Question

2. A car is traveling with a speed of 22 m/s. The brakes can produce a maximum deceleration of 8.0 m/s. What is the minimum stopping distance for the car?​

Answer 1

Answer:

30.25m is the required answer

Explanation:

According to the Question

it is given that

• Initial velocity ,u = 22m/s
• Final velocity ,v = 0m/s
• Deceleration ,a = 8m/s²

we need to calculate the minimum stopping distance for the car . So we apply here the kinematics equation .

• v² = u² + 2as

where,

v is the final velocity

a is the acceleration

u is the initial velocity

s is the minimum stopping distance

substitute the value we get

➾ 0² = 22² + 2×(-8) × s { we take acceleration negative.}

➾ 0 = 484 -16s

➾ -484 = -16s

➾ 484 = 16s

➾ s = 484/16

➾ s = 30.25m

• Hence, the minimum stopping distance for the car is 30.25metres.

Answer 2

Given :-

A car is traveling with a speed of 22 m/s. The brakes can produce a maximum deceleration of 8.0 m/s.

To Find :-

What is the minimum stopping distance for the car?​

Solution :-

We know that

v² - u² = 2as

(0)² - (22)² = 2(-8)(s)

0 - 484 = -16s

-484 = -16s

-484/-16 = s

484/16 = s

30.25 = s

∴ 30.25 m is the minimum distance for stopping