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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 ​

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221 cents Tommasa
Answer:

Explanation:Initial velocity u=30m/s

Explanation:Initial velocity u=30m/sFinal velocity, v=10m/s

Distance traveled, s=240m

v

2

−u

2

=2as where 'a' is acceleration

100−900=2as

a=

2×240

−800

=−

6

10

m/s

2

F=ma=−m

6

10

N

If force is increased by 12.5%,new force

F

=(100+12.5)% ×F

=112.5 % of F

F

F

=

100

112.5

=1.125

ma

ma

=1.125

a

=1.125a=1.125×

6

−10

=−1.875m/s

2

New distance traveled when v

=0

v

′2

−u

2

=2a

s

0−900=2×(−1.875)×s

s

=

2×1.875

900

=240m

I hope it helps you ✌️✌️
324
Tommasa
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