Question

A body moving with uniform acceleration covers 24m in the 4th
second and 36 m in the 6
th second.calculate the
accelelation and
initial velocity​

Answer 1

Answer:

 Formula to be used: distance covered in nth second : $s=u + \frac{1}{2} a(2n-1)$ where u=intial velocity and n=the no. of second.

  $\\ 24=u+\frac{1}{2}a[2(4)-1] \\\\ 24=u+\frac{7}{2}a$---------(1)

now,

$\\ 36=u+\frac{1}{2}a[2(6)-1] \\\\36= u+\frac{11}{2} a$-----------(2)

Subtracting (1) from (2), we get

$\\ \frac{11}{2}a-\frac{7}{2}a=36-24\\ \\ \frac{4}{2}a= 12 \\\\ 2a=12\\\\ a=6$

Now, put the value of a in (1).

$24=u+\frac{7}{2} [6]\\24=u+ 21\\u=3 m/s$