Question

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

Answer 1

distance covered in nth second


S=4+6n .......1


also


S=u+a(2n-1)/2 .........2


differentiating equation 1 with respect to n


dS/dn=0+6


dS/dn=6 ......3


differentiating equation second with respect to n


dS/dn=0+a


dS/dn=a


putting value of dS/dn


a=6m/s^2


put the value of a in equation 2


and equate equation 1 and 2


4+6n=u+6(2n-1)/2


4+6n=6n-3+u


u=7m/s