A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calc

1.	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» ✬ 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.

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»

✬ Acceleration = 0.5 m/s² ✬

✬ Distance = 175 m ✬

Explanation:

Given:

To Find:

Formula to be used:

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.