Question

A point moving in a straight line travels in its second, fifth and eleventh seconds of motion 16m, 28m and 52m respectively. Prove that the point is moving with constant acceleration. Also find the total distance moved in 10 seconds?
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Answer 1

Answer:

As S

nth

=u+an−

2

1

a=u+

2

a

(2n−1)

x=u+

2

a

(2p−1)... (i)

y=u+

2

a

(2q−1)... (ii)

z=u+

2

a

(2r−1)... (iii)

Subtracting Eq. (iii) from Eq. (ii),

y−z=

2

a

(2q−r) or q−r=

a

y−z

or (q−r)x=

a

1

(yx−zx) ..(iv)

Similrly, we can show that

(r−p)y=

a

1

(yx−zx) ..(v)

and (p−q)z=

a

1

(xz−yz) ..(vi)

Adding Eqs. (iv), (v) and (vi), we get

(q−r)x+(r−p)y+(p−q)z=0

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