If the distance travelled by a body in the $n^{t h}$ second is given by (4 + 6 n) m, then find the initial velocity and acceleration of the body respectively.

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Solution

$s_{n t h} = u + \frac{a}{2} (2 n - 1)$
$4 + 6 n = u + \frac{a}{2} (2 n - 1)$
$= (u - \frac{a}{2}) + a n$
$a = 6 m / s^{2}$
$u - \frac{a}{2} = 4$
$u - 3 = 4$
$u = 7 m / s$

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If the distance travelled by a body in the nth second is given by (4 + 6 n) m, then find the initial velocity and acceleration of the body respectively.

snth=u+a2(2n−1)4+6n=u+a2(2n−1)=(u−a2)+ana=6m/s2u−a2=4u−3=4u=7m/s

If the distance travelled by a body in the $n^{t h}$ second is given by (4 + 6 n) m, then find the initial velocity and acceleration of the body respectively.

$s_{n t h} = u + \frac{a}{2} (2 n - 1)$
$4 + 6 n = u + \frac{a}{2} (2 n - 1)$
$= (u - \frac{a}{2}) + a n$
$a = 6 m / s^{2}$
$u - \frac{a}{2} = 4$
$u - 3 = 4$
$u = 7 m / s$