Question

30. A body starts from rest from the origin with an acceleration
of 6 m/s along the x-axis and 8 m/s' along the y-axis. Its
distance from the origin after 4 seconds will be
(a) 56 m
(b) 64 m
(c) 80 m
(d) 128 m​

Answer 1

Answer:-

s = 80m

Explanation:-

Given:-

• $a_x = 6m{s}^{-2}$
• $a_y = 8m{s}^{-2}$
• t = 4sec
• u = 0

To Find:-

Distance travelled by the particle

Solution:-

V = u + at

X - direction:

$V_x = u_x + a_xt$

$V_x = 0+ 6(4)$

$V_x = 24 m {s}^{-1}$.

Now,

v² - u² = 2as

${V_x}^{2}-{ u_x }^{2}= 2a_xs_x$

${24}^{2}= 2(6)s_x$

$576 = 12s_x$

$\bold{ s_x = 48m }$

Y - Direction:-

$V_y = u_y + a_yt$

$V_y = 0 + 8(4)$

$V_y= 32m{s}^{-1}$

Now,

${V_y}^{2}-{ u_y }^{2}= 2a_ys_y$

${32}^{2} = 2(8)s_y$

$1024= 16s_y$

$\bold{s_y = 64}$

Hence,

$S = s_x + s_y$

$S = 48i + 64j$

$|S| = \sqrt{{48}^{2}+ {64}^{2}}$

$|S|= \sqrt{2304+4096}$

$|S|= \sqrt{6400}$

$\bold{|S|= 80m}$______option(c)