_{p} (0) ∙ a_{p} (0) = α \times 0 + 0 \times 0 + 0 \times α = 0$ $S i n c e v_{p} (0) ∙ a_{p} (0) = 0, t h e v e l o c i t y$ $a n d a c c e l e r a t i o n a r e p e r p e n d i c u l a r$ $t o e a c h o t h e r a t t = 0 .$ $a (t) = (\begin{matrix} 0 \\ α \\ 0 \end{matrix})$ $D i r e c t i o n v e c t o r o f l i n e A P i s p a r a l l e l t o$ $b (t) = p (t) - a (t) = (\begin{matrix} α t a n t \\ - α \\ α s e c t \end{matrix}) = α (\begin{matrix} t a n t \\ - 1 \\ s e c t \end{matrix})$ $S o, t h e v e c t o r e q u a t i o n o f t h e l i n e A P i s$ $r = (\begin{matrix} 0 \\ α \\ 0 \end{matrix}) + μ (\begin{matrix} t a n t \\ - 1 \\ s e c t \end{matrix}) (μ \in R) .$ $F o r Q d i v i d i n g A P i n t h e r a t i o (c o s t) : (1 - c o s t),$ $μ = c o s t (O Q = O A + c o s t (A P))$ $∴ q (t) = (\begin{matrix} 0 \\ α \\ 0 \end{matrix}) + c o s t (\begin{matrix} t a n t \\ - 1 \\ s e c t \end{matrix})$ $q (t) = (\begin{matrix} s i n t \\ α - c o s t \\ 1 \end{matrix})$ $v_{q} (t) = \frac{d q (t)}{d t} = (\begin{matrix} c o s t \\ s i n t \\ 0 \end{matrix})$ $a_{q} (t) = \frac{d v_{q} (t)}{d t} = (\begin{matrix} - s i n t \\ c o s t \\ 0 \end{matrix})$ $\Rightarrow∣ a_{q} (t) ∣= \sqrt{{(- s i n t)}^{2} + {(c o s t)}^{2} + 0^{2}} = 1$ $S i n c e ∣ a_{q} (t) ∣= 1 f o r a l l t \in [0, \frac{π}{2}], t h e$ $a c c e l e r a t i o n o f t h e p o i n t Q i s c o n s t a n t i n$ $m a g n i t u d e .$ $S i n c e t h e a c c e l e r a t i o n o f Q h a s$ $n o c o m p o n e n t p a r a l l e l t o t h e u n i t v e c t o r k,$ $b u t v a r i a b l e c o m p o n e n t s p a r a l l e l t o i a n d j,$ $a_{q} (t) a c t s e n t i r e l y i n t h e p l a n e c o n t a i n i n g$ $b o t h t h e v e c t o r s i a n d j .$ $O b s e r v e t h a t v_{q} (t) ∙ a_{q} (t) = 0$ $\Rightarrow v e l o c i t y a n d a c c l e r a t i o n o f Q$ $a r e n o r m a l t o e a c h o t h e r .$
Commented by Tawakalitu. last updated on 30/Jul/16

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