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Question Number 191091 by otchereabdullai last updated on 18/Apr/23

T h e f r o n t o f a t r a i n 80 m l o n g p a s s e s

a s i g n a l a t a s p e e d o f 72 k m / h . I f t h e

r e a r o f t h e t r a i n p a s s e s t h e s i g n a l

5 s e c o n d s l a t e r, f i n d

(a) t h e a c c e l e r a t i o n o f t h e t r a i n

(b) t h e s p e e d a t w h i c h t h e r e a r o f t h e

t r a i n p a s s e s t h e s i g n a l .

Answered by mr W last updated on 18/Apr/23

v_{1} = 72 k m / h = 20 m / s

s = v_{1} t + \frac{{a t}^{2}}{2}

80 = 20 \times 5 + \frac{a \times 5^{2}}{2}

\Rightarrow a = - \frac{8}{5} = - 1.6 m / s^{2}

v_{2} = v_{1} + a t = 20 - 1.6 \times 5 = 12 m / s = 43.2 k m / s

Commented by otchereabdullai last updated on 18/Apr/23

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