Question

6.
A train moving with a velocity of 48 km h-' is brought to
rest in 11 second to avoid collision. What is the
retardation of the train in ms-2 and km h-2?
(Ans. 1.21 m 8-2, 1.57 x 104 km h-21
​

Answer 1

Given:-

• Initial velocity ,u = 48km/hr
• Final velocity ,v = 0m/s
• Time taken ,t = 11

To Find:-

• Retardation ,a

Solution:-

According to the Question

Conversion of unit here .

We change the velocity into m/s by multiplying with 5/18.

→ u = 48×5/18

→ u = 13.33

we need to calculate the retardation of the train .

As we know that acceleration is defined as the rate of change in velocity. Negative Acceleration is called retardation.

• a = v-u/t

where,

• v is the final velocity
• a is the acceleration or retardation
• u is the initial velocity
• t is the time taken

Substitute the value we get

$:\implies$ a = 0-15/11

$:\implies$ a = -13.33/11

$:\implies$ a = -1.21

$:\implies$ a = 1.21 m/s²

• Hence, deceleration in m/s² is 1.21m/s²

Now, calculating the acceleration into km/h²

As we know that

• 1m = 1/1000 km = 0.001km

• 1s = 1/3600 hr

So,

1.21m/s² = 1.21×0.001/1/(3600)²

= 1.21×0.001 ÷ 1/12960000

= 1.21×0.001 × 12960000

= 15681.6 km/h²

• Hence, the acceleration in km/h² is 15681.6km/h².

Answer 2

$\large\bold{\mathfrak{\underline{\underline{Given \: information:-}}}}$

• Velocity at which train is moving = 48 km per hr
• Whats the time taken by the train? = 11 s

$\large\bold{\mathfrak{\underline{\underline{Need \: to \: find?}}}}$

• Retardation of the train in ms-² and km/h-²

$\large\bold{\mathfrak{\underline{\underline{Which \: formula \: should \: be \: used?}}}}$

• $\pink{ \boxed{ \bf{Acceleration = \dfrac{final \: velocity - initial \: velocity }{Time} }}}$

$\large\bold{\mathfrak{\underline{\underline{Step \: by \: step \: explaination:-}}}}$

____________________

$\underbrace{ \bold{understanding \: the \: question \: the \: solving \: way}}$

Here in this question it is already mentioned about the initial velocity that had been taken by the train. We know the initial velocity is in kilometres per hour taken therefore first we have to convert them into metre per second taken. After that we would apply the given formula which we wrote it above and solve it. After we came to knew about the a acceleration in m/s² then we would convert it into km/h².

Time is came to solve it!!

$\star \: \underline \mathfrak{We \: are \:converting...}$

• $\hookrightarrow \: \bf{u = 48 \times \dfrac{5}{18} \: (we \: already \: came \: to \: knew)}$

$\star \: \underline \mathfrak{cancelling \: terms \: for \: further \: calculation...}$

• $\hookrightarrow \: \bf{u = \cancel{48} \times \dfrac{5}{ \cancel{18}} } \\ \rightarrow \: \bf{u = 13.3}$

$\star \: \underline{ \mathfrak{applying \: the \: formula...}}$

• $\hookrightarrow \: \bf\dfrac{0 - 13.33}{11}$
• $\hookrightarrow \: \bf\dfrac{ - 13.33}{11} </li><li>[tex]\hookrightarrow \: \bf\dfrac{ \cancel{- 13.33}}{ \cancel{11}}$
• $\hookrightarrow \: \bf{ u = 1.21}$

Here we came to knew about the acceleration in m/s². Thus after this we will be calculating the acceleration or retardation in km/h².

$\star \: \underline \mathfrak{now \: here \: calculation \: in \: km/h {}^{2} }$

★ We generally know 1 metre consists 1/1000 km and 1 second consists 1/3600 hours

$\star \: \underline \mathfrak{using \: the \: given \: formula \: again...}$

• $\pink{ \boxed{ \bf{Acceleration = \dfrac{final \: velocity - initial \: velocity }{Time} }}}$

$\star \: \underline \mathfrak{evaluating \: the \: new \: values... }$

• $\hookrightarrow \: \bf 1.21 = \bf\: \dfrac{1.21 \times 0.001}{ \dfrac{1}{(3600) {}^{2} } }$
• $\hookrightarrow \: \bf { \dfrac{1.21 \times 0.001}{ \dfrac{1}{12960000} } }$

• $\hookrightarrow \: \bf15681.6$

$\large\bold{\mathfrak{\underline{\underline{Answer:-}}}}$

• Retardation of train in ms-² is 1.21 and retardation of train in km/h² is 15681.6.