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If the car continues to slow down in this way, after what further distance will it be brought to rest?

2006-11-13 21:58:18 · 9 answers · asked by Alexander M 1 in Science & Mathematics Physics

9 answers

It will not be another 75m, that is for sure.
retardation is acceleration backwards - no not accelerating backwards, but the reverse of acceleration in a forwards drection. Slowing down.

The addition of a hippopotumus will affect the retardation, there is no doubt, whether the hippo is educationally astute, or indeed is retarded.

I think the car will (Under the assumption that no animals jump on whilst it is being retarded in the uniform department) stop in around 27 meters.

I have my sources, OK?

2006-11-13 22:22:27 · answer #1 · answered by superman in disguise 4 · 0 0

Here we have to apply equation of motion.
Initial velocity(U)=30 m/s
Final velocity(V)=15 m/s
Distance covered(S)=75 m
Let 'a' be the retardation or you can say acceleration is negative.

V^2 - U^2 = -2aS
or 15 x15 - 30 x 30 = -2a x 75
or 225-900 = - 150a
or 150a = 675
or a = 675/150 = 4.5 m/s^2

If it car continues to slowdown then at rest, final velocity is zero.
Again applying equation of motion. And taking V=0 m/s
Finding value of S.

V^2 - U^2 = -2aS
or 0 x 0 - 30 x 30 = - 2 x 4.5 x S
or -900 = -9S
or 9S=900
or S=900/9 = 100 m

So further distance he has to travel is 100 m - 75 m = 25 m

2006-11-13 22:22:54 · answer #2 · answered by lingaraja b 2 · 0 1

First the decceleration needs to be calculated

Using an equation of motion:

Final speed^2 = initial speed^2 + 2xaccelerationxdisplacement

Rearranging we get:

Acceleration = (Final speed^2 - initial speed^2)/ 2xdisplacement
= (15^2 - 30^2) / 2x75
= (225 - 900) / 150
= -4.5ms^-2

Assuming the decelaration is constant:

Displacement = (Final speed^2 - initial speed^2)/ 2xacceleration
= (0^2 - 15^2) / 2x-4.5
= -225 / -9
= 25m

Hence, the car will travel a further 25m.

2006-11-14 08:30:18 · answer #3 · answered by PaulK24 1 · 0 0

Does the car contain a hippopotamus? If the mean mass decelerating relates inversely to the unicorn in the trunk, then three asymptotic spiders might relate to the web. Alternatively, retardation might require asbestos clothing and a special needs course of education. To rest, perchance to dream.

2006-11-13 22:10:46 · answer #4 · answered by PhD 3 · 1 0

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2016-10-22 01:39:37 · answer #5 · answered by ? 4 · 0 0

a = (900 - 225)/150 = 4.5 m/s²
s = 225/9 = 25 m

2006-11-13 22:07:46 · answer #6 · answered by Helmut 7 · 1 0

If you mean by (uniform) you mean analog, then it's 75m.
BUT if the value of retardation is equated with this then the car will never stop moving in this reality

2006-11-15 12:58:06 · answer #7 · answered by Anonymous · 0 0

I think that the answer is another 75m. because 15 is half of 30 so it would be the same again... ????

2006-11-13 22:04:02 · answer #8 · answered by Robert W 5 · 0 0

hmmmmm......37.5 m methinks....or summat

2006-11-13 22:03:19 · answer #9 · answered by Anonymous · 0 0

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