First slide

Rectilinear Motion

Question

A cyclist starts from rest and moves with a constant acceleration of 1 m/ $s^{2}$ . A boy who is 48 m behind the cyclist starts moving with a constant velocity of 10 m/s. After how much time the boy meets the cyclist?

Moderate

A

8 sec
B

12 sec
C

10 sec
D

both 8sec and 12 sec

Solution

$L e t t h e b o y m e e t t h e c y c l i s t a t B . B o y h a s u n i f o r m m o t i o n : u t = t o t a l d i s \tan c e 10 t = 48 + d - - - (1) C y c l i s t h a s a c c e l e r a t i o n m o t i o n : \to i n i t i a l v e l o c i t y u = 0; a c c e l e r a t i o n a = 1 m / s^{2}; d i s \tan c e s = d; s u b s t i t u t e i n, s = u t + \frac{1}{2} a t^{2} d = \frac{1}{2} \times 1 \times t^{2} - - - (2) s u b s t i t u t e d v a l u e f r o m e q u a t i o n (1) 10 t - 48 = \frac{1}{2} t^{2} \Rightarrow t^{2} - 20 t + 96 = 0 (t - 8) (t - 12) = 0 T h e b o y m e e t s t h e c y c l i s t a t t = 8 s e c, 12 s e c A t t = 8 s e c, v e l o c i t y o f c y c l i s t = 1 \times 8 = 8 m / s v e l o c i t y o f b o y > v e l o c i t y o f c y c l i s t, b o y o v e r t a k e s c y c l i s t A t t = 12 s e c, v e l o c i t y o f c y c l i s t = 1 \times 12 = 12 m / s v e l o c i t y o f c y c l i s t > v e l o c i t y o f b o y, c y c l i s t o v e r t a k e s b o y$

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