Question

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Between two stations a train accelerates uniformly at first, then moves with constant speed and finally retards uniformly if the ratio of the time taken are 1 : 8 : 1 for and the greatest speed is 60 km/h. Then the average speed over the whole journey --
(A) 45 km/h
(B) 54 km/h
(C) 35 km/h
(D) 53 km/h
$\large \bf \mathbb{\underline{ \bf{ \it{ \bf{ANSWER }}}}} - (B) 54 km/h$
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Answer 1

$\huge\mathfrak{Hello}$

Let it accelerates with say ‘a’ in time ‘x’ then it reaches a velocity ‘v’ and continues motion for time ‘8x’ abd finally then retards with ‘b’ in time ‘x’.

v = 0 + ax

=> v = ax

=> 60 = ax

=> a = 60/x

And,

0 = v – bx

=> b = 60/x

Distance covered in time of acceleration,

S1 = 0 + ½ at2

=> S1 = ½ ax2 = ½ × 60/x × x2 = 30x

Distance covered while moving with constant velocity is,

S2 = vt = 60 × 8x = 480x

Distance covered in time of retardation,

02 = v2 - 2bS3

=> S3 = 602/(2 × 60/x)

=> S3 = 30x

So, total distance = 30x + 480x + 30x = 540x

Total time = x + 8x + x = 10x

So, average velocity = 540x/10x = 54 km/h

$\huge\mathfrak{follow \:me}$

Answer 2

Answer:-

$A_v = 54 km/hr$

Option → B

Given :-

The ratio of time taken is 1 : 8 : 1.

Maximum velocity = 60 km/hr

To find :-

The average speed of whole journey.

Solution:-

Let the ratio of the time taken be x : 8x : x

Case :- 1

When train moves with constant

Velocity.

Then, it's acceleration will be :-

→$v = u + at$

→$60 = 0 + a\times x$

→$60 = ax$

→$a = \dfrac{60}{x}$

Now,

Distance covered by train will be :-

→$s = ut + \dfrac{1}{2}at^2$

→$s = 0 \times t + \dfrac{1}{2}\times \dfrac{60}{x}\times (x) ^2$

→$s = \dfrac{30}{x}\times x^2$

→$s = 30 x$

Case :- 2

Now,

The train moves with uniform velocity.

So, a = 0 m/s²

t = 8x

v = 60 km / hr

Distance covered by train is given by :-

→$s' = v t$

→$s' = 60 \times 8x$

→$s' = 480 x$

Case :- 3

When train retarded uniformly,

u = 60 km/hr

v = 0

t = x

The retardation produced in train will be :-

→$a = \dfrac{v-u}{t}$

→$a = \dfrac{0-60}{x}$

→$a = \dfrac{-60}{x}$

Distance covered by train is :-

→$s" = ut + \dfrac{1}{2}at^2$

→$s" = 0 \times t + \dfrac{1}{2}\times \dfrac{-60}{x} \times (x) ^2$

→$s" = \dfrac{1}{2} \times \dfrac{-60}{x}\times x^2$

→$s" = \dfrac{-30}{x}\times x^2$

→$s" = -30 x$

Since, we have to calculate distance and distance can never be in negative so neglect -ve sign.

Now,

Total distance travelled :-

→$S = s + s' + s"$

→$S = 30x + 480 x +30x$

→$S = 540 x$

Total time taken :-

→$T = x + 8x + x$

→$T = 10 x$

Average speed of train is given by :-

→$A_v = \dfrac{\text{Total distance travelled}}{\text{Total time taken }}$

→$A_v = \dfrac{540x}{10x}$

→$A_v = 54 km/hr$

hence,

Average speed of train is 54 km/hr.