Question

A body travels $2m$ in the first two second and $2.20m$ in the next $4$ second with uniform deceleration. The velocity of the body at the end of $9$ second is

• A. $-10m/s$
• B. $-0.20m/s$
• C. $-0.40m/s$
• D. $-0.80m/s$

Answer 1

The correct option is B $-0.20m/s$

A to B
$2=u\times 2+\frac{1}{2}\times a\times 2\times 2\Rightarrow 1=u+a$
A to C
$4.20=u\times 6+\frac{1}{2}\times a\times 6\times 6\Rightarrow 0.7=u+3a$

Let the Body travel from $A$ to $B$ in $2s$ for a distance of $2m$
Let the Body travel from $B$ to $C$ for next $4s$ for a distance of $2.20m$
Velocity after $9s=?$
For $AtoB$
WKT, $S=ut+\frac{1}{2}a{t}^2\Rightarrow 2=2u+\frac{1}{2}a\ast 2\ast 2\Rightarrow 1=u+a...\left(1\right)$
For $B$ to $C$
WKT, $S=ut+\frac{1}{2}a{t}^2\Rightarrow 4.20=6u+\frac{1}{2}a\ast 6\ast 6\Rightarrow 0.7=u+3a...\left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get,
$2a=-0.3\Rightarrow a=-0.15,-ve$ as it is decreasing acceleration .
$u=1-a\Rightarrow u=1+0.15=1.15$
Now, velocity at $t=9s=v=u+at\Rightarrow v=1.15-0.15\ast 9=1.15-1.35=-0.2m/s$
velocity is negative as it is decreasing.