First slide

Rectilinear Motion

Question

A body travels for 15sec starting from rest with a constant acceleration . If it travels distances x, y and z in the first 5 sec,second 5 sec and the next 5 sec respectively. The relation between x, y and z is

Moderate

A

x = y = z
B

5x = 3y = z
C

$x = \frac{y}{3} = \frac{z}{5}$
D

$x = \frac{y}{5} = \frac{z}{3}$

Solution

: $g i v e n i n i t i a l v e l o c i t y u = 0; A t o B \to t i m e t = 5 s e c; d i s p l a c e m e n t s = x; s u b s t i t u t e i n S = u t + \frac{1}{2} a t^{2} x = \frac{1}{2} a {(5)}^{2} -- \to (1)$
$A t o C \to g i v e n S = x + y; u = 0; t = 5 + 5; s u b s t i t u t e i n S = u t + \frac{1}{2} a t^{2} x + y = \frac{1}{2} a {(5 + 5)}^{2} -- \to (2)$
$A t o D \to S = x + y + z; u = 0; t = 5 + 5 + 5 s e c; s u b s t i t u t e i n S = u t + \frac{1}{2} a t^{2} x + y + z = \frac{1}{2} a {(5 + 5 + 5)}^{2} -- \to (3) d i v i d e e q u a t i o n (2) b y e q u a t i o n (1)$
Dividing equation (2) by equation (1), we get,
$\frac{x + y}{x} = 4$
$\Rightarrow y = 3 x x = \frac{y}{3} - - - (4) d i v i d e e q u a t i o n (3) b y e q u a t i o n (2)$
$\frac{x + y + z}{x + y} = \frac{9}{4} substitute y=3x in above equation$
$\frac{4 x + z}{4 x} = \frac{9}{4} \Rightarrow z=5x \Rightarrow x = \frac{z}{5} - - - (5)$
$e q u a t e e q u a t i o n s (4) a n d (5) x = \frac{y}{3} = \frac{z}{5}$

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