Question

A train starting from rest attains a Velocity of 72 km h $\tt ^{-1}$ in 5 minutes. assuming that the acceleration is uniform, find (I) the Acceleration and (ii) the Distance travelled by this train for attaining this Velocity.

Answer 1

Given,

$\tt u = 0 \ km \ h^{-1}$

$\tt v = 72 \ km \ h^{-1}$

$\tt Converting \ in \ m/s = 20 m s^{-1}$

$\tt t = 5 \ minutes$ = 300s

$\huge \boxed { \bf a = \dfrac{v - u}{t} }$

${ \red \longrightarrow} \: \dfrac{\tt 20 \: m \: {s}^{ - 1} - 0 \: m \: {s}^{ - 1}} { \tt \: 300 \: s}$

${ \red \longrightarrow} \tt \dfrac{20 \: m \: {s}^{ - 1} }{300s} \\ \\ { \red \longrightarrow} \tt\dfrac{1}{15} m \: {s}^{ - 2}$

$\huge \boxed{ \bf {v}^{2} - {u}^{2} = 2as }$

Thus,

$\tt s = \dfrac{ {v}^{2} }{2a}$

${ \red \longrightarrow} \tt \dfrac{(20 \: m \: {s}^{ - 1 } ) ^{2} }{2 \times \dfrac{1}{15} \: m \: {s}^{ - 2} } \\ \\ { \red \longrightarrow} \tt \: 3000 \: m \\ { \red \longrightarrow} \tt \: 3km$

Hence, acceleration of train is $\sf \dfrac{1}{15} \: m \: {s}^{ - 2}$ and the distance is $\sf 3 \: km$

Additional Information

• Rate of Change of Distance is known as Speed.

• The speed of body which is applied in beginning is known as Initial Speed and is denoted by u.

• The speed of body which is acquired after the body starts Moving is know as Final Speed and is represented by v.

• Rate of change of Velocity is known as Acceleration.

• Negative Acceleration is known as Retardation.

Important Formula

$\tt Speed = \dfrac{Distance}{Time}$

$\tt Average \: Speed = \dfrac{Total \: distance}{Total \: time}$

$\tt v = u + at$

$\tt v^{2} - u^{2} = 2as$

$\tt S = ut + 1/2at^{2}$

Answer 2

$\huge{\boxed{\sf{\red{Corr}\green{ect}\ \orange{Ques}\pink{tion}\purple{:}}}}$

A train starting from rest attains a Velocity of 72 km/h in 5 minutes. assuming that the acceleration is uniform, find

(I) the Acceleration, and

(ii) the distance traveled by this train for attaining this Velocity.

$\huge{\boxed{\rm{\pink{Ans}\green{wer}:}}}$

$\star \underline{\bf The\ train\ starts\ from\ rest}$

$\red{\tt{Initial\ velocity\ ( u ) } }\sf = 0\ km/h$

$\orange{\tt{Final\ velocity\ ( v ) } }\sf = 72\ km/h$

The time taken is 5 minutes

converting into seconds we multiply 60

= 5 × 60 = 300 seconds

Now converting initial and final velocity in m/s

$\tt Initial\ velocity = u = 0\ km/h \\ \\ \tt = 0 \times \frac{5}{18} m/s \\ \\ \tt = 0\ m/s$

And

$\tt Final\ velocity = v = 72\ km/h \\ \\ \tt = \tt 72 \times \frac{5}{18} m/s \\ \\ \tt = 20\ m/s$

$\boxed{\boxed{\begin{minipage}{6 cm} \bf We know that, \\ 1km = 1000m \\ 1hour - 60 minutes = 3600 sec \\ \\ so, \frac{1km}{1h} = \frac{1000\ m}{3600\ s} \\ \\ 1km/h = \frac{5}{18} m/s\end{minipage}}}$

$\rule{200}2$

(I) The Acceleration

$\boxed{\sf \red{Acceleration} = \dfrac{\pink{final\ velocity}-\green{initial\ velocity}}{\orange{Time\ taken}}}$

$\bf = \dfrac{v-u}{t} \\ \\ \\ \bf = \dfrac{20-0}{300} \\ \\ \\ \bf = \boxed{\rm \dfrac{1}{15}\ m/s^{2}}$

$\rule{200}2$

(II) Distance traveled

Using the 3rd equation of motion

$\red{\boxed{\sf{v^2 - u^2 = 2as}}}$

$\bf (20)^2 - (0)^2 = 2 \times \frac{1}{15} \times s \\ \\ \bf = \bf 400 - 0 = \frac{2}{15} \times s \\ \\ \bf = 400 = \frac{2}{15} \times s \\ \\ \bf = 400 \div \frac{2}{15} = s \\ \\ \bf = 400 \times \frac{15}{2} \\ \\ \bf = 200 \times 15 = 3000 m$

1000 m - 1 km

3000m = 3km

So,

Distance = 3km

Hence

Acceleration = $\frac{1}{15}$ m/s²

and

Distance = 3km