Question

A train is moving at a speed of 50m/s and reduces its speed to 10m/s by covering a distance of 240m then find the retardation of the train ​

Answer 1

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

Retardation = -5m/s^2

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

It is given that :

Initial velocity of train = 50m/s

Final velocity of train = 10m/s

Distance travelled in decreasing speed = 240m

Retardation = ?

Now according to third equation of motion :-

Square of final velocity of a body is equals to sum of square of its initial velocity and double of its product of of acceleration and distance covered.

Mathematically we represent it by :-

${v}^{2} = {u}^{2} + 2as$

In this equation :

• v = final velocity
• u = initial velocity
• a = acceleration or retardation
• s = distance or displacement

Now after inputting the known values in this equation we get :-

${10}^{2} = {50}^{2} + 2 \times 240 \times a$

$100 = 2500 + 480a$

$2500 - 100 = 480a$

$2400 = 480a$

$a = \frac{2400}{480}$

$a = 5$

Thus the retardation of body is -5m/s^2

Here (-) represents the retardation.

______________________________________

BY√\________ SCIVIBHANSHU

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Answer 2

Final velocity = 10 m/s

Initial velocity = 50 m/s

Distance travelled = 240 m

We know,

2as = v² - u²

⇒a = (v² - u²)/(2s)

⇒a = (100 m²/s² - 2500 m²/s²) ÷ 480 m

⇒a = - 2400 m²/s² ÷ 480 m

⇒a = - 5 m/s² {Answer}