Question

In a certain time interval the velocity of a car reduces from 72 km per hour to 36 km per hour . If the car covers a distance of 150 m during this time interval find the retardation of the car ?

Answer 1

$u = 72 \times \frac{5}{18 } \\ u = 4 \times 5 \\ u = 20ms {}^{ - 1} \\ \\ v = 36 \times \frac{5}{18} \\ v = 10 \: m {s}^{ - 1}$
By third equation of motion,
${v}^{2} - {u}^{2} = 2as \\ (10) {}^{2} - ( {20})^{2} = 2(a)(120) \\ - 300 = 240a \\ a = - \frac{ 300}{240} \\ a = - \frac{5}{4} \\ a = - 1.25 \: m {s}^{ - 2}$