Math Is Fun Forum

MechanicsStudent

Guest

Quick Mechanics Question

Hey, I have a quick question, for some reason I can't work out how to do this.

A car comes to a stop from a speed of 30m/s in a distance of 804m. The driver brakes as to produce a deceleration of 0.5m/s² to begin with, and then brakes harder to produce a deceleration of 1.5m/s². Fine the speed of the car at the instant when the deceleration is increased.

Please could you tell me how to go about this?

Thank you very much.

Mech Student

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quick Mechanics Question

I'd use s = (v² - u²)/2a twice.

For the lightly braking part, you have s = (30² - v²).
Here, s is the distance travelled when the hard braking starts, and v is the speed when that starts happening.
s and v are unknown so far, but then the equation for the remaining part looks like:

804 - s = 804 - (30² - v²) = (v²)/3

Now you can solve for v.

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Why did the vector cross the road?
It wanted to be normal.