Question

A train is moving at a speed of 50m/s and reduces its speed to 10m/s by covering a distance of 240m then find the retardation of the train? ​

Answer 1

Explanation:

Answer:

Question

a train is moving at a speed of 50m/s and reduces it's speed to 10m/s by covering a distance of 240m then find retardation of the train

Given

\small \: {initial \: velocity \: = 50m/s}initialvelocity=50m/s

final \: velocity \: = 10 \: \: m \: in \: sfinalvelocity=10mins

distance \: = 240 \: mdistance=240m

Solution

\longrightarrow \: {v}^{2} = {u}^{2} + 2as⟶v

2

=u

2

+2as

\longrightarrow \: {10}^{2} = {20}^{2} + 2 × a × 240⟶10

2

=20

2

+2×a×240

\longrightarrow \: 100 = 2500 + 480a⟶100=2500+480a

\longrightarrow \: 100-2500 = 480a⟶100−2500=480a

\longrightarrow \: -2400 = 480a⟶−2400=480a

\longrightarrow \: a = \frac {-2400}{480}⟶a=

480

−2400

\longrightarrow \: a = -5m / s⟶a=−5m/s

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Answer 2

Answer:

$\huge \: answer$

Initial velocity of train = 50m/s

Final velocity of train = 10m/s

Distance travelled in decreasing speed = 240m

Retardation = ?

Now according to third equation of motion :-

Square of final velocity of a body is equals to sum of square of its initial velocity and double of its product of of acceleration and distance covered.

Mathematically we represent it by :-

${u}^{2} = {u}^{2} + 2as$

In this equation :

•v = final velocity

•u = initial velocity

•a = acceleration or retardation

•s = distance or displacement

Now after inputting the known values in this equation we get :-

${10}^{2} = {50}^{2} + 2 \times 240 \times a \\ 100 = 2500 + 480a \\ 2500 - 100 = 480a \\ 2400 = 480a \\ a = \frac{2400}{480} \\ a = 5$

Thus the retardation of body is -5m/s^2

Here (-) represents the retardation.

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