Question

A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/hr in 10 minutes. Find its acceleration....CAN ANYONE GIVE ME ITS ANSWER STEP BY STEP??????

Answer 1

Answer:

0.0185m/s²

Explanation:

Given :

• Initial velocity,u => 0m/s [As it is starting from the rest]

• Final velocity,v => 40km/hr = 11.1m/s    

[1km/hr = 1000/3600m/s]

• Time,t = 10min = 10*60 = 600s               [1min = 60s]

Find :

• Acceleration,a

Solution :

According to the 1st equation of motion:-

$\sf{}v=u+at$

Put their values and solve it

$\implies \sf{} 11.1=0+(a)(600)$

$\implies \sf{} 11.1=600a$

$\implies \sf{} \dfrac{11.1}{600}=a$

$\therefore \sf{} a= 0.0185m/s^2$

Hence,acceleration is equal to 0.0185m/s²

☆Additional Information☆

• 2nd equation of motion

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

• 3rd equation of motion

$\sf{}v^2-u^2=2as$

Answer 2

Answer:

0.0185 m/s²

GIVEN :

Initial velocity, u = 0

Final velocity, v = 45 km/hr.

Converting km/hr into m/s, we get v = 11.1 m/s.

Time, t = 10 min

Converting time from min to sec, 10×60 = 600 sec.

TO FIND :

Acceleration = ?

SOLUTION :

According to the 1st equation of motion,

$\red{\boxed{\bold{V \: = \: u \: + \: at}}}$

Put the given values in formula,

→ 11.1 = 0 + (a) (600)

→ 11.1 = 600 a

→ a = $\frac {11.1}{600}$

$\therefore \boxed {a \: = \: 0.0185 \: m/s²}$

$\therefore$ $The \: acceleration \: is \: 0.0185 {m/s}^{2}$