Question

A truck moving at a speed of 72kmph applies brakes and stops after 5seconds. The distance covered by the truck during this time period is

Answer 1

Explanation:

Initial speed = 72km/h = 20m/s

Final speed = 0 (zero)

Time taken by truck to stop = 5s

We have to find Distance covered by truck before it is brought to rest

First we have to find acceleration of truck.

➝ v = u + at

➝ 0 = 20 + a(5)

➝ 5a = -20

➝ a = -4m/s²

◈ Negative sign of a shows retardation of truck

Now let's caluculate distance :)

⇒ v² - u² = 2as

⇒ 0² - 20² = 2(-4)s

⇒ -400 = -8s

⇒ s = -400/-8

⇒ s = 50m

Hope it helps you