Question

A train starting from rest attains a velocity of 62 km/h in 5min. Assuming the acceleration is uniform, find. a) the acceleration. b) the distance travelled by train for attaining this velocity​

Answer 1

Answer:

Initial Velocity=u=0m/s

Final velocity= V= 72km/h

Time =5 min =5/60 h =1/12 h

acceleration=a=?

As we know that , a=v-u/t

⇒a=72-0/1/12

=72*12

=864km/h²

Distance travelled  =s=ut+1/2at²

=0x1/12 +1/2(864)(1/12)²

=(1/2)x864x(1/144)

=3km

∴Distance travelled by the train for attaining this velocity 3km.

:

Answer 2

Answer:

a)0.05m/s^2

b)2.25km

Explanation:

Given :-

Initial velocity,u = 0 m/s (As it started from rest)

Final velocity,v =62 km/h = 17.22 m/s        [1 km/hr = (1000)/(60 × 60) m/s]

Time taken, t = 5 min = 300 s                     [1 min = 60s]

To Find :-

Accelaration, a = ?

Distance travelled, s = ?

Formula to be use :-

$\sf{}v=u+at$

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

Solution :-

According  to the 1st equation of motion :-

$\Rightarrow \sf{}v=u+at$

Put thier values and solve it.

$\implies \sf{}17.22=0+a(300)$

$\implies \sf{}17.22=300a$

$\implies \sf{}\dfrac{17.22}{300}=a$

$\implies \sf{}a=0.05m/s^2$

According  to the 2nd equation of motion :-

$\Rightarrow \sf{}s=ut +\dfrac{1}{2}at^2$

Put their values and solve it.

$\Rightarrow \sf{}s=(0)(300) +\dfrac{1}{2}(0.05)(300)^2$

$\Rightarrow \sf{}s=0+\dfrac{1}{2}(0.05)(90000)$

$\Rightarrow \sf{}s=\dfrac{1}{2}\times 4500$

$\Rightarrow \sf{}s= 2250m =2.25km\ \ \ \ \ \ [1m=\dfrac{1}{1000}km]$

Therefore, acceleration = 0.05m/s^2 and distance traveled => 2.25km

Equation of motion :-

Relation among velocity,time,distance,acceleration is called equation of motion.

There are three equation of motion :-

$\sf{}First\ equation\ of\ motion:v=u+at$

$\sf{}Second\ equation\ of\ motion: s=ut +\dfrac{1}{2}at^2$

$\sf{}Third\ equation\ of\ motion: v^2 -u^2 =2as$