Question

A bus is moving with a constant speed of 75kmph the driver observes red signal and applies break if the speed of the bus reduces to 21kmph in 3s find the reterdation produced in the bus

Answer 1

Answer:

-5 m/s²

Explanation:

Given:

Initial velocity = u = 75 km/h

75 km/h = $75 \times \frac{5}{18}$ = $\frac{125}{6}$ m/s

Final velocity = v = 21 km/h

21 km/h = $21 \times \frac{5}{18}$ = $\frac{35}{6} \ m/s$

Time = t = 3 seconds

To find:

Retardation of the bus

Retardation = $\frac{v-u}{t}$ : where v=final velocity, u =initial velocity, t=time

Retardation = $\frac{\frac{35}{6}-\frac{125}{6} }{3}$

Retardation = $\frac{-90}{6}{3}$

Retardation = -5 m/s²

The retardation of the bus is equal to -5m/s²

Note: Decceleration will be a negative value

Answer 2

Given

● Initial speed of bus = 75 km/h

● Final speed of bus = 21 km/h

● Time = 3 s

To find

● Retardation of bus

Solution

Initial speed = 75 km/h

⇒ Initial speed = 75 × (5/18)

⇒ Initial speed = 20.83 m/s

Final speed = 21 km/h

⇒ Final speed = 21 × (5/18) m/s

⇒ Final speed = 5.83 m/s

Time = 3 seconds.

Using 1st equation of motion :

⋆ v = u + at

Where,

● v = final speed

● u = initial speed

● a = acceleration

● t = time

Putting values we get :

⇒ 5.83 = 20.83 + a × 3

⇒ 5.83 = 20.83 + 3a

⇒ 3a = 5.83 - 20.83

⇒ 3a = -15

⇒ a = -15/3

⇒ a = -5 m/s²

∵ Retardation = - Acceleration

⇒ Retardation = 5 m/s²

∴ Retardation produced in bus = 5 m/s²