A car of mass m moving at a speed v is stopped in distance x by friction between tyres and road. If kinetic energy of the car is doubled. its stopping distance will be
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By the third equation of motion,
0^2 - v^2 = 2*a*x
So,
a = -v^2/2x
Now, this will be the maximum acceleration that can be given
Now,
Since the kinetic energy is doubled , that means the velocity is increased to √2v
So,
by third equation of motion,
0^2 - 2v^2 = 2*(-v^2/2x)*y
Where, y is the new stopping distance.
So,
2v^2 = v^2/x * y
Therefore,
y = 2x will be the new stopping distance
0^2 - v^2 = 2*a*x
So,
a = -v^2/2x
Now, this will be the maximum acceleration that can be given
Now,
Since the kinetic energy is doubled , that means the velocity is increased to √2v
So,
by third equation of motion,
0^2 - 2v^2 = 2*(-v^2/2x)*y
Where, y is the new stopping distance.
So,
2v^2 = v^2/x * y
Therefore,
y = 2x will be the new stopping distance
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0
Answer:
Explanation: acceleration is constant, then related with KE
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