Question

A bus moving at a speed of 24 m s begins to slow at a rate oof 3m/sec^2 each sec how far does it goo before stopping

Answer 1

Given:

• Initial velocity, u = 24 m/s
• final velocity, v = 0 m/s ( °•° we are assuming the car to be stopped after travelling a specific distance )
• acceleration, a = -3m/s² ( °•° it's negative because the car is coming to rest )

We know,

2as = v² - u²

Putting values in the above formula.

2 * -3 * s = (0)² - (24)²

- 6 * s = - 576

Cancel the minus , and we are left with only

6 * s = 576

s = 576/6

s = 96 m

Hence, The car will cover 96 m before it stops.

Answer 2

$\Huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

Given :

Initial Speed of bus (u) = 24 m/s

Acceleration (a) = -3 m/s²

[Acceleration is negative because it is applied in opposing direction of motion]

As the Bus applies Break, So Final Speed

(v) = 0 m/s

$\rule{200}{2}$

To Find :

Distance Travelled by Bus after applying brakes.

$\rule{200}{2}$

Solution :

We have a formula :

$\LARGE {\red{\implies}}{\boxed{\boxed{\green{\sf{v^2 \: - \: u^2 \: = \: 2as}}}}}$

Putting Values ⇒(0)² - (24)² = 2(-3)(s)

⇒0 - 576 = -6s

⇒6s = 576

⇒s = 576/6

⇒ s = 96

$\Large{\boxed{\red{\sf{Distance \: = \: 96 \: m}}}}$