the same retarding force is applied to stop a train. The train stops after 80m.If the speed is doubled, then the distance will be
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Answered by
40
Let us assume that the train is moving along a straigtht line which a train often does with a constant initial velocity of u m/s constant speed
Now,F=ma
Where m is mass of train... F is force applied and a is acceleration.
V^2 - u ^2 = 2aS
Here... The train stops after 80m.. So final velocity is 0 =v
U^2 = 2aS. =>S = U^2/2a
Now,
Velocity/speed is doubled so
Initial velocity (increased) = 2U
Let S1 be the dist. Covered now.
As force remains same and the mass cannot change... So acc. Of the train will also remain same.
So, 4u^2 = 2aS1
=>2u^2/a = S1
=>2S = S1
=> S1= 160m.
So now train will stop at a distance of 160m. (ans). [Yeah i know physics is boring.]
Now,F=ma
Where m is mass of train... F is force applied and a is acceleration.
V^2 - u ^2 = 2aS
Here... The train stops after 80m.. So final velocity is 0 =v
U^2 = 2aS. =>S = U^2/2a
Now,
Velocity/speed is doubled so
Initial velocity (increased) = 2U
Let S1 be the dist. Covered now.
As force remains same and the mass cannot change... So acc. Of the train will also remain same.
So, 4u^2 = 2aS1
=>2u^2/a = S1
=>2S = S1
=> S1= 160m.
So now train will stop at a distance of 160m. (ans). [Yeah i know physics is boring.]
Answered by
36
Answer :- 320
Explanation :- stopping distance
s is directly proportional to u^2.
If the speed is doubled then ,
the stopping distance will be
four times
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