Three-villages-A-B-and-C-are-on-a-straight-road-and-B-is-the-mid-way-between-A-and-C-A-motor-cyclist-moving-with-a-uniform-acceleration-passes-A-B-and-C-The-speeds-with-which-the-motorcyclist-

Question Number 191410 by otchereabdullai last updated on 23/Apr/23

T h r e e v i l l a g e s A, B a n d C a r e o n a

s t r a i g h t r o a d a n d B i s t h e m i d - w a y

b e t w e e n A a n d C . A m o t o r c y c l i s t

m o v i n g w i t h a u n i f o r m a c c e l e r a t i o n

p a s s e s A, B a n d C . T h e s p e e d s w i t h

w h i c h t h e m o t o r c y c l i s t p a s s e s A a n d

C a r e 20 {m s}^{- 1} a n d 40 {m s}^{-} r e s p e c t i v e l y .

f i n d t h e s p e e d w i t h w h i c h t h e m o t o r

c y c l i s t p a s s e s B .

Answered by mahdipoor last updated on 23/Apr/23

A B = B C = d

v_{C} = 40 v_{A} = 20 m / s

v_{2}^{2} - v_{1}^{2} = 2 {a d}_{1 \to 2} (w h e n a = c t e)

{\begin{cases} v_{C}^{2} - v_{B}^{2} = 2 a d \\ v_{B}^{2} - v_{A}^{2} = 2 a d \end{cases} \Rightarrow 2 v_{B}^{2} = v_{A}^{2} + v_{C}^{2}

\Rightarrow v_{B} = \sqrt{\frac{40^{2} + 20^{2}}{2}} = 20 \sqrt{2.5}

Commented by otchereabdullai last updated on 23/Apr/23

G o d b l e s s y o u s i r!