Question

the train starting from increase its speed to 90km/h in 5hrs. calculate its acceleration​

Answer 1

Given:

Change in velocity of train,Δv= 90 km/h

Time taken by train,t= 5 h

To Find:

Acceleration of train

Solution:

We know that,

• $\pink{\underline{\boxed{\bf{Acceleration=\dfrac{\triangle v}{t}}}}}$

where,

Δv is change in velocity

t is time taken

$\rule{190}{1}$

Let the acceleration of train be a

So,

$\longrightarrow\rm{a=\dfrac{\triangle v}{t}}$

$\longrightarrow\rm{a=\dfrac{90}{5}}$

$\longrightarrow\rm\green{a=18\ km/h^{2}}$

Hence, the acceleration of train is 18 km/h².

$\rule{190}{1}$

Additional Information

Speed

The distance covered by moving body in unit time is called speed of the body.

$\red{\boxed{\bf{Speed=\frac{Distance}{Time}}}}$

Average Speed

Average speed is defined as the ratio of total distance covered and total time taken.

$\green{\boxed{\bf{Average\ Speed=\frac{Total\ Distance\ covered}{Total\ Time\ Taken}}}}$

Velocity

The change in position or displacement divided by the time interval is called velocity.

Acceleration

The rate of change of velocity is called acceleration.

Momentum

The product of mass and velocity of body at given time is called the momentum of body at that time.

Answer 2

Answer:

Explanation:

Correct Question,

The train starting from rest, increase its speed to 90km/h in 5hrs. calculate its acceleration​ .

Solution,

Here, we have initial and final velocity and time taken,

Initial velocity will be 0 as the train starts from rest.

Final velocity is given as 90 km/h, So we have to convert it into m/s,

For that we have to multiply it with 5/18,

So, 90 × 5/18 = 25 m/s

Now, time taken is given as 5 hrs, So we have to convert it into seconds,

For that we have to multiply it with 3600

So, 5 × 3600 = 18000 seconds

Here, we have to find,

Acceleration, a

According to the first equation of motion,

We know that,

v = u + at

So, putting all the values, we get

⇒ v = u + at

⇒ 25 = 0 + a × 18000

⇒ 25 = 18000a

⇒ 25/18000 = a

⇒ a = 0.00138 m/s²

Hence, the acceleration of the train is 0.00138 m/s².