Question

a bus starts from rest and attain a velocity of 45 metre per second after 10 seconds find the acceleration and distance travelled in 5 seconds and in 10 seconds​

Answer 1

Answer:

u = 0 m/s [ because bus is starting from rest ]

v = 45 m/s

t1 = 5s

t2 = 10s

a1 = v - u / t1

a1 = 45 m/s - 0 m/s / 5s

a1 = 45m/s / 5s

a1 = 9 m/s²

a2 = v - u /t2

a2 = 45 m/s - 0 m/s / 10s

a2 = 45 m/s / 10s

a2 = 4.5 m/s²

The distance travelled in 5s

s1 = v² - u² / 2a1

s1 = 2025 m²/s² - 0 m²/s² / 2( 9 m/s² )

s1 = 2025 m²/s² / 18 m/s²

s1 = 112.5 m

The distance travelled in 10s

s2 = v² - u² / 2a2

s2 = 2025 m²/s² -  0 m²/s² / 2(4.5 m/s²)

s2 = 2025 m²/s² / 9 m/s²

s2 = 225 m

∴ The acceleration and distance travelled in 5 seconds is 9 m/s² and 112.5 m, and the acceleration and distance travelled in 10 seconds​ are 4.5 m/s² and 225 m.

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Answer 2

Given :

Initial velocity, u = 0

Final velocity, v = 45

Time taken, t = 10 sec

To Find :

The Acceleration and the distance travelled by the bus in 5 seconds and in 10 seconds.

Solution :

Using first equation of motion that is,

v = u + at

By substituting all the given values in the formula,

➛ 45 = 0 + a(10)

➛ 45 = a(10)

➛ a = 45/10

➛ a = 4.5 m/s²

Thus, the acceleration of the bus is 4.5 m/s²

Now

Using second equation of motion that is,

s = ut + 1/2at²

Distance travelled in 5 seconds,

➛ s = ut + 1/2at²

➛ s = (0)(5) + 1/2(4.5)(5)²

➛ s = 1/2(4.5)(25)

➛ s = 112.5/2

➛ s = 56.25 m

Thus, the distance travelled in 5 sec is 56.25m.

Distance travelled in 10 second,

➛ s = ut + 1/2(a)(t)²

➛ s = (0)(10) + 1/2(4.5)(10)²

➛ s = 1/2(4.5)(100)

➛ s = 4.5 × 50

➛ s = 225 m

Thus, the distance travelled in 5 sec is 225m