How do you calculate the braking distance

Physics

1 answer:

Anettt [7]3 years ago

6 0

The braking distance is given by $s=\frac{-u^2}{2a}$

Explanation:

When the driver of a car hits the pedal of the brakes, the car starts decelerating until it stops. Assuming the deceleration is constant, then the motion is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

u is the initial speed of the car

v is the final speed of the car, which is zero because the car comes to rest:

v = 0

a is the acceleration of the car

s is the distance travelled by the car during the deceleration, so it is the braking distance

Therefore, re-arranging the equation for s, we find an expression for the braking distance:

$s=\frac{-u^2}{2a}$

Note that the sign of $a$ is negative since the car is decelerating, therefore the final sign of $s$ is positive.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$