Question

a bus acceleration uniformly from 54 km/h to 72 km/h in 10 seconds. calculate 1) acceleration in m/s2 . 2) distance covered by the bus in meters during this interval.


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Answer 1

Acceleration = Change in velocity / Time

                     = v - u / t

72km/h = 72 × 5 / 18 m/s = 20m/s

54km/h = 54 × 5 / 18m/s  = 15m/s

     Acceleration = 20-15 / 10 

                           = 5 /10 

                           = 0.5 m/s²

To find distance we use the formula 

              v² = u² + 2aS   here v is the final velocitiy (72km/h)

                                              u is the initial velocity(54km/h)

                                               a is the acceleration

                                              S is the distance

             20² = 15² + 2 × 0.5 × S 

             400 = 225 + 1×S

             400 - 225 = S

           S = 175 m

Therefore distance = 175m

Answer 2

It is given that initial velocity of car is 54km/h , final velocity is 72 km/h in 10 seconds accelerating uniformly.

We have to find i) Acceleration in m/s² ii) Distance covered by bus in 10 seconds.

Solution

Here, initial velocity of bus = 54 km/h

→ Initial velocity (u) = 54 × (5/18)

→ Initial velocity (u) = 15 m/s

Again, final velocity of bus = 72 km/h

→ Final velocity (v) = 72 × (5/18)

→ Final velocity (v) = 20 km/h

Using 1st equation of motion :

→ v = u + at

→ 20 = 15 + a × 10

→ 20 = 15 + 10a

→ 10a = 20 - 15

→ 10a = 5

→ a = 5/10

→ a = 0.5 m/s²

So, acceleration of bus = 0.5 m/s²

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Again,

Using 3rd equation of motion :

→ v² - u² = 2as

→ 20² - 15² = 2 × 0.5 × s

→ 400 - 225 = s

→ s = 175 m

So, distance covered by bus = 175 m