Question

The speed of a car is reduced from $90$km/hr to $36$ km/hr in $5$ s. What is the distance travelled by car during this time interval?

Answer 1

$v=36\times \frac{5}{18}=10m/s$

$u=90\times \frac{5}{18}=25m/s$

$T=5s$

$v^2=u^2+2as$

${10}^2=(2s{)}^2-2\times as$

$s=\frac{{25}^2-{10}^2}{2a}$

$v=u+at$

$10=25-a\times 5$

$5a=15$

$a=3m/s^2$

$s=\frac{625-100}{2\times 3}=\frac{525}{6}=87.5m$