Question

A body start from rest from the origin with an acceleration of 6 m/s^2 along the x axis and 8 m/s^2 along the y axis its distance from the origin after 4 sec will be

Answer 1

S= ut + (1/2)at²

u = 0

t =  4 sec

S = 0 + (1/2)a(4)²

S = 8a

a = 6m/s² x & 8m/s² y

| a | = √(6² + 8²) = √(36+64) = √100 = 10

S = 8*10 = 80 m

80 m is the distance

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Answer 2

Explanation:

body starts acceleration from origin so its intial velocity is u=0

now, u=0

t=4sec

applying equation...

s = ut +1/2at^2

s = 0×4 + 1/2 a (4×4)

s= 0 + 1/2 ×16×a

s= 8a

now ,, we have

a1( alonv x-axis) = 6m/s^2

a2 (along y-axis) = 8m/s^2

taking mag. of acceleration

|a| =

$\sqrt{6 {}^{2} } + \sqrt{ } 8 {}^{2}$

$\sqrt{36 + 64}$

$\sqrt{100}$

=10

put this value in s=8a

s= 8×10

s= 80 ans.