Question

stating from statiinary position , Rahul paddels bicycle to attion a velocity of 60m/s in 30s then he applies break such that the velocity of the bicycle comes down to 40m/s in 5 sec . Calculate the acceleration of the bicycle in both the cases .

Answer 1

so acceleration in the first case
$u=0(since \:he \:starts \:from \:rest)$
$v=60$
$t=30$

now equation would be

$\frac{v-u}{t}=a$

$\frac{60-0}{30}=a$

$now \:acceleration \:would \:be$
$a=2ms^{-2}$

acceleration in second case

$u=60(since \:he \:is \:moving \:continues)$
$v=40$
$t=5$

$\frac{v-u}{t}=a$

$\frac{40-60}{5}=a$

$\frac{-20}{5}=a$

$a=-4ms^{-2}$

$acceleration \:in \:both \:case \:are$

$5ms^{-2} \:and \:-4ms^{-2} \:respectively$

$-4ms^{-2} \:because \:it \:retards$

Answer 2

Answer:

I hope it helpful answer.