Question

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

Answer 1

✬ Acceleration = 0.5 m/s² ✬

✬ Distance = 175 m ✬

Explanation:

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 ?

Formula to be used:

• v = u + at ( For acceleration )
• s = ut + 1/2at² ( For distance )

Solution: 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,put the values on formula

$\implies{\rm }$ v = u + at

$\implies{\rm }$ 20 = 15 + a(10)

$\implies{\rm }$ 20 – 15 = 10a

$\implies{\rm }$ 5 = 10a

$\implies{\rm }$ 5/10 = a

$\implies{\rm }$ 0.5 m/s² = a

Hence, Acceleration of the bus is 0.5 m/s².

Now, for Distance use the second formula

$\implies{\rm }$ s = ut + 1/2at²

$\implies{\rm }$ s = 15 $\times$ 10 + 1/2 $\times$ 0.5 $\times$ 10²

$\implies{\rm }$ s = 150 + 0.5/2 $\times$ 100

$\implies{\rm }$ s = 150 + 0.5(50)

$\implies{\rm }$ s = 150 + 25

$\implies{\rm }$ s = 175 m

Hence, the distance covered by bus is 175 m.

Answer 2

Answer:

Explanation:

Given :-

Initial velocity, u = 72 km/h = 72 (5/18) m/s = 20 m/s  

Final velocity, v = 54 km/h = 54 (5/18) m/s = 15 m/s  

Time in which velocity changes is, t = 10 seconds  

To Find :-

(a) Acceleration

(b) Distance traveled

Formula to be used :-

a = (v - u)/t  and v² = u² + 2aS  

Solution :-

(a) Acceleration

Acceleration, a = (v - u)/t  

⇒ a = (20-15)/10

⇒ a = 5/10

⇒ a = 0.5 m/s²

(b) Distance traveled by the bus  

v² = u² + 2aS  

⇒ 20² = 15² + 2 × 0.5 × S  

⇒ 400 = 225 + 1×S

⇒ 400 - 225 = S

⇒ S = 175 m

Hence, the acceleration and distance traveled​ by bus is 0.5 m/s² and 175 m.