Question

17) 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=distance/time

=(72000-54000)/10...(1km=1000m)

=18000/10

=1800m/s

distance=speed×time

=1800×10

=18000m

Answer 2

✬ 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

⟹ v = u + at

⟹ 20 = 15 + a(10)

⟹ 20 – 15 = 10a

⟹ 5 = 10a

⟹ 5/10 = a

⟹ 0.5 m/s² = a

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

Now, for Distance use the second formula

⟹ s = ut + 1/2at²

⟹ s = 15 \times× 10 + 1/2 \times× 0.5 × 10²

⟹ s = 150 + 0.5/2 × 100

⟹ s = 150 + 0.5(50)

⟹ s = 150 + 25

⟹ s = 175 m

Hence, the distance covered by bus is 175 m.