Question

A bus starts from rest and moving with uniform acceleration attains a velocity of 50 km/h in 3 minutes. Find the acceleration and the distance travelled.​​

Answer 1

Answer :-

→ Acceleration is 0.077 m/s² .

→ Distance travelled is 1247.4 m .

Explanation :-

We have :-

• Initial velocity (u) = 0 m/s

• Final velocity (v) = 50 km/h

• Time taken (t) = 3 mins = 180 sec

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Firstly, let's convert the unit of final velocity of the bus from km/h to m/s .

⇒ 1 km/h = 5/18 m/s

⇒ 50 km/h = 50(5/18)

⇒ 250/18 m/s

⇒ 13.89 m/s

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Now, let's calculate acceleration of the bus by using the 1st equation of motion .

v = u + at

⇒ 13.89 = 0 + a(180)

⇒ 13.89 = 180a

⇒ a = 13.89/180

⇒ a = 0.077 m/s²

Finally let's calculate the distance travelled by the bus using the 2nd equation of motion .

s = ut + ½at²

⇒ s = 0(180) + ½ × 0.077 × (180)²

⇒ s = 0 + (0.5 × 0.077 × 32400)

⇒ s = 0 + 1247.4

⇒ s = 1247.4 m

HOPE THIS HELPS!!

Answer 2

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Given:-

A bus starts from rest and moving with uniform acceleration attains a velocity of 50 km/h in 3 minutes.

Initial Velocity= 0

Final Velocity= 50×5/18=13.8m/s=14m/s

Time Taken= 3×60s=180s

To Find:-

Acceleration of the bus.

Distance traveled by the bus.

Solution:-

Now ,For finding acceleration we use first equation of motion

i.e

v = u + atv = u+at

where,

V=Final Velocity

u= initial Velocity

a= acceleration

t= time taken

Now, putting the values in this equation

$= > a = \frac{14}{180}$

$= > a = \frac{7}{90} ms { - }^{2}$

$= > a = 0.07ms { - }^{2}$

It is given that the bus is travelling in Uniform acceleration.

∴The Acceleration of the bus is 0.07m/s²

Now,for finding the distance we use 2nd equation of motion

i.e

$s = ut + \frac{1}{2} {at}^{2}$

where,

s=distance

u=initial Velocity

t= time taken

a=acceleration

$s = 0 \times 180 + \frac{1}{2} \times (0.07) \times {180}^{2}$

$= > s = \frac{1}{2} \times (0.07) \times 3240$

$= > 0.07 \times 1620$

$s = 113.4m$

∴The distance traveled by the bus is 113.4m

Hope it will help you

Good morning