Question

In a certain time interval, the velocity of a car
reduces from 72 km/h to 36 km Ir. If the car
covers a distance of 150 m during this time
interval, find the retardation of the car​

Answer 1

Explanation:

Given: U = 72 × 5

18

U = 4 × 5

U = 20 ms–¹

V = 36 × 5

18

V = 10ms–¹

By third equation of motion,

= v² – u² =2as

= (10)² – (20)² = 2(a)(120)

= –300 = 240a

= a = —300

240

= a = — 5

4

= a = —1.25 ms–²

Hence, the retardation of the car is -1.25ms-² .

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