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Problem Solving of Rectilinear Equation

T h e p a t h t a k e n b y a p a r t i c l e i s a h o r i z o n t a l l i n e a s i t

S t e p 1 : G e t t h e D e r i v a t i v e o f s = \frac{1}{6} t^{4} - \frac{7}{6} t^{3} - 7 t^{2} + \frac{1}{2} t + 1

v = \frac{2}{3} t^{3} - \frac{7}{2} t^{2} - 14 t + \frac{1}{2}

S t e p 2 : G e t t h e D e r i v a t i v e o f v = \frac{2}{3} t^{3} - \frac{7}{2} t^{2} - 14 t + \frac{1}{2}

a = 2 t^{2} - 7 t - 14

S t e p 3 : E q u a t e t h e a c c e l e r a t i o n t o o n e

2 t^{2} - 7 t - 14 = 1

S t e p 4 : W h e n t = 5 a t v = \frac{2}{3} t^{3} - \frac{7}{2} t^{2} - 14 t + \frac{1}{2}

a (5) = \frac{2}{3} (5)^{3} - \frac{7}{2} (5)^{2} - 14 (5) + \frac{1}{2}

F i n a l A n s w e r : t = 5, a (5) = - \frac{221}{3}

Solving a Rectilinear Motion

A p a r t i c l e m o v e s h o r i z o n t a l l y a c c o r d i n g t o t h e l a w s = \frac{1}{3} t^{3} - \frac{1}{2} t^{2} - 12 t + 1.

A t w h a t v e l o c i t y i s t h e p a r t i c l e m o v i n g w h e n t = 1 ? w h e n w i l l t h e p a r t i c l e

c o m e t o r e s t a n d w h a t i s i t s a c c e l e r a t i o n a t t h a t t i m e ?

S t e p 1 : G e t t h e d e r i v a t i v e o f s = \frac{1}{3} t^{3} - \frac{1}{2} t^{2} - 12 t + 1 t o g e t t h e v e l o c i t y

S t e p 2 : A t w h a t v e l o c i t y i s p a r t i c l e m o v i n g w h e n t = 1 ?

S t e p 3 : G e t t h e f a c t o r o f t h e v e l o c i t y e q u a t i o n,

S t e p 4 : W h a t i s i t s a c c e l e r a t i o n a t t h a t t i m e ?

t^{2} - t - 12

when t = 4

F i n a l A n s w e r : v = - 12, t = 4, a = 7

y = \frac{x^{2} - 4}{x + 1} a t t h e p o i n t (0, - 4)

S t e p 1 : G e t t h e D e r i v a t i v e b y A p p l y i n g Q u o t i e n t R u l e

Given :

\frac{d y}{dx} = \frac{x + 1 (2 x) - x^{2} - 4 (1)}{(x + 1)^{2}}

S t e p 2 : G e t t h e S l o p e a t p o i n t (0, - 4)

S t e p 3 : A p p l y t h e e q u a t i o n o f a N o r m a l L i n e at point (0, - 4)

F i n a l A n s w e r : x + 4 y + 16 = 0

y = (x^{2} - 1)^{2} a t t h e p o i n t (- 2, 9)

S t e p 1 : G e t t h e D e r i v a t i v e b y A p p l y i n g t h e G e n e r a l P o w e r R u l e

S t e p 2 : G e t t h e s l o p e a t p o i n t (- 2, 9)

S t e p 3 : A p p l y t h e E q u a t i o n o f N o r m a l L i n e a t p o i n t (- 2, 9)

y-216=x+2

24y-216-x-2=0

F i n a l A n s w e r : x - 24 y + 218 = 0

y = \sqrt{x^{2} - 25} a t t h e p o i n t (13, 12)

S t e p 1 : G e t t h e d e r i v a t i v e o f y = \sqrt{x^{2} - 25}

A p p l y t h e D e r i v a t i v e o f S q u a r e R o o t R u l e

S t e p 2 : G e t t h e s l o p e a t p o i n t (13, 12)

S t e p 3 : U s i n g t h e p o i n t - s l o p e f o r m a t p o i n t (13, 12)

F i n a l A n s w e r : 13 x - 12 y - 25 = 0

y = x^{3} (x - 1)^{4} a t t h e p o i n t (2, 8)

S t e p 1 : F i n d t h e D e r i v a t i v e b y a p p l y i n g t h e P r o d u c t o f T w o F a c t o r s R u l e

S t e p 2 : U s i n g t h e p o i n t s l o p e f o r m a t p o i n t (2, 8)

F i n a l A n s w e r : 44 x - y - 80 = 0