Question

a train starting from rest attains a velocity of 72kmh¹ in 5 minutes. assuming that the acceleration is uniform, find (1) the acceleration and (2) the distance travelled by the train for attaining this velocity.​

Answer 1

To Find:

1. Acceleration of train:?
2. Distance travelled by train:?

$\bf{\pink{GIVEN}}$

• Initial Velocity (u) =0 m/s
• Final velocity (v)=72 km/h=$72\times \frac{5}{18}$=20 m/s
• Time (t)=5 mins=5×60=300 secs

$\bf{\fbox{Formula}}$

For acceleration we use 1st Kinematical Equation

• v=u+at .........(i)

For distance travelled we use 2 nd Kinematical Equation

• s=ut+$\frac{1}{2} at^{2}$ ........(ii)

$\bf{\fbox{SOLUTION}}$

$a=\frac{(v-u)}{t} from (i)$

$a= \frac{(20-0)}{300}=\frac{1}{15}m/s^{2}$

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

s=0+$(\frac{1}{2}) (\frac{1}{15})(300)^{2}$

s=3 kms

Acceleration of train 20 m/s²

Distance travelled by train 3 kms

Answer 2

Answer:

your answer in given pic

Step-by-step explanation:

I hope it's help you