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a car moving with speed v on a straight road can be stopped with in distance d on applying brakes. if the same car is moving with speed 3v and brakes provide half retardation , then car will stop after travelling distance [ option:1)6 d 2)3 d 3) 9 d 4)18 d ] sir, i tired a lot and i got 3d but the correct option is 18 d

Hello saroj,

You need to use the principle of conservation of energy. When moving at speed v, the kinetic energy of the car is

KE = (1/2)*m*v^2

The work the car's brake system does while stopping the car is the braking force, F, times the distance, d, the car travels while stopping.

W = F*d

To stop the car, the brakes have to do an amount of work equal to the car's kinetic energy. So

KE = W

(1/2)*m*v^2 = F*d

So you can write an expression for d. I'll leave that to you: solve for distance d.

In the second case, when moving at speed 3*v, the kinetic energy of the car is

KE = (1/2)*m*(3*v)^2 = (1/2)*m*9*v^2

The braking system's braking force is now F/2. Therefore work the car's brake system does while stopping the car is the braking force, F/2, times the distance, d', the car travels while stopping in this case.

W = (F/2)*d

To stop the car, the brakes have to do an amount of work equal to the car's kinetic energy. So

KE = W

(1/2)*m*9*v^2 = F*d'/2

Now solve for new distance d' and see by what factor this distance is different.

I hope this helps,

Steve

Edit: OK, trouble with the math. I can do the math.

1st case:

(1/2)*m*v^2 = F*d ...... Solve for distance d.

d = (1/F)*(1/2)*m*v^2 .......... Eq 1

2nd case:

(1/2)*m*9*v^2 = F*d'/2

Now solve for new distance d'.

d' = 2*9*(1/F)*(1/2)*m*v^2 = 18*(1/F)*(1/2)*m*v^2 .......... Eq 2

I've grouped the string of things in Equations 1 and 2 above to make this next step easier. Notice in Equation 2 that everything from (1/F) to the end is bit for bit what Equation 2 says d is equal to.

So, you can substitute d for (1/F)*(1/2)*m*v^2. And that gives you

d' = 18*d

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Comment | thanks u sir. can you do the calculation and send . |

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