First slide

Rectilinear Motion

Question

Two particles A and B are at separation of 100 m, particle A moves with constant acceleration 4 $m / s^{2}$ with initial speed 5m/s and B moves with uniform speed 12m/s, towards each other. When and where the particles meet?

Moderate

A

t = 3 sec, d = 55m
B

t = 4sec, d = 52m
C

t = 4 sec ,d = 57m
D

t = 2sec, d = 60m

Solution

$F o r A \to : d = 5 t + \frac{1}{2} \times 4 t^{2} d = 5 t + 2 t^{2} (A c c e l e r a t e d m o t i o n) - - - (1) F o r B \to : 100 - d = 12 t (u n i f o r m m o t i o n) - - - (2) A d d i n g (1) a n d (2) w e g e t 100 = 2 t^{2} + 17 t 2 t^{2} + 17 t - 100 = 0 t = \frac{- 17 \pm \sqrt{{(17)}^{2} - 4 (2) (- 100)}}{2 \times 2}$
= $\frac{- 17 \pm \sqrt{289 + 800}}{4}$
$t = \frac{- 17 \pm 33}{4} c o n s i d e r + 33 a s t i m e c a n n o t b e n e g a t i v e t = 4 s e c s u b s t i t t u t e t = 4 i n e q n (2) d = 52 p a r t i c l e s m e e t a f t e r 4 s e c, a f t e r c o v e r i n g a d i s \tan c e o f 52 m$

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