Question

A body in uniformly accelerated motion travels 200 cm in the first two seconds and 220 cm in the next four seconds. The velocity at the end of seventh second from start is

• A. $5cm{s}^{-1}$
• B. $10cm{s}^{-1}$
• C. $15cm{s}^{-1}$
• D. $20cm{s}^{-1}$

Answer 1

The correct option is B $10cm{s}^{-1}$

We have, in the first 2 seconds, distance covered, $s_1$ = 200 cm, and in the next 4 seconds, the distance covered, $s_2$ = 420 cm;
time,$t_1$ = 2s and $t_2$ = 6s.
From the second equation of motion,
$s=ut+\frac{1}{2}a{t}^2$
Case I :
$s_1=u{t}_1+\frac{1}{2}a{t}_1^2$
$\Rightarrow 200=2u+\frac{1}{2}a\times 4$
$\Rightarrow$ u+a=100 …(1)
Case II :
$s_2=u{t}_2+\frac{1}{2}a{t}_2^2$
$\Rightarrow 420=u\times 6+\frac{1}{2}a\times 6^2$
$\Rightarrow$ 420=6u+18a
$\Rightarrow$ u+3a=70 …(2)
Solving (1) and (2), we get
$a=-15m{s}^{-2}$
and $u=115cm{s}^{-1}$.
From the first equation of motion, v = u + at
V = 115 - 15 $\times$ 7 = 10 $cm{s}^{-1}$.