Question

This question is from Physics bolo of class 11, CBSE board It is a numerical 2) On a 60km track, a train travels the first 30km at a uniform speed of 360 km/hr. How fast much the train travel the next 30 km km so as to average 40 km/h for the entire trip?

Answer 1

$\huge \bf {Given}$

$\tt Total \: Distance (d) = 60 \: km$

$\tt Speed (v_{1} ) \: during \: the \: first \: half \: journey = 360 \: km \: {h}^{ - 1}$

$\tt Average \: Speed(v_{av} ) = 40 \: km \: {h}^{ - 1}$

We have to Calculate, the speed $\bf (v_{2} )$ for the second half distance

Here,

$\tt t_{1} = \dfrac{30}{360} = \dfrac{1}{12} \: h$

Now,

$\tt t_2 = \dfrac{30}{v_2}$

Now, using the formula

$\huge \boxed{ \boxed {\tt v_{av} = \dfrac{Total \: distance}{Total \: time} }}$

$\bf 40 = \dfrac{60} {( \frac{1}{2} + \frac{30}{v_2} )}$

$\bf 40 = \dfrac{60}{ \frac{ (v_2 + 360) }{12v _2} }$

$\bf 40 = 60 \times \frac{12v _2}{v _2 + 360 }$

$\bf 40 \: v_2 + 14400 = 720v_2$

$\bf 720 \: v_2 - 40 \: v_2 = 14400$

$\bf680 \: v_2 = 14400$

$\bf v_2 = \dfrac{14400}{680}$

$\bf v_2 = 21.18 \: km \: {h}^{ - 1}$

Therefore, the Final Speed of the train is 21.18 km/h

$\huge \orange {\underline{ \bf Additional \: Information} }$

• The fixed point with respect to which the position of any object changes is known as Origin.

• Distance Travelled is always greater than or equal to Displacement.

• Distance Travelled is always Postitive.

• Displacement may be Positive, Negative or Zero.

• Rate of Change of distance is known as Speed.

• Distance Travelled, divided by the total time taken is said to be Average Speed.

• The speed at which a body starts Motion is known as Initial Speed. It is denoted by u.

• The speed which a body gains after its start, is known as Final speed and is denoted by v.

• Rate of Change if Velocity is known as Acceleration.

• If the final speed of a body is lower than initial speed, the body is said to be in Retardation.

$\huge \red{ \underline{ \bf Important \: Formulas}}$

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

• $\tt v = u + at$

• $\tt v^{2} - u^{2} = 2aS$

• $\tt S = ut + \dfrac{1}{2} at ^{2}$

• $\tt Force = Mass \times Acceleration$

• $\tt Momentum = Mass \times Speed$

Answer 2

Given

☞ Total distance = 60 km

☞ First speed (half path) = 360 km/h

☞ Average speed has to be = 40 km/h

To find

☞ Last speed of train to attain that average speed.

Solution

From the data we have ;

☛ Total distance (d) = 60 km/h

☛ First speed (v₁) = 360 km/h

☛ Average speed (v) = 40 km/h

Now we know that,

☞ Speed = Distance/Time

☞ 360 = 30/Time

☞ 360 × Time = 30

☞ Time = 30/360

☞ Time₁ = 1/12 h

Now,

☞ Time₂ = 30/v₂

Now we know that,

☛ Average speed = Total distance/Total time

☞ 40 = 60/(1/12 + 30/v₂)

☞ 40 = 60/{(v₂ + 360)/12v₂}

☞ 40 = 60 × (12v₂/ v₂ + 360)

☞ 40v₂ + 14400 = 720v₂

☞ 720v₂ - 40v₂ = 14400

☞ 680v₂ = 14400

☞ v₂ = 14400/680

☞ v₂ = 21.18 km/h

∴ Final speed of train should be = 21.18 km/h