Question

answer it............​

Answer 1

⚛ Given :-

• A train is moving at a speed of 50m/s and reduces it's speed to 10m/s by covering a distance of 240m.

⚛ To find :-

• Retardation of the train.

⚛ Solution :-

• Initial velocity (u) = 50m/s

• Final velocity (v) = 10m/s

• Distance (s) = 240 m

According to third equation of motion

→ v² = u² + 2as

• Where " v " is final velocity, " u " is initial velocity, " a " is acceleration and " s " is distance covered by object.

According to the question;

• → v² = u² + 2as

• → (10)² = (50)² + 2 × a × 240

• → 100 = 2500 + 480a

• → 100 - 2500 = 480a

• → -2400 = 480a

• → a = -2400/480

• → a = - 5m/s

• Negative sign shows retardation.

Hence,

• Retardation of train is 5 m/s

☢ More to know ☢

• v = u + at (First equation of motion)

• s = ut + ½ at² (second equation of motion)

Answer 2

Given:-

• initial velocity (u) = 50m/s
• final velocity (v) = 10m/s
• distance = 240m.

To find:-

• retardation of this train.

Formula to be used:-

• $\underline{\boxed {\sf v²=u²+2as}}$

• This is called The Third Equation of Motion.

here,

• v = final velocity
• u = initial velocity
• a = acceleration
• s = distance covered by the train.

Solution:-

$\longrightarrow$ (10)² = (50)²+2(a)(240)

$\longrightarrow$ 100 = 2500 + 480a

$\longrightarrow$ -2400 = 480a

$\longrightarrow$ a = $\sf\dfrac{-2400}{480}$

$\longrightarrow$ a = -5m/s²

hence, the retardation is of 5m/s²

Additional information:-

1st equation of motion:-

• $\underline{\boxed{\sf v = u+at}}$

here,

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time taken

2nd equation of motion:-

• $\underline{\boxed {\sf S = ut + \dfrac{1}{2} at²}}$

here,

• s = distance covered by the object
• u = initial velocity
• t = time taken
• a = acceleration