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 train was moving at speed of 50m/s and decreases it's speed to 10m/s after travelling 240m.

Retardation of train = ?

Now, some keywords :

• Final velocity of train = 10m/s
• Initial velocity of train = 50m/s
• Distance covered = 240
• Retardation = ?

Coming to question :-

Now, we know according to third equation of motion the square, of final velocity of a body is equal to sum of square, of its initial velocity with double of product of its displacement or distance and acceleration or retardation. Mathematically it is represented by :

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

Where :

• 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 :

$a = \frac{ {v}^{2} - {u}^{2} }{2s}$

$a = \frac{ {10}^{2} - {50}^{2} }{2 \times 240} \\ \\ a = \frac{100 - 2500}{480} \\ \\ a = \frac{ - 2400}{480} \\ a = - 5$

Here {-) represents opposite direction.

Now we got retardation is -(-5)

= 5m/s^2.

______________________________________

BY SCIVIBHANSHU

THANK YOU

STAY CURIOUS

Answer 2

Initial velocity = 50 m/s {u}

Final velocity = 10 m/s {v}

Distance covered = 240 m {s}

We know,

2as = v² - u²

⇒a = (v² - u²)/2s

⇒a = ½[(10 m/s + 50 m/s)(10 m/s - 50 m/s)]/(2 • 240 m)

⇒a = [(60 m/s)(- 40 m/s)]/480 m

⇒a = (- 2400 m²/s²)/(480 m)

⇒a = - 5 m/s²