Question

A body travels 200 cm in the first two seconds and 220 cm in the next four seconds with constant acceleration. Find the velocity at the end of the seventh second from the start.

Answer 1

$s=\frac{1}{2}a{t}^2+ut$

$s=20cm,t=2s$

$(200\times {10}^{-2}{)}^2=\frac{1}{2}\times a{t}^2+ut$

$2u+2a=4$

$u+a=2$ ..(1)

Now, $s=420cm,t=65$

$420=ut+\frac{1}{2}a{t}^2$

$420=6u+18a$

$u+3a=70$ ...(2)

Solving (1) and (2)

$a=34m/s^2$

$u=-32m/s$

$v_{7th}=u_7+at$

$=-32t+34\times 7$

$=206m/s$