Question

3) A bus is moving with relocity 28 km the driven
appiled the brkes so that it came to nest onor
to secondes lind the diclare tred during the time​

Answer 1

Correct Question

A bus is moving with velocity 28 km/hr the drive appiled the brkes so that it came to rest in two seconds. Find the deceleration produced during the time.

Solution

Give that, the initial velocity of the bus is 28 km/hr. When brakes are applied it comes to rest in two seconds. (As brakes are applied means the final velocity is 0 m/s and time is 2 sec).

We have to find the declaration produced in the bus.

Firstly convert the km/hr into m/s. To convert km/hr into m/s, multiply the given value by 5/18.

= 28 × 5/18 = 7.8 m/s

First equation of motion:

v = u + at

Second equation of motion:

s = ut + 1/2 at²

Third equation of motion:

v² - u² = 2as

From above data we have value of v, u and t. So,

Using the First Equation Of Motion:

v = u + at

Substitute the known values in the above formula,

0 = 7.8 + a(2)

-7.8 = 2a

Divide by 2 on both sides,

-7.8/2 = 2a/2

-3.9 = a

(Negative sign shows retardation or deceleration)

Therefore, the declaration produced in bus is 3.9 m/s².

Answer 2

$\bf{\blue{\bigstar}}$ Correct Question

A bus is moving with velocity 28 km/hr the driver applied the brakes so that it came to rest in two seconds. Find the deceleration produced during the time.

$\bf{\blue{\bigstar}}$ Answer

Here the bus is moving with velocity 28 km/hr then the driver applied break so that it comes to rest in 2 seconds.

Given ,

Initial velocity , u = 28 km/hr = 28 * (5/18) = 7.8 m/s

Time , t = 2 sec

Final velocity , v = 0

Use first equation of motion .

⇒ v = u + at

⇒ 0 = 7.8 + a (2)

⇒ 2a = -7.8

⇒ a = -3.9 m/s²

Here negative sign denotes the deceleration.

Hence the deceleration produced during the time is 3.9 m/s²