Question

a body describe 10 m in a third second of the motion and 28 m in the sixth second . what distance will be covered by the body in the seventh second of the motion?

Answer 1

Answer:

Check answer in the given attachment

Answer 2

Answer:

34m will be covered by the body in the seventh second.

Explanation:

The distance traveled in nth second is given as,

$S_n=u+\frac{1}{2}a(2n-1)$     (1)

Where,

$S_n$=distance traveled in nth second

u=intial velocity

a=acceleration with which the body is moving

The distance traveled in the third second=10m

The distance traveled in sixth second=28m

Distance traveled in the third second

$10=u+\frac{1}{2}a(2\times 3-1)$

$10=u+\frac{5}{2}a$   (2)

Distance traveled in the sixth second

$28=u+\frac{1}{2}a(2\times 6-1)$

$28=u+\frac{1}{2}a(12-1)$

$28=u+\frac{11}{2}a$  (3)

On subtracting equation (2) from (3) we get;

$28-10=u+\frac{11}{2}a -u-\frac{5}{2}a$

$18=\frac{11}{2}a -\frac{5}{2}a$

$18=\frac{6}{2}a$

$18=3a$

$a=6m/s^2$   (4)

Now let's find "u" using equation (2) i.e.

$10=u+\frac{5}{2}\times 6$

$10=u+15$

$u=-5m/s$   (5)

Distance traveled in the seventh second

$S_7=-5+\frac{1}{2}\times 6\times (2\times 7-1)$

$S_7=-5+\frac{1}{2}\times 6\times (14-1)$

$S_7=-5+3\times 13$

$S_7=-5+39$

$S_7=34m$

Hence, 34m will be covered by the body in the seventh second.

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