Question

A car moving with uniform motion at 33 m/s is brought to rest at a dista
of 200 m what is the acceleration in m/s 2? *​

Answer 1

Given:-

• Initial Velocity = 33m/s

• Final Velocity = 0m/s (as it comes to rest)

• Distance Travelled = 200m

To Find:-

• Acceleration - ?

FORMULAE USED:-

• v² - u² = 2as

Where,

v = Final Velocity

u = Initial Velocity

a = Acceleration

s = Distance.

Now,

v² - u² = 2as

(0)² - (33)² = 2 × a × 200

0 - 1089 = 400a

- 1089 = 400a

a = -1089/400

a = -2.72m/s²

Hence, The declaration is -2.72m/s².

• Velocity is defined as the displacement of body in per unit time.

• The Rate of the change in the Velocity is known as Acceleration.

• Velocity is a vector quantity

• Acceleration is a vector quantity

Answer 2

Answer:

-2.7 ms^2

Explanation:

u = 33m/s

v = 0 m/s

s = 200m

2as = v^2 - u^2

2*a*200 = (0^2) - (33^2)

a = -1089/400

a = -2.7 m/s