Question

a body is moving with uniform velocity of 50km/h come to rest after 10 second find retardation and distance travelled by body before it come to rest​

Answer 1

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf \green{Retardation \: is \: 1.38 \: m/s^2}} }$ $\red\dashrightarrow\underline{\underline{\sf \green{Distance\: travelled \: is \: 69.8 \: m}} }$

$\large \dag$ Step by step Explanation :-

❒ Converting Initial Velocity in m/s :-

We Have, $\text{Initial Velocity= 50 km/h}$

$= \rm\bigg( 50\times \frac{5}{18}\bigg)second$

$\red{:\longmapsto \text{Initial velocity = 13.88 \:m/s}}$

Now Here we have :

• Initial Velocity = u = 13.88 m/s

• Final Velocity = v = 0 m/s

• Time = t = 10 s

1 》Calculating Retardation :-)

Let acceleration be = a m/s²

❒ We Have 1st Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{\blue{v = u + at}}}}$

⏩ Applying 1st Equation of Motion ;

$\\ :\longmapsto \rm 0 = 13.88 + a \times 10$

$:\longmapsto \rm - 13.88 = 10a$

$:\longmapsto \rm a = \frac{ - 13.88}{ \: \:\: 10}$

$\purple{ \large :\longmapsto \underline {\boxed{{\bf a = - 1.38} }}}$

⇝ Acclerration of body is - 1.38 m/s²

Therefore,

⇝ Retardation of body is 1.38 m/s²

2 》Calculating Distance Traveled :-)

As we have,

• Initial Velocity = u = 13.88 m/s

• Time = t = 10 s

• Acceleration = a = - 1.38 m/s²

Let distance travelled be = s m

❒ We Have 2nd Equation of Motion as :

$\large \bf \red \bigstar \: \: \orange{ \underbrace{ \underline{s=ut+\dfrac{1}{2}at^2}}}$

⏩ Applying 2nd Equation of Motion :

$\\:\longmapsto \rm s = 13.88 \times 10 + \frac{1}{2} \times (-1.38) \times {10}^{2}$

$:\longmapsto \rm s= 138.8 + \frac{1}{2} \times (-1.38) \times 100$

$:\longmapsto \rm s=138.8 + \frac{1}{2} \times (-138)$

$:\longmapsto \rm s=138.8 - 69$

$\purple{ \large :\longmapsto \underline {\boxed{{\bf s = 69.8 \: m} }}}$

⇝ Distance Traveled is 69.8 meter

Answer 2

Answer:

• Acceleration a = 1 . 38 m/s

• Speed v = 69 . 8 m

step by step explaination :

• Using equation of motion,

• Vu+ at

Where,

v = final velocity a = acceleration

u = initial velocity

• t = time

• Put the value into the equation

Finally train will be rest so, final velocity,

:0 13.88 + a × 10

-13.88 = 10a

t = -13.88/10

t = -1.38

Again, using equation of motion,

S = ut + 1/2 at

Where, s = distance

Where, s = distancev = final velocity u = initial velocity

Where, s = distancev = final velocity u = initial velocitya = acceleration

Where, s = distancev = final velocity u = initial velocitya = accelerationt = time

Put the value into the equation

Where S is distance travelled before stop

s = 13.88 × 10 + 1/2 × (-1.38) × 10

s = 138.8 + 1/2 x (-1.38) x 100

s =138.8 + 1/2 × (-138)

s = 138.8 - 69

S= 69.8 m

So, the train will go before it is brought to rest is 69.8