The displacement (in meters) of a particle moving along x-axis is given by x = 18t + 5t2
.
Calculate:
(i) the instantaneous velocity at t = 2s
(ii) average velocity between t = 2s and t = 3s.
(iii) instantaneous acceleration.

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Solution

Dear student,
$c o n s i d e r i n g t h e d i s p l a c e m e n t (i n m e t e r s) o f a p a r t i c l e m o v i n g a l o n g x - a x i s i s g i v e n b y x = 18 t + 5 t^{2} t h e i n s \tan t e o u s v e l o c i t y a t t = 2 s e c c a n b e c a l c u l a t e d a s; v e l o c i t y v = \frac{d x}{d t} = \frac{d}{d t} (18 t + 5 t^{2}) = 18 + 10 t a t t = 2 s e c i n s \tan t e o u s v e l o c i t y = 18 + 2 \times 10 = 38 m / s e c (i) D i s p l a c e m e n t a t t_{1} = 2 s e c i s x_{1} = 18 \times 2 + 5 \times 2^{2} = 36 + 20 = 56 m a n d, D i s p l a c e m e n t a t t_{2} = 3 s e c i s x_{2} = 18 \times 3 + 5 \times 3^{2} = 99 m a v e r a g e v e l o c i t y = t o t a l d i s p l a c e m e n t i n b e t w e e n 2 s e c a n d 3 s e c / t i m e g a p a v e r a g e v e l o c i t y = (99 - 56) / (3 - 2) = 43 m / s (i i) i n s \tan t e o u s a c c e l e r a t i o n c a n b e c a l c u l a t e d a s a = \frac{d v}{d t} = \frac{d}{d t} (18 + 10 t) = 10 m / s^{2} t h e v a l u e o f a c c l e r a t i o n i s c o n s \tan t o v e r t i m e; a s i t d o e s n o t d e p e n d s o n t i m e s o, i n s \tan t e o u s a c c e l e r a t i o n = 10 m / s^{2} (i i i)$

Regards,

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