Question

If a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance d40 compared to the stopping distance d25 if the driver were going at the posted speed limit of 25 mph?

Answer 1

Assume that the deceleration due to braking is a ft/s².

Note that
40 mph = (40/60)*88 = 58.667 ft/s
25 mph = (25/60)*88 = 36.667 ft/s

The final velocity is zero when the car stops, therefore
v² - 2ad = 0, or d = v²/(2a)
where
v = initial speed
a = deceleration
d = stopping distance.

The stopping distance, d₄₀, at 40 mph is
d₄₀ = 58.667²/(2a)
The stopping distance, d₂₅, at 25 mph is
d₂₅ = 36.667²/(2a)

Therefore
d₄₀/d₂₅ = 58.667²/(2a) ÷ 36.667²/(2a)
           = (58.667/36.667)²
           = 2.56

Answer:
The stopping distance at 40 mph is 2.56 times the stopping distance at 25 mph.

Answer 2

The stopping distance of a car going at 40 miles an hour in comparison to that of a car going at 25 miles an hour, for the same deceleration, is 2.56 times greater.

Further explanation

When bodies move they have what is called "linear momentum". Because of Newton's laws of motion, when a force goes against the body's movement, it will tend to slow it down (thus making its momentum smaller). When we apply these concepts to the car, it's easier to understand, since if a car is going at a certain speed and it breaks, due to the force applied on the brakes (which goes against the car's movement) the car slows down.

A particular issue, is how the car breaks (sometimes the car can break harder than at other moments), however due to the simplicity of the problem we will assume that the deceleration of both cars (the one going at 40 mph and the other going at 20 mph) is constant and are the same (in case this is not assumed, then we're missing important data to solve the problem). Based on these assumptions, we can easily compute the stopping distances from the following equation:

$V_f ^2 = V_i ^2 + 2 \cdot a \cdot d$

Where $V_f$ is the final velocity, $V_i$ is the initial velocity, a is the car's acceleration, and d is the stopping distance. Appliying the above equation to both cars (in which the final velocity is 0 since the car stops), we can get:

$0 = (40) ^2 + 2 \cdot a \cdot d_{40}$

$0 = (25)^2 + 2 \cdot a \cdot d_{25}$

Solving for the stopping distance we can get:

$d_{40} = - \frac{(40) ^2}{2 \cdot a}$

$d_{25} = - \frac{(25) ^2}{2 \cdot a}$

Dividing both equations (in order to compare them), we get:

$\frac{d_{40}}{d_{25}} = (\frac{40}{25})^2 = 2.56$

Therefor the stopping distance for the car that goes at 40 mph is 2.56 greater than that which goes at 25 mph (provided their deceleration is constant and the same).

Learn more

• Other problems on acceleration: brainly.com/question/659054
• Parabolic motion: brainly.com/question/1597396
• How to identify forces on a car: brainly.com/question/5461619

Keywords

Acceleration, forces, car, stopping distance