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:- 0.077m/s²

Step by step explanation::--

As bus starts from rest.. hence:-

u=0

Given that..

v= 50km/hr =50 ×5/18 = 13.88 m/s

t =3 min= (3×60)seconds =180 sec

A.T.Q, FORMULA...

v=u+at

so ,

13.88=0 + a(180)

a= 13.88/180

a= 0.077m/s²

Answer 2

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:-

1. Acceleration of the bus.
2. Distance traveled by the bus.

Solution:-

Now ,For finding acceleration we use first equation of motion

i.e

$\sf\huge \: v = u + at$

where,

• V=Final Velocity
• u= initial Velocity
• a= acceleration
• t= time taken

Now, putting the values in this equation

$14 = 0 + a \times 180$

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

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

$= > a = 0.07m {s}^{ - 2}$

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

$\sf\huge \: s = ut + \frac{1}{2} a {t}^{2}$

where,

• s=distance
• u=initial Velocity
• t= time taken
• a=acceleration

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

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

$= > s = 0.07 \times 1620$

$= > s = 113.4m$

∴The distance traveled by the bus is 113.4m