When Maggie uses the brakes of her car, the car slows uniformly indigenous 15.6 m/s come 0 m/s in 2.30 s. How far ahead of a stop sign have to she apply her brakes in stimulate to avoid at the sign?
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A third means can be provided to calculation the distance the car needs come stop.

For one dimensional movement with a constant acceleration the is additionally true that

(Vi + Vf)/2 = D/T

Doing one action of algebra produces

D = T(Vi + Vf)/2

D = 2.30s(15.6m/s + 0m/s)/2

D...


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A third way can be provided to calculate the street the car needs come stop.

For one dimensional movement with a continuous acceleration it is also true that

(Vi + Vf)/2 = D/T

Doing one action of algebra produces

D = T(Vi + Vf)/2

D = 2.30s(15.6m/s + 0m/s)/2

D = 17.94 m which again rounds to 17.9 m together the previous answer produced.



come answer this inquiry one must be acquainted with the equations of one dimensional movement for an item undergoing uniform acceleration.

Acceleration is defined as the price of change in one object"s velocity. The is, acceleration occurs when an object"s velocity changes and the amount of acceleration is proportional come the change in velocity and also inversely proportional to the time it takes to reason the change. In one dimensional motion, the acceleration will result in either rise in speed or a diminish in speed (commonly referred to as "deceleration").

The equation because that acceleration in one-D movement is

a = (Vf - Vi)/T wherein Vf is the last speed, through is the early stage speed, and T is the time it takes to cause the change.

From this we can calculate the acceleration the the automobile required to cause it to stop:

a =(0 m/s - 15.6m/s)/2.30s = -6.7826 m/s^2

With the acceleration the is now feasible to calculation the street it will take the auto to stop. There is an ext than one equation that will certainly work:

D = ViT + (1/2)aT^2

or

Vf^2 = Vi^2 + 2aD

The very first equation requires less algebra.

D = (0m/s)X2.30s + (1/2)(-6.7826m/s^2)(2.30s)^2

D = -17.94 m

At this point one must round the answer to the appropriate number of far-ranging digits. Rounding must be done together the last step prior to stating the last answer. From the provided information, we see that each measured value had actually only 3 far-reaching digits. Thus the answer requirements to be rounded to 3 digits.

D = -17.9 m or one can say the driver must use her brakes 17.9 meters before the prevent sign in bespeak to get the vehicle to stop in time.

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If one chooses to use the 2nd equation to identify the street it is an initial necessary to do algebra an isolate the variable D.