Question

A cat running with constant acceleration in one direction is at rest initially and has velocity $v\left(t\right)=1.00m/s$ at $2.00$ seconds later.
At $t=4.00$ seconds, how far away from the starting point will the cat be?

• A. $0.00m$
• B. $1.00m$
• C. $2.00m$
• D. $4.00m$
• E. $8.00m$

Answer 1

The correct option is D $4.00m$

Initial speed of the cat $u=0$ m/s

Speed of cat after $t=2$ s $v=1.00$ m/s

Let the acceleration of cat be $a$.

Using $v=u+at$ where $t=2$ s

$\therefore$ $1.00=0+a\times 2$ $⟹a=0.5$ $m/s^2$

Thus distance covered by cat in 4 s, $S=u{t}^{\prime }+\frac{1}{2}a(t^{\prime }{)}^2$ where $t^{\prime }=4$ s

$\therefore$ $S=0+\frac{1}{2}\times 0.5\times 4^2=4.00$ m