Question

2- A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calculate
(i) acceleration in m/s2
(ii) distance covered by the bus in metres during this interval. ​

Answer 1

Answer :-

Given :-

• Initial velocity = 54 km/h
• Final velocity = 72 km/h
• Time = 10 seconds

To Find :-

• Acceleration
• Distance travelled

Solution :-

Converting initial and final velocity to m/s :-

• Initial velocity = 54 km/h = 54 × 5/18 = 15 m/s

• Final velocity = 72 km/h = 72 × 5/18 = 20 m/s

Calculating Acceleration :-

We have initial velocity, final velocity and time taken. Substituting the value in 1st equation of motion :-

→ v = u + at

→ 20 = 15 + 10a

→ 10a = 20 - 15

→ 10a = 5

→ a = 5 / 10

→ a = 2

Acceleration = 2 m/s²

Calculating Distance travelled :-

Substituting the values of initial velocity, acceleration and time in 2nd equation of motion :-

→ s = ut + ½ at²

→ s = 15 × 10 + ½ × 2 × 10²

→ s = 150 + 100

→ s = 250

Distance travelled = 250 m

Answer 2

A N S W E R :

• Acceleration of the bus is 2 m/s².

• Distance covered by bus is 250 m.

Given :

• Uniform velocity of bus is 54 km/h

• Final velocity of bus is 72 km/h

• Time taken to reach final velocity 10 seconds

To find :

• Acceleration of bus is m/s ?

• Distance covered by bus ?

Solution :

As we know that,

Formula's Used :

★ v = u + at (For acceleration)

★ s = ut + 1/2at² (For distance)

• First changing the uniform and final velocity into meter/second. To change multiply by 5/18

=> Uniform velocity (u) = 54(5/18)

=> u = 15 m/s

=> Final velocity (v) = 72(5/18)

=> v = 20 m/s

Now,

Substituting the values,

=> v = u + at

=> 20 = 15 + a(10)

=> 20 - 15 = 10a

=> 5 = 10a

=> 5/10 = a

=> a = 2

Therefore,

• Acceleration of the bus is 2 m/s²

Now,

• Distance use the second formula

=> s = ut + 1/2 at²

=> s = 15 × 10 + 1/2 × 2 × 10²

=> s = 150 + 2/2 × 100

=> s = 150 + 2(50)

=> s = 150 + 100

=> s = 250

Hence,

• Acceleration of the bus is 2 m/s².

• Distance covered by bus is 250 m.