A dog initially running at a speed of 2 m/s accelerates at constant rate of 1.5 m/s2 for 5 s.Calculate:(1) distance covered by it from start (2) the final velocity of the dog?

 

A dog initially running at a speed of 2 m/s accelerates at constant rate of 1.5 m/s2 for 5 s.Calculate:(1) distance covered by it from start (2) the final velocity of the dog?

Read more:"Life Processes Class X Chapter One Solved Questions and Answers" Part-1

Solution: We know the formula,

$v=u+at$ 

where $v$ is final velocity, 

$u$ is initial velocity, $a$ is acceleration &

 $t$ is time.
Here we get, $v=?,u=2m{s}^{-1},a=1.5m{s}^{-2},t=5s$
Hence,(2) Final velocity of the dog is $(2+1.5\times 5)=9.5m{s}^{-1}$
And the another formula to determine the covered distance is
$v^2=u^2+2as$ where $s$ is displacement.
Hence,here displacement is
$\frac{v^2-u^2}{2a}=\frac{{9.5}^2-2^2}{2\times 1.5}=\mathrm{28.75\; m.}\mathrm{(1)\; Total\; distance\; covered\; is}$$28.75m$ and (2) the final velocity is $9.5m/s$